User sergey melikhov - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T01:53:19Z http://mathoverflow.net/feeds/user/10819 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115567/covering-maps-in-real-life-that-can-be-demonstrated-to-students/124486#124486 Answer by Sergey Melikhov for Covering maps in real life that can be demonstrated to students Sergey Melikhov 2013-03-14T08:52:16Z 2013-03-14T08:52:16Z <p>Not only a covering, but every piecewise linear map $f$ between finite graphs can be realized in $\Bbb R^3$ in your sense. To see this, pick a triangulation $T$ of the mapping cylinder of $f$ that projects simplicially to some triangulation of $I=[0,1]$. Then pick a generic level-preserving map $g$ of the mapping cylinder in $\Bbb R^3\times I$ that is linear on each simplex of $T$. (Generic here means that the images of any $k+1$ vertices span a $k$-dimensional affine subspace in $\Bbb R^4$ as long as either $k\le 3$, or $k\le 4$ and not all of the $k+1$ vertices have the same $I$-coordinate.)</p> <p>Then $g$ has only finitely many double points, all in $\Bbb R^3\times (0,1)$. (Say, if a pair of 2-simplices with at most one vertex in common have a 1-dimensional intersection, then their 5 or 6 vertices aren't affinely independent.) If $s$ is the minimal $I$-coordinate of a double point, then the part of $g$ that lands in $\Bbb R^3\times [0,s/2]$ yields the desired homotopy.</p> <p>I'm quite amazed that this rather standard argument didn't occur to anyone having seen this question in 3 months. Maybe a tag like "geometric topology" could have helped. There is some literature on realizing maps in your sense, which goes under "isotopic realization". </p> http://mathoverflow.net/questions/118833/making-cw-complexes-metrizable/124038#124038 Answer by Sergey Melikhov for Making CW-complexes metrizable Sergey Melikhov 2013-03-09T02:30:53Z 2013-03-09T20:04:16Z <p>This addresses the modified question in Jeremy's comments, on keeping the preferred CW-structure.</p> <p>1) If the CW complex happens to be regular and PL (i.e. the attaching maps are injective and piecewise-linear), its barycentric subdivision is a simplicial complex (namely, the order complex of the poset of nonempty faces of the CW complex), which can be endowed with the usual barycentric metric. The identity map will then be a homotopy equivalence (proofs can be found in some old textbooks, including the Appendix of Dold's Algebraic topology, or "Theory of retracts" by Hu Sze-Tsen). </p> <p>2) For a general (countable) CW complex, one can inductively homotop the attaching maps of $(n+1)$-cells by a homotopy with values in the $n$-skeleton so that the modified CW complex $K$ admits a barycentric subdivision $K'$ that is a regular simplicial set, in the sense that cells of $K$ are identified with the unions of simplices of $K'$ whose first vertex is a fixed $0$-simplex of $K'$. (Regular means that the representing map of every non-degenerate simplex only makes identifications along the last facet of the simplex.) The geometric realization of a regular simplicial set is a regular CW-complex, so the previous construction applies. (In more detail, the order complex $K''$ of the poset of nonempty nondegenerage simplices of $K'$, ordered by inclusion, is a simplicial complex.) An enlightening overview of subdivisions of simplicial sets can be found <a href="http://mathoverflow.net/questions/46827/number-of-subdivisions-needed-to-compare-simplicial-sets-to-simplicial-complexes/116867#116867" rel="nofollow">here</a>.</p> <p>The homotopies of attaching maps can be constructed using <a href="http://en.wikipedia.org/wiki/Simplicial_approximation_theorem" rel="nofollow">Brouwer's simplicial approximation theorem</a>, which implies that any continuous map $|L|\to |X|$ between geometric realizations of finite simplicial sets is homotopic, upon precomposing with the geometric realization of an iterate $L^{(n)}\to L$ of the last vertex map $L'\to L$, to the geometric realization of a morphism $f:L^{(n)}\to X$ of simplicial sets (see Corollary 3.2 <a href="http://www.tac.mta.ca/tac/volumes/12/2/12-02abs.html" rel="nofollow">here</a>). Here $L$ is any triangulation $S^n$ by a non-singular simplicial set (i.e. a subcomplex of the order complex of a poset), and $X$ is the $n$-skeleton of $K'$, which is a regular simplicial set. Then the mapping cone of $f$ is again a regular simplical set.</p> <p>In light of the <a href="http://mathoverflow.net/questions/61948/are-there-two-non-homeomorphic-finite-regular-cw-complexes-with-isomorphic-face-p/61961#61961" rel="nofollow">combinatorial view</a> of regular PL CW complexes, one could try to homotop the attaching maps of a general CW complex so as to achieve a more rigid combinatorial structure of the simplicial set $K'$. However, a homotopy class is not generally representable by a non-degenerate map (in the sense of collapsing no simplices). Because of this, $K'$ cannot be generally chosen to be the nerve of a category, nor even a quasi-category. </p> <p>3) One drawback of the metric topology on simplicial complexes is that it indeed is incompatible with quotients (they are non-metrizable, unless each equivalence class is compact). This difficulty <a href="http://arxiv.org/abs/1106.3249" rel="nofollow">can be avoided</a> by endowing the simplicial complexes with a <a href="http://arxiv.org/abs/1109.0346" rel="nofollow">"cubical" l_infty metric</a> and working uniformly. This applies to regular PL CW complexes, as well as to those CW complexes whose attaching maps are jointly uniformly continuous (using iterates of canonical, rather than barycentric, subdivision and Theorem 7.4 <a href="http://arxiv.org/abs/1109.0346v2" rel="nofollow">here</a>).</p> http://mathoverflow.net/questions/118185/where-should-i-learn-about-immersion-theory/124055#124055 Answer by Sergey Melikhov for Where should I learn about immersion theory? Sergey Melikhov 2013-03-09T09:04:17Z 2013-03-09T09:04:17Z <p>Immersion theory has been "explained" by the Compression Theorem, with new proofs arguably being much more elementary and intuitive:</p> <p><a href="http://arxiv.org/abs/math/0301356" rel="nofollow">Rourke and Sanderson</a>, </p> <p><a href="http://dspace.mit.edu/handle/1721.1/35136" rel="nofollow">A master's thesis with another exposition</a></p> http://mathoverflow.net/questions/123513/the-role-of-anr-in-modern-topology/123585#123585 Answer by Sergey Melikhov for The role of ANR in modern topology Sergey Melikhov 2013-03-05T02:18:39Z 2013-03-05T02:18:39Z <p>ANRs are (and have always been) irrelevant as long as homotopy-invariant properties of spaces homotopy equivalent to CW-complexes are concerned. But modern algebraic topologists do not seem to be really interested in (or anyway have real tools to deal with) more general spaces AFAIK. (Of course, "general nonsense" like simplicial model categories works for general spaces, but if you are using any invariants like homotopy groups or singular (co)homology theories to get substantinal results that do not mention those invariants, you'll probably need theorems such as Whitehead's - which means restricting to spaces homotopic to CW-complexes.)</p> <p>Shape theory did go beyond spaces homotopic to CW-complexes. But being an ANR is not a shape-invariant property. It is an invariant of local shape (which Ferry, Quinn, Hughes and their collaborators do touch upon in their works) and indeed Quinn once wrote an expository paper on "Local algebraic topology". I don't think these "local" developments have ever been of interest for (mainstream) algebraic topology, but they have very good applications in geometric topology so are usually associated with the latter.</p> <p>This area of geometric topology, where ANRs and topological manifolds natuarally belong, has been steadily falling out of fashion with younger generations (since the 80s I would say, not 60s), apparently because it's tough enough, but not nearly as attractive for an outsider as knots, say. That might as well be a problem of the generations rather than a "flaw" in ANRs.</p> http://mathoverflow.net/questions/121161/original-proof-of-the-existence-of-seifert-surfaces/122024#122024 Answer by Sergey Melikhov for Original proof of the existence of Seifert surfaces Sergey Melikhov 2013-02-16T21:35:33Z 2013-02-16T21:52:10Z <p>Here is a picture of the Frankl-Pontrjagin construction, from a paper by Lucien Guillou and Alexis Marin:</p> <p><img src="http://www.freeimagehosting.net/newuploads/6qg1x.png" width="700" height="737"></p> <p>The inscription reads: "The Frankl-Pontryagin surface of the trefoil knot is depicted with two 'windows' showing what happens inside."</p> <p>The picture is accompanied in the paper by an actual exposition of a generalization of the Frankl-Pontrjagin construction. This is section 4 of Guillou and Marin's comments 'On the second paper' by Rokhlin in their book "A la Recherche de la Topologie Perdue" (in French; the scan above is from the Russian translation).</p> http://mathoverflow.net/questions/99463/how-to-see-that-double-suspension-of-homology-3-sphere-is-homeomorphic-to-a-sph/99485#99485 Answer by Sergey Melikhov for How to "see" that double suspension of homology 3-sphere is homeomorphic to a sphere? Sergey Melikhov 2012-06-13T18:54:05Z 2012-06-14T01:04:35Z <p>To get some intuition behind the double suspension theorem, you can try three pages with a lot of pictures in <a href="http://www.math.rutgers.edu/~sferry/ps/geotop.pdf" rel="nofollow">Ferry's notes</a> (starting with p.166, in Chapter 26). He gives a rough sketch of proof in the case of one particular homology sphere, the one for which the theorem was first proved by Edwards.</p> <p>The double suspension theorem boils down to showing that a certain non-manifold (namely, the single suspension over a homology sphere) becomes a manifold when multiplied by $\Bbb R$. When Milnor conjectured the double suspension theorem in early 60s, he must have been aware of the existence of other non-manifolds with this property (found earlier by Bing). It is fortunate that they also exist in a lower dimension so it's easier to visualize what's going on. One such example is $(S^3/W)\times\Bbb R\cong S^3\times\Bbb R$, where $W$ is the Whitehead continuum, and there is a rather explicit construction of this homeomorphism in Ferry's Chapter 4 (p.15).</p> <p><strong>Added later:</strong> Another example is $(M/D)\times\Bbb R\cong M\times\Bbb R$, where $D$ is a wild copy of the $n$-disk contained in the interior of the manifold $M$. The case $n=2$ is actually used the above-mentioned proof of the double suspension theorem in Ferry's notes, but is not proved there; a proof of the case $n=1$ with some pictures can be found in the <a href="http://www.calvin.edu/~venema/embeddingsbook/" rel="nofollow">Daverman-Venema book</a>, Section 2.6.</p> <hr> <p>To address the specific question about neighborhoods, the open star (in the original double-suspension triangulation) of any vertex in the suspension circle is homeomorphic to $\Bbb R^5$ $-$ at least in the case of one particular homology sphere, the boundary of the Mazur manifold $W$. Indeed, the closed star of this vertex is the suspension over $cone(\partial W)$. As explained in Ferry's notes, $cone(\partial W)\times\Bbb R$ is homeomorphic to $W\times\Bbb R$. Hence the open star of the vertex is homeomorphic to $(W\setminus\partial W)\times\Bbb R$. The latter can be identified with the interior of $W\times I$. But it is <a href="http://mathoverflow.net/questions/26385/when-factors-may-be-cancelled-in-homeomorphic-products/48658#48658" rel="nofollow">easy to see</a> that $W\times I$ is homeomorphic to the $5$-ball. So its interior is homeomorphic to $\Bbb R^5$.</p> http://mathoverflow.net/questions/98750/can-the-alexander-horned-sphere-arise-as-a-cell-boundary-in-a-finite-cw-sphere/98762#98762 Answer by Sergey Melikhov for Can the Alexander horned sphere arise as a cell boundary in a finite CW-sphere? Sergey Melikhov 2012-06-04T12:23:52Z 2012-06-04T12:58:29Z <blockquote> <p>Specific question: is there a finite CW complex homeomorphic to a sphere such that one of its maximal cells has as its closure a ball whose boundary is embedded in the CW-sphere as an Alexander horned sphere?</p> </blockquote> <p>1) No. If $K$ is the $2$-skeleton of the CW complex, then $K$ has a mapping cylinder neighborhood in $S^3$, and hence by <a href="http://www.ams.org/journals/tran/1969-143-00/S0002-9947-1969-0248788-4/" rel="nofollow">Nicholson's theorem</a> it is tame, that is, equivalent to a subpolyhedron of $S^3$ by a homeomorphism $h$ of $S^3$. Since $K$ is $2$-dimensional, it is not hard to show that $h$ also takes any $2$-sphere in $K$ onto a subpolyhedron of $S^3$. So $K$ cannot contain the horned sphere.</p> <p>2) Yes, if you allow an unspecified wild codimension one sphere in place of the Alexander horned sphere. By Example 7.11.2 on p.419 in the <a href="http://www.calvin.edu/~venema/embeddingsbook/" rel="nofollow">Daverman-Venema book</a> (this seems to be among a few original results in the book, perhaps the most important one), $S^n$ for $n\ge 6$ contains a wildly embedded sphere $\Sigma$ with a mapping cylinder neighborhood $N$. By construction, the complement to the interior of $N$ consists of two closed $n$-balls. It follows that the closures of the complementary domains of $\Sigma$ are the mapping cones of some self-maps of $S^{n-1}$. So we get a CW-complex with one $0$-cell, one $(n-1)$-cell and two $n$-cells, which is homeomorphic to $S^n$, and has a wild $(n-1)$-skeleton.</p> <blockquote> <p>Follow-up question: if the answer is no, is there a finite CW complex homeomorphic to a sphere such that the closure of one of the maximal cells is a ball, but the closure of its complement is not a ball?</p> </blockquote> <p>1) Yes. You can glue one of the complementary domains of $\Sigma$ and an $n$-ball along their boundary sphere. The result is again $S^n$, according to Proposition 7.10.1 in Daverman-Venema.</p> <p>2) I should also mention a simpler but somewhat similar example. Using the Edwards-Cannon theorem, it is not hard to construct a finite regular CW-complex $K$ that is homeomorphic to $S^5$, even though the boundary of some $2$-cell (in fact, of each $2$-cell) of $K$ is wild, viewed as a copy of $S^1$ in $S^5$. In more detail, if $H$ is a traingulation of a non-simply-connected homology $3$-sphere, then the double suspension $S^0*S^0*H$ is a simplicial complex homeomorphic to $S^5$; the desired CW-complex is the 'prejoin' $(S^0*S^0)+H$, which is PL homeomorphic to $S^0*S^0*H$ and has all its $2$-cells attached to the suspension circle $S^0*S^0$. On identifying regular CW-complexes with their posets of nonempty faces, the prejoin $P+Q$ of two posets is defined by placing all the elements of $P$ below all the elements of $Q$ in the Hasse diagram, and keeping the original order within $P$ and within $Q$. The order complex of $P+Q$ is easily seen to be isomorphic to the join of the order complexes of $P$ and of $Q$. As an example, $S^0*S^0*pt$ is a simplicial complex with $4$ of $2$-simplices, whereas $(S^0*S^0)+pt$ is a cell complex with only one (quadrilateral) $2$-cell. </p> <p>Beware that $K$ itself is a PL CW-complex, with PL attaching maps; the only trouble is with the homeomorphism between $K$ and $S^5$.</p> http://mathoverflow.net/questions/97464/how-to-tackle-the-smooth-poincare-conjecture/97627#97627 Answer by Sergey Melikhov for How to Tackle the Smooth Poincare Conjecture Sergey Melikhov 2012-05-22T04:58:26Z 2012-05-22T07:46:03Z <p>The conjecture in question can also be thought of as the $4$-dimensional PL Poincare conjecture (because low-dimensional PL manifolds, including those of dimension $4$, carry a unique smooth structure) and this is how it is understood in most references mentioned below.</p> <p>Some interesting approaches to the conjecture and its special cases can be found in several papers by Frank Quinn (some based on TQFTs and others in more classical spirit) and in some papers by Robert Craggs. </p> <p>Much of the effort has been focused on the group-theoretic <a href="http://www.ams.org/journals/proc/1965-016-02/S0002-9939-1965-0173241-8" rel="nofollow">Andrews-Curtis</a> conjecture, whose validity would imply that PL (or smooth) homotopy $4$-spheres given as handlebodies without $3$-handles are PL (or smoothly) standard. The latter assertion would also follow from the <a href="http://arxiv.org/abs/math/0603511" rel="nofollow">"Generalized Property R"</a> conjecture. Then there's a separate industry of finding handlebody presentations of simply-connected $4$-manifolds without $3$-handles (see Problems 4.18 and 4.73 in <a href="http://math.berkeley.edu/~kirby/problems.ps.gz" rel="nofollow">Kirby's list</a>, Section 6 <a href="http://arxiv.org/abs/math/0006003" rel="nofollow">here</a>, Gadgil's <a href="http://math.iisc.ernet.in/~gadgil/papers/OnehndlCK.pdf" rel="nofollow">preprint</a> and Quinn's <a href="http://www.google.com/url?sa=t&amp;rct=j&amp;q=&amp;esrc=s&amp;source=web&amp;cd=9&amp;sqi=2&amp;ved=0CHYQFjAI&amp;url=http%253A%252F%252Fwww.ams.org%252Ftran%252F2002-354-04%252FS0002-9947-01-02940-3%252FS0002-9947-01-02940-3.pdf&amp;ei=TT27T8PbJoyMOq7P5YMP&amp;usg=AFQjCNEepa51_VMFmfcjORDUfTrtpS_-xA" rel="nofollow">Corollary 3.2</a>; note also Rasmussen's withdrawn paper arxiv.org/abs/1005.4674).</p> <p>As observed by Curtis in an earlier paper (in "Topology of 3-manifolds and related topics"), every compact contractible $2$-polyhedron PL embeds in some PL homotopy $4$-sphere; so if you find one that doesn't PL embed in $S^4$, you're done with the 4D PL Poincare conjecture. This line of attack inspired some literature on PL embeddings of acyclic $2$-polyhedra in $S^4$ starting I guess with Zeeman's dunce hat paper; see <a href="http://www.ams.org/mathscinet-getitem?mr=MR1978374" rel="nofollow">this review</a> for additional references.</p> <p>There are numerous other approaches and related techniques, e.g. "Gluck twists" and "Akbulut corks". <a href="http://math.berkeley.edu/~kirby/problems.ps.gz" rel="nofollow">Kirby's problem list</a> is a good source of further references prior to mid-90s; some other basic references on the Andrews-Curtis conjecture are collected <a href="http://www.sci.ccny.cuny.edu/~shpil/gworld/problems/Back.html" rel="nofollow">here</a> under (O1).</p> http://mathoverflow.net/questions/96883/morse-theory-in-top-and-pl-categories/96926#96926 Answer by Sergey Melikhov for Morse theory in TOP and PL categories? Sergey Melikhov 2012-05-14T17:05:52Z 2012-05-16T15:19:06Z <p><a href="http://www.emis.de/journals/SLC/wpapers/s48forman.pdf" rel="nofollow">Forman</a>'s papers and the books by Kozlov and Orlik-Welker that you see cited at Daniel's Wikipedia link are good starting points for the more combinatorial tradition of PL Morse theory. </p> <p>For more geometric approaches see Bestvina's <a href="http://www.math.utah.edu/~bestvina/eprints/minicourse.pdf" rel="nofollow"><em>PL Morse theory</em></a> and the ancient <a href="http://www.ams.org/journals/tran/1972-170-00/S0002-9947-1972-0310899-2/home.html" rel="nofollow"><em>Piecewise linear critical levels and collapsing</em></a> by Kearton and Lickorish. Unfortunately, the literature on discrete Morse theory seems to be unaware of the Kearton-Lickorish 1972 paper and the still earlier papers by Kosinski and Kuiper (cited by Kearton and Lickorish) which also constructed some kinds of PL Morse functions. Even if they did it with heavier notation and by less illuminating arguments, their alternative understanding of PL Morse functions is still worth to be aware of.</p> http://mathoverflow.net/questions/87557/a-simplicial-complex-which-is-not-collapsible-but-whose-barycentric-subdivision/87666#87666 Answer by Sergey Melikhov for A simplicial complex which is not collapsible, but whose barycentric subdivision is Sergey Melikhov 2012-02-06T14:09:00Z 2012-05-15T00:02:37Z <p><a href="http://www.ams.org/journals/tran/1969-137-00/S0002-9947-1969-0238288-X/S0002-9947-1969-0238288-X.pdf" rel="nofollow">Lickorish and Martin</a> constructed, for each $r$, a triangulation of the $3$-ball whose $r$th barycentric subdivision collapses, but $(r-1)$th doesn't. The basic idea, going back to <a href="http://infoshako.sk.tsukuba.ac.jp/~HACHI/math/library/knot_eng.html" rel="nofollow">Furch</a> and Bing, is to triangulate a cube with a knotted hole, where the missing knot has bridge index $2^r+1$, and then fill back a small part of the hole - so that topologically, no hole remains, but the $3$-ball now contains a knot triangulated by a single edge.</p> <p><strong>Added later:</strong> <a href="http://www.ams.org/journals/tran/1972-170-00/S0002-9947-1972-0310899-2/" rel="nofollow">Kearton and Lickorish</a> also constructed triangulations of the $n$-ball, $n\ge 3$, whose $r$th barycentric subdivision is not collapsible. On the other hand, every triangulation of a ball becomes collapsible after some number of barycentric subdivisions, according to a recent preprint by <a href="http://arxiv.org/abs/1202.6606" rel="nofollow">Adiprassito and Benedetti</a> (see their Corollary 3.5).</p> http://mathoverflow.net/questions/96069/status-of-pl-topology/96403#96403 Answer by Sergey Melikhov for Status of PL topology Sergey Melikhov 2012-05-09T03:58:22Z 2012-05-09T11:24:53Z <p>Disclaimer: What follows is probably a bit off-topic for this site, but no more than the original questions, numbered one and two. In fact I suspect that this answer attempts to address just what the OP really wanted to ask ("isn't PL topology useless?") by posting those two lightly euphemistic questions. If there was an active meta thread for closing this question, I'd rather put this answer there. </p> <p>Some topologists, perhaps the majority, tend to think that smooth and topological manifolds are "present in nature" and are the genuine objects of study in geometric topology, while PL topology is a somewhat artificial, unnatural construct, and matters just as long as it is helpful for the "real" topology. I've heard this opinion stated explicitly once, and I see a lot of this kind of attitude in this thread. In fact I think this philosophy/intuition is sufficiently familiar to nearly everyone that I don't need to elaborate on it. Moreover, I suspect that a lot of people are not even aware that it is not the only possible religion for a topologist, or else they would be more considerate to the heretics in stating their strong opinions.</p> <p>I'd like to discuss one other philosophy/intuition then, according to which both smooth and topological manifolds are obviously artificial, highly deficient models for what could be "present in nature", whereas the PL world is much "closer to the reality". I don't consider myself a practitioner of this or any other religion; what follows should be regarded as said by a fictional character, not by the author.</p> <p>1) As is well-known, the predisposition to seeing continuous and smooth as more natural than discrete is historical, following centuries of preoccupation with derivatives and (later) limits. Quantum physics and computer science may be changing the tide, but they don't usually compete with Calculus in a mathematician's education, at least not in the initial years. </p> <p>Here is a simple test. When you fold a sheet of paper, what is the intuitive model in your imagination: is it a smooth surface (when you look with a loupe at the fold), a cusp-like singularity (generic smooth singularity), or an an angle-like singularity (PL singularity)? No matter what is your subconscious preference, I bet you didn't base it on considerations of individual photons detected by the eye. But you could have based it on your previous experience with abstract models of surfaces, which is not independent of the historically biased education. (Just for fun, I wonder if your intuitive model would change if the paper sheet is folded second time so as to make a corner - which is unstable as a singularity of a smooth map $\Bbb R^2\to\Bbb R^2$, but has a stable singularity in the link.)</p> <p>2) On a molecular scale, the sheet of paper of course doesn't fit the model of a smooth surface, and although it is arguably not "discrete" or "PL" on a subatomic scale, the smooth surface model isn't restored either. Similarly, as is well-known, Maxwell equations and general relativity (which I guess are among the best reasons to study smooth topology) don't work at very small scales. The problem is that this "imperfection" of matter doesn't usually shake one's belief in "perfect" physical space. But it is perfectly consistent with modern physics (for those who don't know) that physical space is kind of discrete at a sub-Planck scale, as in <a href="http://en.wikipedia.org/wiki/Loop_quantum_gravity" rel="nofollow">loop quantum gravity</a> (which is somewhat reminiscent of PL topology!). It is also consistent with the present day knowledge, and indeed derivable in variants of the competing string theory, that a finite volume of physical space can only contain a finite amount of information, as with the <a href="http://en.wikipedia.org/wiki/Holographic_principle" rel="nofollow">holographic principle</a>. (In fact I didn't see much discussion of possible alternatives to this principle, many physicists appear to take it for granted.) I'm getting on a slippery slope, but finite information does not sound like it could be compatible with limits that occur in derivatives (which returns us to MacPherson's program on combinatorial differential manifolds) and especially with Casson handles that occur in topological manifolds.</p> <p>The fictional character is now saying that his religion teaches him to avoid concepts based on inherently infinitary constructions, because they are likely to be unnatural, in the sense of the physical nature which might simply have no room for them (and even the question of whether it does is not obviously meaningful!). Ironically, this is quite in line with Poincare's philosophical writings, where he argued at length that the principle of mathematical induction is not an empirical fact.</p> <p>3) The fictional character goes on to say that this is not just the crazy metaphysics that displays the warning, but also Grothendieck with his <a href="http://kolmogorov.unex.es/~navarro/res/esquisseeng.pdf" rel="nofollow">"tame topology"</a> which inspired a whole area in logic (initiated by van den Dries' book <a href="http://books.google.ru/books?id=CLnElinpjOgC&amp;lpg=PP1&amp;hl=ru&amp;pg=PP1#v=onepage&amp;q&amp;f=false" rel="nofollow"><em>Tame topology and o-minimal structures</em></a>). Here is a short quote from Grothendieck:</p> <blockquote> <p>It is this [inertia of mind] which explains why the rigid framework of general topology is patiently dragged along by generation after generation of topologists for whom "wildness" is a fatal necessity, rooted in the nature of things.</p> <p>My approach toward possible foundations for a tame topology has been an axiomatic one. Rather than declaring [what] the desired “tame spaces” are ... I preferred to work on extracting which exactly, among the geometrical properties of the semianalytic sets in a space $\Bbb R^n$, make it possible to use these as local "models" for a notion of "tame space" (here semianalytic), and what (hopefully!) makes this notion flexible enough to use it effectively as the fundamental notion for a “tame topology” which would express with ease the topological intuition of shapes. </p> </blockquote> <p>Grothendieck dismisses from the start PL and smooth topology as possible forms of tame topology, because </p> <p>(i) they're "not stable under the most obvious topological operations, such as contraction-glueing operations", and</p> <p>(ii) they're not closed under constructions such as mapping spaces, "which oblige one to leave the paradise of finite dimensional spaces".</p> <p>I'm not familiar with "contraction-glueing operations", nor is Google. Perhaps someone fluent in French could explain what (i) is supposed to mean? My first guess would be that this could refer to mapping cylinder, mapping cone or other forms of homotopy colimit, but PL topology is closed under those (finite homotopy colimits). </p> <p><strong>Edit:</strong> Indeed, it is clear from the preceding pages that by "gluing" Grothendieck means the <a href="http://en.wikipedia.org/wiki/Adjunction_space" rel="nofollow">adjunction space</a>, which he also calls "amalgamated sum". In particular, he says:</p> <blockquote> <p>It was also clear that the contexts of the most rigid structures which existed then, such as the "piece-wise linear" context were equally inadequate – one common disadvantage consisting in the fact that they do not make it possible, given a pair $(U,S)$ of a "space" $U$ and a closed subspace $S$, and a glueing map $f: S\to T$, to build the corresponding amalgamated sum.</p> </blockquote> <p>There is, of course, no problem with forming adjunction spaces in the PL context. Perhaps Grothendieck was just not aware of pseudo-radial projection or something. <strong>End of edit</strong></p> <p>As to (ii), there now <a href="http://front.math.ucdavis.edu/1109.0346" rel="nofollow">exists</a> some kind of an infinite-dimensional extension of PL topology, which includes mapping spaces and infinite homotopy colimits up to homotopy equivalence (and hopefully up to uniform homotopy equivalence, which would be more appropriate in that setup). Besides, there are, of course, Kan sets, which are closed under Hom, but they arguably don't belong to tame topology in any reasonable sense because they quickly get uncountable (in every dimension, in particular, there are uncountably many vertices) and even of larger cardinality.</p> <p>In any case, logicians, who tried to set up Grothendieck's aspiration in a rigorous framework of <a href="http://mathoverflow.net/questions/44129" rel="nofollow">definability</a> (see <a href="http://eprints.ma.man.ac.uk/1745/01/covered/MIMS_ep2012_3.pdf" rel="nofollow">Wilkie's survey</a>), do now <a href="http://arxiv.org/abs/1002.1508" rel="nofollow">have</a> the "o-minimal tringulability and Hauptvermutung" theorem, saying roughly that tame topology (as they understood it) is the same as PL topology. Still more roughly (perhaps, too roughly) is could be restated as "topology without infinite constructions is the same as PL topology".</p> <p>Even if smooth topology will some day be reformulated in purely combinatorial terms, it is highly unlikely that it can be characterized by purely logical constraints. From this viewpoint, smooth topology is primarily justified by its role in applied math and natural sciences, but is no less and no more fundamental than symplectic topology or topology of hyperbolic manifolds. </p> http://mathoverflow.net/questions/96069/status-of-pl-topology/96080#96080 Answer by Sergey Melikhov for Status of PL topology Sergey Melikhov 2012-05-05T18:59:17Z 2012-05-05T21:22:03Z <blockquote> <p>Maybe I should put the question this way: is or is not PL topology an integral part of the education of every geometric topologist today?</p> </blockquote> <p>According to a recent poll by the Central Planning Commitee for Universal Education Standards, some geometric topologists don't have a clue about regular neighborhoods, while others haven't heard of multijet transversality; but they all tend to be equally excited when it comes to Hilbert cube manifolds.</p> <blockquote> <p>some recommended references (textbooks) for a beginner</p> </blockquote> <p><a href="http://www.google.com/url?sa=t&amp;rct=j&amp;q=&amp;esrc=s&amp;source=web&amp;cd=2&amp;ved=0CGUQFjAB&amp;url=http%253A%252F%252Fbooks.google.com%252Fbooks%252Fabout%252FIntroduction_to_piecewise_linear_topolog.html%253Fid%253DNkPvAAAAMAAJ&amp;ei=42ylT_2WCoLo-gbj_uDxAg&amp;usg=AFQjCNF0d5u_0pu5t0cadrDK98fQh3BlYw" rel="nofollow">Rourke-Sanderson</a>, <a href="http://www.maths.ed.ac.uk/~aar/papers/zeemanpl.pdf" rel="nofollow">Zeeman</a>, <a href="http://www.maths.ed.ac.uk/~aar/surgery/stallpl.pdf" rel="nofollow">Stallings</a>, <a href="http://www.maths.ed.ac.uk/~aar/papers/hudson.pdf" rel="nofollow">Hudson</a>, </p> <p>L. C. Glaser, <em>Geometrical combinatorial topology</em> (2 volumes)</p> <blockquote> <p>Is there any more modern textbook on the subject?</p> </blockquote> <p>Not really (as far as I know), but some more recent books related to PL topology include:</p> <p><a href="http://books.google.ru/books?id=w7dActmezxQC&amp;lpg=PP1&amp;hl=ru&amp;pg=PP1#v=onepage&amp;q&amp;f=false" rel="nofollow">Turaev, <em>Quantum invariants of knots and 3-manifolds</em></a> (chapters on the shadow world)</p> <p><a href="http://books.google.ru/books?id=f3OGV4BR-6QC&amp;lpg=PR3&amp;hl=ru&amp;pg=PR3#v=onepage&amp;q&amp;f=false" rel="nofollow">Kozlov, <em>Combinatorial algebraic topology</em></a> (chapters on discrete Morse theory, lexicographic shellability, etc.)</p> <p><a href="http://books.google.ca/books?id=vFLgAyeVSqAC&amp;lpg=PP1&amp;hl=ru&amp;pg=PP1#v=onepage&amp;q&amp;f=false" rel="nofollow">Matveev, <em>Algorithmic topology and classification of 3-manifolds</em></a></p> <p><a href="http://books.google.com/books?id=9_9QHM7hQRcC&amp;lpg=PP1&amp;ots=ZbkctTLaut&amp;dq=homotopy%252520and%252520combinatorial%252520group%252520theory%252522&amp;pg=PP1#v=onepage&amp;q&amp;f=false" rel="nofollow"><em>2D homotopy and combinatorial group theory</em></a></p> <p><a href="http://www.calvin.edu/~venema/embeddingsbook/embeddings-noprint.pdf" rel="nofollow">Daverman-Venema, <em>Embeddings in manifolds</em></a> (about a third of the book is on PL embedding theory)</p> <p>Benedetti-Petronio, <em>Branched standard spines of 3-manifolds</em></p> <p><a href="http://www.google.com/url?sa=t&amp;rct=j&amp;q=&amp;esrc=s&amp;source=web&amp;cd=5&amp;ved=0CGcQFjAE&amp;url=http%253A%252F%252Fmath.stanford.edu%252F~anton%252FTorusActions_BP.pdf&amp;ei=qGalT4m1Ko_n-gbZrbnmAg&amp;usg=AFQjCNFwXcrrMj4qnhIBrzRhtzjiXxHTJQ" rel="nofollow">Buchstaber-Panov, <em>Torus actions and their applications in topology and combinatorics</em></a> </p> <p><a href="http://books.google.ru/books?id=fBA8AAAAIAAJ&amp;lpg=PP1&amp;hl=ru&amp;pg=PP1#v=onepage&amp;q&amp;f=false" rel="nofollow">Buoncristiano, Rourke, and Sanderson, <em>A geometric approach to homology theory</em></a> (includes the PL transversality theorem)</p> <p><a href="http://www.maths.ed.ac.uk/~aar/books/haupt.pdf" rel="nofollow"><em>The Hauptvermutung book</em></a></p> <p><a href="http://www.maths.ed.ac.uk/~aar/haupt/sandro.pdf" rel="nofollow">Buoncristiano, <em>Fragments of geometric topology from the sixties</em></a></p> <blockquote> <p>Also I would like to know some important open problems in the area, in what problems are mathematicians working in this field nowadays</p> </blockquote> <p>I'll mention two problems.</p> <p>1) Alexander's 80-year old problem of whether any two triangulations of a polyhedron have a common iterated-stellar subdivision. They are known to be related by a sequence of stellar subdivisions and inverse operations (Alexander), and to have a common subdivision (Whitehead). However the notion of an arbitrary subdivision is an affine, and not a purely combinatorial notion. It would be great if one could show at least that for some family of subdivisions definable in purely combinatorial terms (e.g. replacing a simplex by a simplicially collapsible or consructible ball), common subdivisions exist. See also remarks on the Alexander problem by <a href="http://msp.warwick.ac.uk/gtm/1999/02/gtm-1999-02-016p.pdf" rel="nofollow">Lickorish</a> and by <a href="http://arxiv.org/abs/0708.4039" rel="nofollow">Mnev</a>, including the story of how this problem was thought to have been solved via algebraic geometry in the 90s.</p> <p>2) MacPherson's <a href="http://www.math.ucdavis.edu/~deloera/MISC/BIBLIOTECA/trunk/MacPherson/MacPherson4.pdf" rel="nofollow">program</a> to develop a purely combinatorial approach to smooth manifold topology, as attempted by <a href="http://www.google.com/url?sa=t&amp;rct=j&amp;q=&amp;esrc=s&amp;source=web&amp;cd=2&amp;ved=0CGUQFjAB&amp;url=http%253A%252F%252Fannals.math.princeton.edu%252Fwp-content%252Fuploads%252Fannals-v158-n3-p04.pdf&amp;ei=onOlT5-IOsSe-wbshbXeAg&amp;usg=AFQjCNHSCTbpY1TJ8pc6dqrj4z9wJF-I4Q" rel="nofollow">Biss</a> and refuted by <a href="http://arxiv.org/abs/0709.1291" rel="nofollow">Mnev</a>.</p> http://mathoverflow.net/questions/95923/category-of-uniform-spaces/95964#95964 Answer by Sergey Melikhov for Category of Uniform spaces Sergey Melikhov 2012-05-04T08:28:06Z 2012-05-04T22:59:40Z <p>There are two possible meanings of "uniform space" in the literature. I'll follow Isbell's terminology where the definition of a <em>uniform</em> space includes the separation axiom, and one speaks of a <em>pre-uniform</em> space when it is not included.</p> <p>The category of pre-uniform spaces and uniformly continuous maps has <em>concrete</em> limits and colimits: you can take the (co)limit of the diagram of underlying sets, in the category of sets, and then endow it with the initial (final) uniform structure. </p> <p>The category of uniform spaces still has concrete limits; and colimits that are generally not concrete, using a standard uniform quotient of a pre-uniform space.</p> <p>Neither category is cartesian closed. The exponential law $Z^{Y\times X}\ne (Z^Y)^X$ fails in general, unless $X$ is compact. For instance, a uniformly continuous map $I\times\Bbb R\to\Bbb R$ (that is, a uniform homotopy) is not the same as a uniformly continuous map $\Bbb R\to \Bbb R^I$ (that is, a homotopy through uniformly continuous maps $\Bbb R\to\Bbb R$). Indeed, $id:\Bbb R\to\Bbb R$ is not uniformly null-homotopic, but is null-homotopic through uniformly continuous maps.</p> <p>All of the above is discussed in some form in Isbell's book "Uniform spaces". For a quick review see also section 2.B <a href="http://front.math.ucdavis.edu/1106.3249" rel="nofollow">here</a>. Beware of the tricky nature of sequential colimits, as noticed by Taras Banakh, <a href="http://arxiv.org/abs/0908.2228" rel="nofollow"><em>The topological structure of direct limits in the category of uniform spaces</em></a>.</p> <p>As for complete uniform spaces, I believe they are closed under limits, but not closed under pushouts.</p> <p><strong>Added later:</strong> If you want something that feels like uniform spaces and is cartesian closed, I suggest the <a href="http://www.sciencedirect.com/science/article/pii/0166864185900069" rel="nofollow"><em>Cartesian closed hull of the category of uniform spaces</em></a> by Jiří Adámek and Jan Reiterman. The objects of this hull are bornological uniform spaces, i.e. uniform spaces endowed with a collection of "bounded" sets; the morphisms are the uniformly continuous maps which preserve the bounded sets. Bornological uniform spaces are really cute: when all sets are designated as bounded, these can be identified with the usual uniform spaces, and when only subsets of compact sets are designated as bounded, these can be identified with compactly generated Tychonoff topological spaces. </p> <p>Unfortunately Adámek and Reiterman have an error in the proof of Lemma 2.3 (in the last line of the proof of assertion (i)). I haven't seen this error discussed in the literature, but I believe it can be remedied by replacing $Hom(A^\ast,I)$ in the statement of the lemma with $Hom(A^\ast,I^\Lambda)$, where $\Lambda$ is a family of pseudo-metrics defining the uniformity of $A$. .</p> <p>Other work in this direction includes <a href="http://dx.doi.org/10.1017/S0004972700022905" rel="nofollow"><em>The category of uniform convergence spaces is cartesian closed</em></a> by R. S. Lee and some related work by her adviser Oswald Wyler; and <em>Metrizable spaces in Cartesian-closed subcategories of uniform spaces</em> and <a href="http://onlinelibrary.wiley.com/doi/10.1111/j.1749-6632.1989.tb22395.x/abstract" rel="nofollow"><em>Productivity of α-Bounded Uniform Spaces</em></a> by Gloria Tashjian, which I think clarifies her previous results with her advisor M. D. Rice. </p> http://mathoverflow.net/questions/95911/classifying-spaces-of-a-profinite-groups/95967#95967 Answer by Sergey Melikhov for Classifying spaces of a profinite groups Sergey Melikhov 2012-05-04T09:06:20Z 2012-05-04T09:06:20Z <p>There's a lot of literature on classifying spaces of the $p$-adic integers. For instance, there are <a href="http://www.ams.org/journals/tran/1967-129-01/S0002-9947-1967-0212127-3/home.html" rel="nofollow">two 45-year old "computations"</a> of the complex K-theory of the $p$-adic integers, resulting in two different answers. An explanation of why they are different, only found recently, is by considering a more refined notion of a classifying space, which is a uniform space well-defined up to uniform homotopy equivalence (see, for lack of a better account, subsection 1.4 <a href="http://arxiv.org/abs/1109.0346v1" rel="nofollow">here</a>).</p> http://mathoverflow.net/questions/94404/singular-fibers-of-generic-smooth-maps-of-negative-codimension/95597#95597 Answer by Sergey Melikhov for Singular fibers of generic smooth maps of negative codimension Sergey Melikhov 2012-04-30T18:07:32Z 2012-04-30T20:49:48Z <p>Fibers of a generic smooth map are polyhedra (so in particular CW-complexes) by the triangulation conjecture of Thom, proved by Andrei Verona [<a href="http://www.calstatela.edu/faculty/averona/Triang.html" rel="nofollow"><em>Stratified Mappings - Structure and Triangulability</em>, Springer LNM vol. 1102</a>]. I'm not sure that the full strength of the triangulation conjecture is really needed here - hopefully Ryan's projected answer will clarify this. A more general triangulation conjecture was proved by Masahiro Shiota [<em>Thom’s conjecture on triangulations of maps</em>, Topology, 39 (2000), 383–399], who also has further results on this subject on the arXiv. </p> <p>Verona's theorem says that any proper, topologically stable smooth map between smooth manifolds $M$ and $N$ is "triangulable", i.e. equivalent to a PL map by a topological change of coordinates in $M$ and in $N$. On the other hand the set of proper topologically stable smooth maps $M\to N$ is dense in $C^\infty(M,N)$ by the Thom-Mather theorem.</p> <p>So don't worry, there are no Hawaiian earrings or other nightmares hiding in the fibers. To ensure that nothing obstructs easy sleep, we want the fibers of a generic smooth map $M^m\to N^n$ to be polyhedra of dimension $\le\max(m-n,0)$. Indeed, forgetting about the "polyhedra" part for a moment, we know the dimension estimate from multijet transversality (as sketched by Tom). We can now achieve "polyhedra" and "of dimension $\le\max(m-n,0)$" to hold simultaneously because maps generic in two ways are still generic. (That is, the intersection of two open and dense subsets of $C^\infty(M,N)$ is open and dense. I guess this needs $M$ to be compact, and the general case follows by writing $M$ as a union of an increasing chain of compact submanifolds and applying Baire's theorem.)</p> http://mathoverflow.net/questions/94462/when-is-the-freudenthal-compactification-an-anr/94524#94524 Answer by Sergey Melikhov for When is the Freudenthal compactification an ANR? Sergey Melikhov 2012-04-19T13:13:53Z 2012-04-19T22:24:59Z <p>Regarding simple examples: take $X$ to be the infinite mapping telescope of an inverse sequence of connected compact polyhedra $P_i$ (so that $X$ has one end) such that for some $j$, the inverse sequence of the groups $G_i=H_j(P_i)$ does not satisfy the Mittag-Leffler condition or its inverse limit does not inject in any $G_i$. Then the Freudenthal compactification, which in this case is the same as the one-point compactification $X^+$, is not an ANR.</p> <p>The Mittag-Leffler condition is not satisfied for instance if each $P_i=S^j$ and each bonding map $P_{i+1}\to P_i$ is a degree $2$ map ($j>0$). The inverse limit does not inject in any $G_i$ for instance if each $P_i$ is the $j$-fold suspension $S^{j-1}*[2^i]$ over the discrete space of cardinality $2^i$ (here $j>0$ in order to have $P_i$ connected), and each bonding map $P_{i+1}\to P_i$ is the suspension over a trivial $2$-fold cover.</p> <p>More generally one can take the $P_i$ to be disconnected, but such that the inverse limit of $\pi_0(P_i)$ injects into some $\pi_0(P_i)$. Then $X$ has finitely many ends (though there might be infinitely many proper homotopy classes of proper maps $[0,\infty)\to X$ if $\pi_1(P_i)$ do not satisify the Mittag-Leffler condition). So if the Freudenthal compactification of $X$ is an ANR, then $X^+$ is an ANR. </p> <p>So why is $X^+$ not an ANR if the $G_i$ do not satisfy the Mittag-Leffler condition or their inverse limit does not inject in any $G_i$? This follows from Dydak's necessary and sufficient condition for $X^+$ to be ANR, where $X$ is the infinite mapping telescope of an inverse sequence of compact polyhedra $P_i$, in terms of the homology groups and the fundamental groups of the $P_i$. I've tried to understand this result, and have written up a somewhat different proof of the homological part (which perhaps also counts as a short exposition that you request), see Theorem 6.12 in <a href="http://arxiv.org/abs/0812.1407" rel="nofollow">http://arxiv.org/abs/0812.1407</a>. Theorem 3.12 also gives a similar necessary and sufficient condition for the mapping telescope $X$ to be forward tame, in terms of the homotopy groups of the $P_i$. See also Lemma 3.4 for a quick explanation of some terms.</p> http://mathoverflow.net/questions/94182/trivial-cobordism-group-in-dimensions-1-3-7-related-to-h-space-structures-on-th/94201#94201 Answer by Sergey Melikhov for Trivial cobordism group in dimensions 1, 3, 7 related to H-space structures on the spheres in these dimensions? Sergey Melikhov 2012-04-16T11:02:11Z 2012-04-16T11:02:11Z <p>How about $n\ne 0,2,6$ instead of $n=1,3,7$? The Adams theorem on the vanishing of the stable Hopf invariant, which implies the non-existence of an H-structure on $S^n$ for $n\ne 1,3,7$, also implies the theorem of <a href="http://projecteuclid.org/euclid.ijm/1256049409" rel="nofollow">R. L. W. Brown</a> and <a href="http://projecteuclid.org/euclid.ijm/1256050807" rel="nofollow">Liulevicius (and Mahowald)</a> that if $M^n$ immerses in $\Bbb R^{n+1}$, then $M^n$ is a boundary or is cobordant to $\Bbb RP^0$, $\Bbb RP^2$ or $\Bbb RP^6$ (and these three do immerse in $\Bbb R^{n+1}$).</p> http://mathoverflow.net/questions/93408/reference-on-geometric-topology/93523#93523 Answer by Sergey Melikhov for Reference on Geometric Topology Sergey Melikhov 2012-04-08T22:10:40Z 2012-04-09T05:57:49Z <p>Geometric topology is not really a unique field in the way that algebraic topology and general topology are. Its various subareas may share something of a common feel (and indeed an arxiv category), but are often too diverse to have any common techniques. Those areas include, for instance:</p> <p>Low-dimensional topology (classical knots, 3-manifolds, 4-manifolds, etc.)</p> <p>Morse theory, simple homotopy theory and algebraic K-theory of spaces</p> <p>Dimension theory (of separable metrizable spaces)</p> <p>Topology of manifolds (surgery theory, codimension two knots, etc.)</p> <p>Singularity theory (of smooth maps), geometric immersion theory (dealing e.g. with 4-tuple points of sphere eversions)</p> <p>PL topology (block bundles, collapsing, bistellar moves, etc.)</p> <p>Generalized manifolds, wild knots, etc.</p> <p>Group actions on manifolds</p> <p>Manifold structures (smoothing/trangulability and the Hauptvermutung; also Lipschitz structures, <a href="http://www.math.ucdavis.edu/~deloera/MISC/BIBLIOTECA/trunk/MacPherson/MacPherson4.pdf" rel="nofollow">CD-manifolds</a>, ... )</p> <p>Embedding theory (smooth embeddings of projective spaces, PL embeddings of polyhedra, etc.)</p> <p>I surely forgot to mention many important subjects here; even the grouping of items in this list is rather arbitrary (and the order is random). The point is, you will probably not get far with diving in some depth into geometric topology unless you're more specific on what you're interested in. If unsure, try some knot theory or low-dimensional manifolds. These now cover more than half of all geometric topology by any count. Ryan and Jim gave some good suggestions of starting points in their math.SE answers, such as Rolfsen's 'Knots and links'. There are also <a href="http://books.google.ru/books?id=vFLgAyeVSqAC&amp;lpg=PP1&amp;hl=ru&amp;pg=PP1#v=onepage&amp;q&amp;f=false" rel="nofollow">other</a> <a href="http://books.google.ru/books?id=IdRsF_mezbIC&amp;lpg=PP1&amp;hl=ru&amp;pg=PP1#v=onepage&amp;q&amp;f=false" rel="nofollow">flavors</a> of <a href="http://books.google.ru/books?id=jPPiur0Dd6UC&amp;lpg=PP1&amp;hl=ru&amp;pg=PP1#v=onepage&amp;q&amp;f=false" rel="nofollow">low-dimensional</a> <a href="http://www.pdmi.ras.ru/~duzhin/papers/cdbook/" rel="nofollow">topology</a>.</p> <p>There exist some books and courses mentioning 'geometric topology' in the title, but they are often specialized and/or advanced. For instance, the 'geometric topology' notes by <a href="http://www.maths.ed.ac.uk/~aar/books/gtop.pdf" rel="nofollow">Sullivan</a> and <a href="http://www.math.harvard.edu/~lurie/937.html" rel="nofollow">Lurie</a> are mostly focused on manifold structures, and are firmly grounded in methods which are very clever and useful, but kind of external to geometric topology (localization, Galois theory and simplicial sets). Likewise, Bing's <a href="http://books.google.ru/books?id=NBCQ0a_cdXcC&amp;lpg=PP1&amp;hl=ru&amp;pg=PP1#v=onepage&amp;q&amp;f=false" rel="nofollow">'Geometric topology of 3-manifolds'</a> and Moise's 'Geometric topology in dimension 2 and 3' are mostly about wild things. (There's definitely a trend in the literature that if geometric topology gets explicitly mentioned, things are likely not all smooth or PL.) </p> <p>Arguably, closer to the point are Fenn's <a href="http://books.google.ru/books?id=M144AAAAIAAJ&amp;lpg=PP1&amp;hl=ru&amp;pg=PP1#v=onepage&amp;q&amp;f=false" rel="nofollow">'Techniques of geometric topology'</a> and Ferry's <a href="http://www.math.rutgers.edu/~sferry/ps/geotop.pdf" rel="nofollow">'Geometric topology notes'</a>. Even these two virtually don't overlap with each other, so they are certainly not equivalents of some canonical algebraic topology text such as Spanier's or Hatcher's. But perhaps closer to such an equivalent than anything else that I can think of.</p> http://mathoverflow.net/questions/87387/well-pointed-space-which-is-not-locally-contractible/88628#88628 Answer by Sergey Melikhov for Well-pointed space which is not locally contractible Sergey Melikhov 2012-02-16T13:44:23Z 2012-02-17T10:14:09Z <p>My initial answer was fully ignorant of the literature on the subject; I'm now recalling that it does exist. In 1934, Borsuk and Mazurkiewicz constructed a $2$-dimensional AR (=contractible, locally contractible $2$-dimensional compactum) $X$ that is not a countable union of smaller ARs (Sur les rétractes absolus indécomposables, C.R.. Acad. Sci. Paris 199 (1934), 110-112; see Borsuk's "Theory of retracts", Section VI.4). The proof actually shows that $X$ is not a countable union of contractible proper subsets (open or not). Thus $X$ contains points that do not have a basis of contractible neighborhoods. </p> <p>A simplified version $Y$ of $X$ (constructed below), which is an AR failing to have a basis of contractible neighborhoods at a certain point $\infty$, is also forward tame at $\infty$ and so answers Ricardo's <a href="https://lists.lehigh.edu/pipermail/algtop-l/2012q1/001331.html" rel="nofollow">revised question</a> at the Algebraic Topology Discussion List. $Y$ looks somewhat like a $2$-skeleton of $W^+$, the example in my previous answer, except that $W^+$ is not forward tame. A google search for "Mazurkiewicz singularity" (this is how Borsuk called the kind of ANRs that do not have arbitrarily fine countable covers by ARs) returns a paper by <a href="http://matwbn.icm.edu.pl/ksiazki/fm/fm69/fm69110.pdf" rel="nofollow">Armentrout</a> which has an example somewhat similar to $W^+$ (also a decomposition space and a manifold factor) though more laborious.</p> <p>Note also a positive result of <a href="http://projecteuclid.org/euclid.bams/1183511682" rel="nofollow">Begle</a> proved using Zorn's lemma.</p> <hr> <p>Here's a construction of $X$ (which I hope is easier to read than Borsuk's account) and also of $Y$. </p> <p>Let $Q=D^2/\{a,b\}$, the quotient of the $2$-disk where two distinct interior points $a,b$ are identified with each other. Let $\partial Q$ be the image of $\partial D^2$ in $Q$, and let $C$ be the image in $Q$ of a path joining $a$ and $b$ in the interior of $D^2$. Let $(Q_1,\partial Q_1)=(D^2,\partial D^2)$, and form $(Q_k,\partial Q_k)$ by attaching $Q_{k-1}$ to $(Q,\partial Q)$ along a homeomorphism $\partial Q_{k-1}\to C$. Then $Q_k$ is contractible, but no proper subset of $Q_k$ containing $\partial Q_k$ is contractible (or even acyclic).</p> <p>Now $Y$ is just the one point compactification $(Q_\infty)^+$ of the direct limit $Q_\infty$ of the chain of inclusions $Q_1\subset Q_2\subset\dots$.</p> <p>To see that $Y$ is locally contractible, observe that $D^2\times [0,1]$ collapses onto a copy of $Q_2$ such that $\partial Q_2=\partial D^2\times\{1\}$ and $Q_1=D^2\times\{0\}$. It follows that $D^2\times [0,\infty)$ admits a proper deformation retraction onto a copy of $Q_\infty$. Thus $Q_\infty$ is proper homotopy equivalent to $[0,\infty)$. Then $Q_\infty$ is forward tame in the sense of Quinn (see Definition 7.1, p.75 and Proposition 9.6, p.99 in <a href="http://www.maths.ed.ac.uk/~aar/books/ends.pdf" rel="nofollow">Hughes-Ranicki</a>), and therefore its one-point compactification is locally contractible.</p> <p>For the construction of $X$ we will need $k$ copies $C_1^{Q_k},\dots,C_k^{Q_k}$ of $S^1$ embedded in $Q_k$, namely $C_1^{Q_k}=\partial Q_k$, and $C_{i+1}^{Q_k}=C_i^{Q_{k-1}}$. Now $X$ is formed by attaching increasingly smaller copies of $Q_1,Q_8,Q_{64},...$ to the <a href="http://en.wikipedia.org/wiki/Sierpinski_carpet" rel="nofollow">Sierpinsky carpet</a> so that for each $k$ of the form $8^m$, the $k$ circles $C_1^{Q_k},\dots,C_k^{Q_k}$ in $Q_k$ are attached to the boundaries of the disks removed at the $m$th stage of the construction of the Sierpinski carpet, and additionally $(k-1)$ arcs connecting $C_i^{Q_k}$ with $C_{i+1}^{Q_{k+1}}$ are identified with analogous arcs in the Sierpinsky carpet. This can, and must be done so that every neighborhood of every point of the Sierpinsky carpet contains $C_1^{Q_k}=\partial Q_k$ for some $k$ (of the form $8^m$).</p> http://mathoverflow.net/questions/87387/well-pointed-space-which-is-not-locally-contractible/87678#87678 Answer by Sergey Melikhov for Well-pointed space which is not locally contractible Sergey Melikhov 2012-02-06T15:46:08Z 2012-02-16T18:17:45Z <p>Edit of edit: I'm adding still more details and references. </p> <p><em> The one-point compactification $W^+=W\cup\{\infty\}$ of the <a href="http://en.wikipedia.org/wiki/Whitehead_manifold" rel="nofollow">Whitehead manifold</a> $W$ is locally contractible (in the sense of geometric topology - see Goodwillie's comment above) and hence well-pointed at $\infty$.</em></p> <p>This can be seen directly, and also follows from the fact $W^+\times\Bbb R$ is homeomorphic to $S^3\times\Bbb R$ (for this homeomorphism see Chapter 4 "Manifold factors" in <a href="http://www.math.rutgers.edu/~sferry/ps/geotop.pdf" rel="nofollow">Ferry's notes</a>). Perhaps this implication works as an argument for judging what definition of "locally contractible" is the "right" one, so I'm including the details.</p> <p>Since $\{\infty\}\times\Bbb R\subset W^+\times\Bbb R$ is just a copy of $\Bbb R$ in $S^3\times\Bbb R$ (a wild copy, though - see my comment below), every neighborhood $U$ of $\{\infty\}\times\Bbb R$ in $W^+\times\Bbb R$ contains a smaller neighborhood $V$ such that the inclusion $i:V\to U$ is homotopic to a map $V\to\Bbb\{\infty\}\times\Bbb R$. (Using that $\Bbb R$ is an ANR to get the map, and that $S^3\times\Bbb R$ is locally contractible to get the homotopy.) Since $\Bbb R$ is contractible, $i$ is null-homotopic. Then every neighborhood $U'$ of $\infty$ in $W^+$ contains a smaller neighborhood $V'$ (namely $V'=V\cap W^+\times\{0\}$) that contracts in $U:=U'\times\Bbb R$. This contraction can be projected down to $W^+$.</p> <p><em> On the other hand $W^+$ has no basis of contractible neighborhoods at $\infty$. </em></p> <p>Indeed $W$ is the union of compact submanifolds $W_1\subset W_2\subset\dots$ of $S^3$, each $W_k$ being the closure of the complement of a solid torus $T_k$. If $U$ is a contractible neighborhood of infinity in $W^+$ contained in $W^+\setminus W_1$, then $U$ contains $W^+\setminus W_k$ for some $k$. Now $W\setminus W_k$ contains a knot $K_k$ isotopic to the core of the solid torus $S^3\setminus W_k$ by an isotopy of that solid torus. Thus $K_k$ is contained in the contractible subset $U$ of $W^+\setminus W_1$. </p> <p>Consider $f:S^3\to W^+$ that is the identity on $W$ and sends $S^3\setminus W$ to $\infty$ (in fact it is a quotient map). This map has acyclic point inverses (in the sense of Cech cohomology) so $f^{-1}(U)$ is an acyclic subset of $S^3\setminus W_1$ containing $K_k$. (This is using the Vietoris-Begle theorem, which is a special case of the Leray spectral sequence for a continuous map - see for instance Bredon's "Sheaf Theory".) But it is not hard to show that there exists no such acyclic subset (see N. Smythe, $n$-linking and $n$-splitting, <a href="http://www.jstor.org/pss/2373507" rel="nofollow">Amer. J. Math. 92 (1970), 272-282</a>). </p> <hr> <p>Smythe's proof of the latter assertion is very lovely. He calls compact subsets $A,B\subset S^3$ $n$-split (this is the revised definition, on p.277), if there exists a sequence of compact subpolyhedra $A\subset P_0\subset\dots\subset P_{n+1}\subset S^3\setminus B$ such that each inclusion $P_i\subset P_{i+1}$ is trivial on reduced integral homology groups. Thus “(−1)-split” means “disjoint”, and a two-component link has 0-split components iff it has zero linking number. It is also easy to see that a two-component link has 1-split components iff it is a boundary link (i.e. the components bound disjoint Seifert surfaces in $S^3$). It is easy to see, using the Alexander duality, that the relation of being $n$-split is symmetric.</p> <p>Returning to the Whitehead manifold, it suffices to show that $K_k$ is not $(k-1)$-split from $W_1$, or equivalently from the core $K_0$ of the solid torus $W_1$. We prove by induction a slightly stronger assertion, that every essential (i.e. not null-homotopic) simple closed curve in the solid torus $T_k$ (=the closure of $S^3\setminus W_k$) is not $(k-1)$-split from $K_0$. Assume that this holds with $k=n$, and let $C$ be an essential simple closed curve in $T_{n+1}$. If $C$ is $n$-split from $K_0$, then $C$ bounds an orientable surface $F$ in $S^3\setminus K_0$ that is $(n-1)$-split from $K_0$. Without loss of generality $D:=F\cap\partial T_n$ is a closed $1$-manifold. By the induction hypothesis, every simple closed curve in $F\cap T_n$ is null-homotopic in $T_n$; in particular, so is each component of $D$.</p> <p>Let $\tilde T_n$ be the universal cover of $T_n$. Since $C$ lies in $T_{n+1}$, it is null-homotopic in $T_n$, and so lifts to a simple closed curve $C_0$ in $\tilde T_n$. Similarly $F\cap T_n$ lifts to a compact surface in $\tilde T_n$ with boundary $\tilde C_0\cup\tilde D_0$. The translate $C_1=t(C_0)$, where $t$ is a generator of the covering translation group $\Bbb Z$, has linking number zero with every component of $\tilde D_0$, and therefore also with $\tilde C_0$. On the other hand if $[C]=m\in\Bbb Z=\pi_1(T_{n+1})$, then $\tilde C_0$ has linking number $m^2$ with $\tilde C_1$. So $m=0$ and $C$ is null-homotopic in $T_{n+1}$, which is a contradiction.</p> http://mathoverflow.net/questions/86436/the-grassmannian-of-a-simplicial-complex/87690#87690 Answer by Sergey Melikhov for The "grassmannian" of a simplicial complex Sergey Melikhov 2012-02-06T17:24:43Z 2012-02-06T17:24:43Z <p>What you describe is similar to something well-known that does satisfy your condition (ie, has the same homotopy as $X$ in dimensions less than $n-k$). I'm speaking of the dual complex to the dual $(n-k)$-skeleton of your simplicial complex. Let me call this $X^{\ast (n-k)\ast}$ (which is short for $((X^\ast)^{(n-k)})^\ast$). This is not a simplicial complex in general, only a cone complex. It has one vertex for each $k$-simplex; vertices corresponding to those $k$-simplices that belong to the same $(k+1)$-simplex are connected by an "edge" (that is, a cone over those vertices - notice that there can be more than two of them); and so on. You can find the discussion of dual complexes in standard texts on PL topology, but cone complexes as such are not clearly exposed there; see references on cone complexes in my answer <a href="http://mathoverflow.net/questions/61948/are-there-two-non-homeomorphic-finite-regular-cw-complexes-with-isomorphic-face-p/61961#61961" rel="nofollow">here</a>.</p> http://mathoverflow.net/questions/82956/cohomology-of-a-space-with-local-coefficients-and-singular-cohomological-dimensio/82992#82992 Answer by Sergey Melikhov for Cohomology of a space with local coefficients and singular cohomological dimension Sergey Melikhov 2011-12-08T18:21:31Z 2011-12-08T18:21:31Z <p>How about $X=\Bbb RP^2\times L^2_3$, where $L^2_3$ is the cone of the $3$-fold cover $S^1\to S^1$. Here $H^4(X;\Bbb Z)\simeq\Bbb Z/2\otimes\Bbb Z/3=0$, but $H^4(X;\mathcal L)\simeq\Bbb Z\otimes\Bbb Z/3$, where $\mathcal L$ is the pullback of the orientation sheaf of $\Bbb RP^2$.</p> <p>As a side remark, "singular cohomological dimension" is not something that people normally do, perhaps because singular cohomology is not Brown representable, or because of the Barratt-Milnor example of a compact subset of $\Bbb R^3$ (in fact it is just the one-point compactification of $\Bbb R^2\times\Bbb Z$) which has infinite "singular cohomological dimension". If you do care about spaces not homotopy equivalent to CW-complexes, you can look up some books on traditional dimension theory, which deal with usual (that is, Cech) cohomological dimension. For spaces homotopy equivalent to CW-complexes, there's no distinction because all ordinary cohomology theories coincide.</p> http://mathoverflow.net/questions/80925/degenerating-immersion/81064#81064 Answer by Sergey Melikhov for degenerating immersion Sergey Melikhov 2011-11-16T13:22:42Z 2011-11-16T14:48:35Z <p>New answer to the generalized question. It's shown in previous answers that for $z^2$, and some other branched coverings, there are no immersions that are $C^1$-close except at the branch points. (I believe this should also imply that there are no immersions that are $C^1$-close except on a finite set.)</p> <p>But $z^3:S^2\to S^2$ is arbitrarily $C^\infty$-close, except at the two branch points, to a $C^\infty$ immersion in $\Bbb R^3$. (Also, any $C^\infty$ map $S^2\to S^2$ that is equivalent to $z^3$ by a $C^0$ change of coordinates is $C^\infty$-close on the entire $S^2$ to an immersion in $\Bbb R^3$). To see this, pick a generic lift $f:S^1\to S^1\times\Bbb R$ of the $3$-fold covering $S^1\to S^1$. It suffices to show that the composition $f':S^1\xrightarrow{f} S^1\times\Bbb R\subset S^2$ bounds an immersion of a $2$-disk in a $3$-ball. Equivalently, we want to find a regular homotopy from $f'$ to an embedding. But it is an exercise that that there are only two regular homotopy classes of immersions $S^1\to S^2$, distinguished by the parity of the number of double points (in the case of self-transverse immersions).</p> http://mathoverflow.net/questions/80925/degenerating-immersion/81004#81004 Answer by Sergey Melikhov for degenerating immersion Sergey Melikhov 2011-11-15T20:00:03Z 2011-11-15T22:02:06Z <p>The idea of Anton Petrunin can be made into an accurate proof. One does not need $C^2$ convergence, $C^1$ convergence is enough. That is, I claim that there is no $C^1$ immersion sufficiently $C^1$-close to the composition $\phi:S^2\xrightarrow{z^2}S^2\subset\Bbb R^3$. (By the way, any map $S^2\to\Bbb R^3$ is $C^0$-close to a $C^\infty$ immersion, according to the $C^0$-dense $h$-principle and using that $S^2$ immerses in $\Bbb R^3$.)</p> <p>Let $f:S^2\to\Bbb R^3$ be a self-transverse map (not necessarily an immersion) that is $C^1$-close to $\phi$. The image of $f$ lies in a tubular neighborhood $S^2\times\Bbb R$ of the image of $\phi$. Consider the composition $\psi:S^2\xrightarrow{f}S^2\times\Bbb R\xrightarrow{\text{projection}}S^2$. It is $C^1$-close to $\phi$, so it is equivalent to $\phi$ by a change of coordinates outside a small neighborhood of the poles (which are the singular points of $\phi$). </p> <p>So we may assume that, outside of a small neighborhood of the poles, $f$ is a vertical lift of $\phi$ (with respect to the projection $S^2\times\Bbb R\to S^2$). Then, in particular, $f$ sends the equator of $S^2$ into the plane $\Pi$ in $\Bbb R^3$ that contains the equator of $S^2$. This equatorial map is a $C^1$-approximation to the composition $S^1\xrightarrow{\text{double covering}}S^1\subset\Pi$, so it is an immersion and has an odd number of double points. But then the double point set of $f$ cannot be a union of closed curves. So $f$ cannot be an immersion.</p> http://mathoverflow.net/questions/80777/what-is-a-continuous-path/80790#80790 Answer by Sergey Melikhov for What is a continuous path? Sergey Melikhov 2011-11-13T01:12:34Z 2011-11-15T00:39:41Z <p>The technique/framework mentioned by Jim Conant and used by Berestovskii and Plaut goes back to the paper </p> <p><a href="http://matwbn.icm.edu.pl/ksiazki/fm/fm104/fm104113.pdf" rel="nofollow">J. Krasinkiewicz and P. Mine, Generalized paths and pointed 1-movability</a></p> <p>For an update on this subject see <a href="http://front.math.ucdavis.edu/0706.3937" rel="nofollow">http://front.math.ucdavis.edu/0706.3937</a> and the last section in <a href="http://front.math.ucdavis.edu/0812.1407" rel="nofollow">http://front.math.ucdavis.edu/0812.1407</a> (which includes a simplified proof of the Krasinkiewicz-Minc result).</p> <p><strong>Edit:</strong> I was writing this in rush for an airplane, and could not elaborate on what the references are about. Now that strong shape (and even related things like Cech cohomology and Steenrod-Sitnikov homology) have been mentioned by others this simplifies my job. </p> <p>What Krasinkiewicz and Minc were doing in that paper is essentially paths in the sense of strong shape. (They don't explicitly speak of "strong shape", but on the other hand it happens that papers and books that originally developed strong shape, including Tim Porter's, and most of subsequent literature under the "strong shape" brand has been incredibly focused on either categorical or general-topology aspects and didn't care to pursue any specific geometric problems, so if you're interested in any kind of substantial results on paths in the sense of strong shape, you have to look for them elsewhere!) </p> <p>In the above-mentioned paper, Krasinkiewicz and Minc proved the following wonderful theorem: If $X$ is a connected (metrizable) compactum that is disconnected in the sense of strong shape (that is, not all strong shape morphisms from a point into $X$ are the same) then there exist distinct strong shape morphisms (in fact, uncountably many ones) from a point into $X$ that are represented by genuine points in $X$. This may sound like it should be either trivial or wrong, but no, it's a deep geometric result.</p> http://mathoverflow.net/questions/79488/is-there-a-knot-theory-for-graphs/79544#79544 Answer by Sergey Melikhov for Is there a "knot theory" for graphs? Sergey Melikhov 2011-10-30T20:28:41Z 2011-10-30T20:41:35Z <blockquote> <p>whether there is a "knot theory" for graphs...</p> <p>or can it be essentially reduced to the study of knots (and links)?</p> </blockquote> <p>As Dror Bar-Natan points out in his interesting answer, it can, "if you totally understand the theory of tangles". If you don't, but you're very generous as to what amounts to a reduction, then it "almost can" (up to about one integer invariant) by a theorem of Roberston, Seymour and Thomas: two knotless, linkless embeddings $f,g$ of a graph $G$ in $\Bbb R^3$ are equivalent (by an isotopy of $\Bbb R^3$) if and only if the restictions of $f$ and $g$ to every subgraph of $G$, homeomorphic to $K_5$ or $K_{3,3}$ are equivalent. Here "knotless" means that every cycle (a subgraph homeomorphic to $S^1$) in $G$ is unknotted, and "linkless" means that every two disjoint subgraphs are separated by an embedded $2$-sphere. To be precise, Robertson, Seymour and Thomas had a slightly different formulation (with "panelled" in place of "knotless and linkless") and the above version is proved in <a href="http://arxiv.org/abs/math/0612082" rel="nofollow">http://arxiv.org/abs/math/0612082</a>.</p> <p>What is the "about one integer invariant"? As Ryan Budney points out in his interesting answer, it helps to study graphs up to weaker equivalence relation than ambient equivalence or non-ambient isotopy (which, incidentally, already kills all local knots). Taniyama (Topol. Appl. 65 (1995), 205-228) has shown that two embeddings of a graph $G$ in $\Bbb R^3$ are "homologous" (=cobound an embedded $G\times I$+(handles) in $\Bbb R^3\times I$, where each handle is a torus attached by a tube to a $2$-cell, (edge)$\times I$) if and only if they have the same Wu invariant (this integer invariant is really just the $1$-parameter version of the van Kampen obstruction). On the other hand, Shinjo and Taniyama (Topol. Appl. 134 (2003), 53-67) have shown that the vanishing of the Wu invariant of a graph is determined by the vanishing of its restriction to subgraphs homeomorphic to $K_5$, $K_{3,3}$ and $S^1\sqcup S^1$.</p> <p>Another interesting relation on embedded graphs in link homotopy, i.e. arbitrary self-intersections of connected components are allowed, but distinct components may not intersect. The link homotopy classification of embeddings in $\Bbb R^3$ of a disjoint union of two $S^1$'s and a wedge of $S^1$ is already <a href="http://mi.mathnet.ru/eng/znsl1145" rel="nofollow">pretty nontrivial</a>.</p> http://mathoverflow.net/questions/79084/representing-rational-homology-by-manifolds/79140#79140 Answer by Sergey Melikhov for Representing Rational Homology by Manifolds Sergey Melikhov 2011-10-26T02:54:49Z 2011-10-26T03:39:48Z <p>A nice, direct combinatorial construction was given by Gaifullin, see his <a href="http://front.math.ucdavis.edu/0712.1709" rel="nofollow">papers</a> <a href="http://front.math.ucdavis.edu/0806.3580" rel="nofollow">on the</a> <a href="http://front.math.ucdavis.edu/0912.3933" rel="nofollow">arXiv</a>. A drawback of this approach is that if you think of it as realizing some multiple $m\alpha$ of the given integral homology class $\alpha$ by an oriented smooth manifold, then $m$ is not bounded in terms of the dimension of $\alpha$.</p> <p>There has also been another geometric approach. Thom also proved that $\bmod2$ homology classes are representable by maps of smooth (possibly unorientable) manifolds. This was reproved geometrically in </p> <blockquote> <p>S. Buoncristiano and D. Hacon, <em>An elementary geometric proof of two theorems of Thom,</em> Topology 20 (1981), no. 1, 97–99</p> </blockquote> <p>The other theorem of their title is that unoriented bordism is determined by Stiefel-Whitney numbers, and it is used in their proof that mod 2 homology classes are representable by smooth manifolds.</p> <p>I believe the same geometric argument should also work to show that rational homology classes are representable by oriented smooth manifolds - <em>modulo</em> the fact that Pontryagin numbers determine oriented bordism tensored by $\Bbb Q$. This fact I'm afraid I don't know how to prove geometrically (for some proof, see e.g. the Milnor-Stasheff book). But note that in a subsequent paper Buouncristiano and Hacon also gave a geometric proof that Chern numbers determine complex bordism (Ann. of Math., 118 (1983), 1-7). Their other papers may also be of interest if you care about geometric proofs of classical results on bordism.</p> http://mathoverflow.net/questions/75896/status-of-the-compact-ar-problem Status of the compact AR problem? Sergey Melikhov 2011-09-20T00:02:46Z 2011-10-24T15:22:12Z <p>The so-called "compact AR Problem" reads: </p> <blockquote> <p>Is every compact convex set in a metrizable topological vector space an absolute retract?</p> </blockquote> <p>It is open according to the chapter by T. Banakh, R. Cauty and M. Zarichnyi in <a href="http://books.google.com/books?id=XdXnQCV5K08C&amp;lpg=PA609&amp;vq=1264&amp;pg=PA609" rel="nofollow">"Open Problems in Topology II"</a> (2007) and according to a <a href="http://www.math.ac.vn/publications/acta/pdf/0903389.pdf" rel="nofollow">2009 paper</a>, which acknowledges discussions of its authors with Nguyen To Nhu.</p> <p>Yet according to Sehie Park's 2004 <a href="http://hdl.handle.net/2433/25349" rel="nofollow">survey</a> and another <a href="http://www.m-hikari.com/imf-password2009/17-20-2009/arandjelovicIMF17-20-2009.pdf" rel="nofollow">2009 paper</a> it has been resolved affirmatively in a 8-page paper by Park published in some conference proceedings in 2004. (If you happen to solve <a href="http://www.math.nmsu.edu/~nnguyen/20paper.ps" rel="nofollow">"one of the most outstanding open problems in infinite dimensional topology"</a>, apparently you may have an urge to ignore top international journals and submit it to Antarctica Journal of Mathematics - especially if you happen to live in Antarctica.)</p> <p>Given that Cauty and Nhu have been long active in the area and have obtained substantial results related to the problem, it would be strange if they missed a correct solution. But it is also strange that they didn't mention Park's paper (whatever they think of it). </p> <blockquote> <p>So what is going on here?</p> </blockquote> <p>(I must admit that so far I have mostly checked out just those sources that are freely accessible on the internet, but given that some confusion already exists in at least one of the two 2009 papers, I thought that even if I'm missing a trivial solution such as MathSciNet it is perhaps already legitimate to ask this question here, so as to make that trivial solution more widely known.)</p> http://mathoverflow.net/questions/78824/realisability-cohomological-class-as-product-or-as-immersed-sphere/78834#78834 Answer by Sergey Melikhov for Realisability cohomological class as product or as immersed sphere Sergey Melikhov 2011-10-22T16:18:20Z 2011-10-23T00:45:38Z <p>I will construct a closed simply-connected $8$-manifold $M$ and an $a\in H^3(M;\Bbb Z)$ such that the Poincare dual $b$ of $a$ is not realizable by a map $S^5\to M$, and a Hom-dual element in $H^5(M;\Bbb Z/2)$ to the $\bmod 2$ reduction of $b$ is not a nontrivial product. </p> <p>Let $K$ be the suspension over $\Bbb C P^2$. Then there is an $\alpha\in H^3(K;\Bbb Z/2)$ with $Sq^2(\alpha)\ne 0$, but the $\bmod 2$ cohomology ring of $K$ is trivial. Let $N$ be a regular neighborhood of a PL copy of $K$ in some $\Bbb R^m$. So $N$ is homotopy equivalent to $K$. A loop in $\partial N$ bounds a disk in $N$, which can be pushed off $K$ as long as $5+2\le m-1$. Thus $\partial N$ is simply-connected. Let $M$ be the double of $N$, i.e. $M=\partial (N\times I)$. So $M$ is a closed $m$-manifold, it is simply-connected by Seifert-van Kampen, and the inclusion $N\subset M$ is split by the projection $\phi:M\subset N\times I\to N$. So the cohomology of $N\simeq K$ is a direct summand in the cohomology of $M$. Let $\beta=\phi^\ast(\alpha)$, then $\gamma:=Sq^2\beta\ne 0$, and $\gamma$ is not a nontrivial product. Since the nonzero element of $H^5(S^5;\Bbb Z/2)$ is not in the image of $Sq^2$, there is no map $f:S^5\to M$ such that $f^*(\gamma)\ne 0$.</p> <p>Let $b\in H_5(M;\Bbb Z)$ be such that $\gamma(b)\ne 0$, and let $a\in H^{m-5}(M;\Bbb Z)$ be the Poincare dual of $b$. If $b$ is realized by an immersion, or just a map, $f:S^5\to M$, then $0\ne\gamma\smallfrown f_\ast[S^5]=f_\ast(f^\ast(\gamma)\smallfrown[S^5])$, contradicting $f^*(\gamma)=0$. </p> <p>As for $m$, $\Bbb C P^2$ is the mapping cone of the Hopf map $h:S^3\to S^2$. The mapping cylinder of $h$ embeds in $S^2*S^3=S^6$, so $\Bbb CP^2$ embeds in $\Bbb R^7$ and $K$ in $\Bbb R^8$.</p> http://mathoverflow.net/questions/25171/how-thinly-connected-can-a-closed-subset-of-hilbert-space-be/78816#78816 Answer by Sergey Melikhov for How thinly connected can a closed subset of Hilbert space be? Sergey Melikhov 2011-10-22T04:03:33Z 2011-10-22T19:58:51Z <p>Some remarks, not an answer.</p> <p>1) As pointed out by BS, the problem is: does there exist a connected Polish space containing no nondegenerate proper closed connected subspace? In the language of continuum theory this sounds suspiciously simple: does there exist a connected Polish space whose all <a href="http://en.wikipedia.org/wiki/Composant" rel="nofollow">composants</a> are singletons? Or (still equivalently): does there exist a connected Polish space that is <a href="http://eom.springer.de/i/i052580.htm" rel="nofollow">irreducible</a> between every pair of distinct points? Surely continuum theorists have thought about this question, haven't they?</p> <p>2) As hinted by Garabed, such a space $X$ cannot be locally compact. Indeed, in locally compact spaces, components coincide with <a href="http://en.wikipedia.org/wiki/Locally_connected_space#Quasicomponents&amp;rct=j&amp;q=quasicomponents&amp;usg=AFQjCNG9QniNCOIL0WBpqafW4JoMC6-Fwg&amp;sa=X&amp;ei=ay-iTv-UJaHY4QTl8vHuBA&amp;ved=0CCcQygQwAQ" rel="nofollow">quasi-components</a>. Let $F$ be a closed ball of some radius about some $x\in X$, so that $F\ne X$, and let $C$ be the component of $x$ in $F$. If $C\ne\{x\}$, then $C$ is a closed connected nondegenerate subset of $X$. If $C=\{x\}$, then $\{x\}$ is also a quasi-component, so $x$ is contained in arbitrarily small clopen sets in $F$. Then they are also clopen in $X$, so $X$ cannot be connected. </p> <p>3) There exists a connected Polish space $X$ containing no nondegenerate <em>compact</em> connected subspace. For instance, the graph of $f(x)=\sum_{n=1}^\infty 2^{-n}\sin(\frac1x-r_n)$, where $r_n$ is the $n$th rational number (in some order) and $\sin(\infty)=0$. See Kuratowski's "Topology" (volume II, section 47.IX in the 1968 edition). <a href="http://www.ams.org/journals/bull/1942-48-02/S0002-9904-1942-07615-4/S0002-9904-1942-07615-4.pdf" rel="nofollow">In fact</a>, such an $X$ can even be locally connected (of course, it cannot be locally path-connected at any point).</p> <p><a href="http://www.ams.org/journals/bull/1942-48-02/S0002-9904-1942-07615-4/S0002-9904-1942-07615-4.pdf" rel="nofollow">4)</a> The graph of a discontinuous function $f:\Bbb R\to\Bbb R$ satisfying $f(x+y)=f(x)+f(y)$ can be connected, and in that case it contains no nondegenerate <em>bounded</em> connected subset.</p> http://mathoverflow.net/questions/115567/covering-maps-in-real-life-that-can-be-demonstrated-to-students Comment by Sergey Melikhov Sergey Melikhov 2013-03-14T09:17:53Z 2013-03-14T09:17:53Z G.C.: The question is not equivalent to asking whether the mapping cylinder embeds in $\Bbb R^3$. Also, there are no problems with point-set topology because quotient topology wasn't mentioned. In fact, the question is equivalent to asking whether the mapping cylinder has a level-preserving embedding in $\Bbb R^3\times [0,1]$, where the mapping cylinder is endowed with the (metrizable) topology of quotient uniformity. If the domain is compact, this is the same as quotient topology, but if it's not, quotient topology is non-metrizable. http://mathoverflow.net/questions/64744/how-should-one-think-about-pushforward-in-cohomology/64785#64785 Comment by Sergey Melikhov Sergey Melikhov 2013-03-14T02:12:16Z 2013-03-14T02:12:16Z It turns out that some people seemingly unaware of B-R-S have developed a similar (but not identical) approach to geometric homology, which involves additional choices, but on the other hand may be more suitable for K-theory: <a href="http://mathoverflow.net/questions/119872/references-for-geometric-k-homology/119874#119874" rel="nofollow" title="references for geometric k homology">mathoverflow.net/questions/119872/&hellip;</a> http://mathoverflow.net/questions/64744/how-should-one-think-about-pushforward-in-cohomology/64785#64785 Comment by Sergey Melikhov Sergey Melikhov 2013-03-14T01:46:49Z 2013-03-14T01:46:49Z $f_!$ can well be non-zero for negative $n$ (in ordinary cohomology). For example, even though the projection $f:S^5\times S^7\to S^5$ represents $[f]=0\in H_{-7}(S^5)$, the pushforward map $f_!:H^7(S^5\times S^7)\to H^0(S^5)$ is an isomorphism. It sends the class of a 7-comanifold $g:S^5\to S^5\times S^7$ to the class of the $0$-comanifold $fg:S^5\to S^5\times S^7\to S^5$. http://mathoverflow.net/questions/118563/seifert-surfaces-via-alexander-duality/118568#118568 Comment by Sergey Melikhov Sergey Melikhov 2013-03-09T09:33:41Z 2013-03-09T09:33:41Z (see also <a href="http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1076924/" rel="nofollow">ncbi.nlm.nih.gov/pmc/articles/PMC1076924</a>) http://mathoverflow.net/questions/118280/constructing-a-deformation-neighborhood Comment by Sergey Melikhov Sergey Melikhov 2013-03-09T08:44:14Z 2013-03-09T08:44:14Z It is easier 1) to find a neighborhood of $X$ in the standard cubical grid with side $1/2^m$, for some $m$, that retracts (but not deformation retracts) onto $X$; 2) to find a neighborhood of $X$ that is a cubical complex (but not a subcomplex of the standard grid) and deformation retracts onto $X$. To get (1), use 4.30 and 4.31 in <a href="http://arxiv.org/abs/1106.3249v4" rel="nofollow">arxiv.org/abs/1106.3249v4</a> . To get (2), use 4.3 in <a href="http://arxiv.org/abs/1109.0346v2" rel="nofollow">arxiv.org/abs/1109.0346v2</a> , or alternatively (this is slightly less effective) use Misha's answer and apply the canonical subdivision to cut simplices into cubes. http://mathoverflow.net/questions/118563/seifert-surfaces-via-alexander-duality/118568#118568 Comment by Sergey Melikhov Sergey Melikhov 2013-03-09T07:55:56Z 2013-03-09T07:55:56Z &quot;Serre's theorem&quot; that $H^1(X)=[X,S^1]$ was proved by Bruschlinsky, Math. Ann. 109 (1934), 525-537. http://mathoverflow.net/questions/123739/is-there-a-conceptual-reason-why-topological-spaces-have-quotient-structures-whil/123753#123753 Comment by Sergey Melikhov Sergey Melikhov 2013-03-07T04:10:46Z 2013-03-07T04:10:46Z Eric, do you mean 1-Lipschitz (non-expansive) maps, or is it strictly distance-decreasing and/or strictly preserving (isometries)? It seems that what you're saying is easy to prove for 1-Lipshitz maps, but anyway are there any good references for this sort of questions? http://mathoverflow.net/questions/123739/is-there-a-conceptual-reason-why-topological-spaces-have-quotient-structures-whil/123809#123809 Comment by Sergey Melikhov Sergey Melikhov 2013-03-07T03:30:17Z 2013-03-07T03:30:17Z Yes, your answer is related to Eric's. The category of metric spaces and uniformly continuous maps has many quotients (though not all) defined by the formula in Eric's answer. (In other words, that formula gives a well-defined metric, in many cases, if the original metric is changed without changing the uniform structure.) For example, you can define mapping cylinder and join in this category (which you can't do for metric spaces and continuous maps). This is done in <a href="http://arxiv.org/abs/1106.3249" rel="nofollow">arxiv.org/abs/1106.3249</a> http://mathoverflow.net/questions/123513/the-role-of-anr-in-modern-topology/123557#123557 Comment by Sergey Melikhov Sergey Melikhov 2013-03-05T04:15:39Z 2013-03-05T04:15:39Z Breadth of applicability is very good; I'm all for model categories (and homotopy type theory). I just don't see what this all has to do with ANRs (and topological manifolds). As you explain, ANRs are not really needed to do homotopy theory; on the other hand, model categories haven't yet helped anyone to do ANRs (and hence topological manifolds), AFAIK. http://mathoverflow.net/questions/123513/the-role-of-anr-in-modern-topology/123557#123557 Comment by Sergey Melikhov Sergey Melikhov 2013-03-05T04:03:57Z 2013-03-05T04:03:57Z Also, by &quot;topology literature&quot; and &quot;early topology papers&quot; you probably mean algebraic topology? Things like Freedman's proof of the topological 4D Poincare Conjecture, Quinn's proof of the Annulus Conjecture, and Edwards and Cannon's proof of Milnor's Double Suspension Conjecture are very much about ANRs. For the record, these include 2 Annals papers from 1979, and a 1986 Fields medal; a further 1975 Annals paper mentions &quot;ANRs&quot; in its title (it's the main ingredient of West's proof that ANRs have finite types). http://mathoverflow.net/questions/123513/the-role-of-anr-in-modern-topology/123557#123557 Comment by Sergey Melikhov Sergey Melikhov 2013-03-05T03:17:34Z 2013-03-05T03:17:34Z Does &quot;more scaleable&quot; mean more axiomatic? But I don't know any nontrivial model category where all objects, or all (co)fibrant objects are ANRs. (One problem is that the cone over a non-compact space is non-metrizable.) So I don't see how any talk about ANRs (or even regular neighborhoods) could be made implicit by cofibrations. If your point is that using ANRs looks dated in some AT textbooks, it's a question of presentation. The essential feature of ANRs is, of course, that they include topological manifolds and are similar enough to them, but more easily manageable. http://mathoverflow.net/questions/122356/how-to-see-the-quaternionic-hopf-map-generates-the-stable-3-stem Comment by Sergey Melikhov Sergey Melikhov 2013-03-02T00:15:10Z 2013-03-02T00:15:10Z A geometric construction of a generator of $\pi_3^{st}$ is discussed in this paper by Eckholm and Takase: <a href="http://arxiv.org/abs/0903.0238" rel="nofollow">arxiv.org/abs/0903.0238</a> Note also that the composition of the 8-fold covering $S^3\to S^3/Q$, where $Q$ is the quaternion group, and a standard embedding $S^3/Q\to S^4$ represents a generator of $3\pi_3^{st}$. http://mathoverflow.net/questions/123160/a-uniformity-with-a-countable-base-is-a-pseudometric-uniformity/123171#123171 Comment by Sergey Melikhov Sergey Melikhov 2013-03-01T23:49:22Z 2013-03-01T23:49:22Z Fran&#231;ois, uniform spaces have some (pre)history independent of topological groups (see the chapter by Bentley, Herrlich, and Husek in &quot;Handbook of the History of General Topology, Vol. 2&quot;). In particular your proof above is originally from a paper by Alexandroff and Urysohn (C. R. Acad. Sci., Paris 177 (1923), 1274-1276) http://mathoverflow.net/questions/76188/is-there-any-fibration-mathbbrn-to-mathbbsn/76251#76251 Comment by Sergey Melikhov Sergey Melikhov 2013-03-01T00:40:53Z 2013-03-01T00:40:53Z This makes it much simpler, thanks! http://mathoverflow.net/questions/112629/has-the-notion-of-space-been-reconsidered-in-20th-century/112643#112643 Comment by Sergey Melikhov Sergey Melikhov 2012-11-17T09:36:23Z 2012-11-17T09:36:23Z ... as well as more recent developments, <a href="http://arxiv.org/pdf/gr-qc/9905087v1.pdf" rel="nofollow">arxiv.org/pdf/gr-qc/9905087v1.pdf</a>