User vcf - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T22:03:39Z http://mathoverflow.net/feeds/user/23204 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115795/extended-integral-in-spivaks-calculus-on-manifolds/115801#115801 Answer by VCF for Extended integral in Spivak’s Calculus on Manifolds VCF 2012-12-08T15:29:42Z 2012-12-08T20:13:23Z <p>Following the notation in the book (see page 65), if it exists, the extended integral is defined as $\int_{A}f:=\sum_{\varphi\in\Phi}\int_{A}\varphi\cdot f$, where each integral $\int_{A}\phi\cdot f$ exists and being of the the usual type it verifies all the corresponding properties for usual integrals. This and a bit of care, as we are dealing with series, is enough to show each and every one of the properties mentioned above for extended integrals.</p> http://mathoverflow.net/questions/115795/extended-integral-in-spivaks-calculus-on-manifolds Extended integral in Spivak’s Calculus on Manifolds VCF 2012-12-08T13:13:24Z 2012-12-08T20:13:23Z <p>On page 48 of Calculus on Manifolds Spivak defines (Riemann) integration over <em>rectangles</em> $[a_{1},b_{1}]\times\cdots\times[a_{n},b_{n}]\subset\mathbb{R}^{n}$. Then on page 55 he extends this integral to bounded subsets $C\subset\mathbb{R}^{n}$ via characteristic functions. This integral is the usual one. </p> <p>Then, on page 63 he defines (smooth) partitions of unity and uses them later on page 65 to define and <em>extended integral</em> over open sets $A\subset\mathbb{R}^{n}$. </p> <p>The usual and extended integrals are not always the same. However, Theorem 3-12 (3) gives us a precise relation between the extended integral and the usual one.</p> <p>Now, mostly via problems, Spivak makes the reader verify all the familiar properties of the usual integral (<em>linearity, comparison, monotonicity, etc</em>). However, there is no mention in either the theory or the exercises of whether these properties hold for the extended integral. Moreover, when doing the problems I found myself making use of them, so it is natural to ask if the extended integral also verifies these properties. That is: </p> <p>Let $A$ be an open subset of $\mathbb{R}^{n}$ and $f,g:A\rightarrow\mathbb{R}$ be continuous functions:</p> <ol> <li><em>Linearity</em>: If $f,g$ integrable over $A$, so is $af+bg$ and $\int_{A}af+bg=a\int_{A}f+b\int_{A}g$.</li> <li><em>Comparison</em>: $f,g$ integrable over $A$ and $f(x)\leq g(x)$ then $\int_{A}f\leq\int_{A}g$. In particular $\left|\int_{A}f\right|\leq\int_{A}|f|$.</li> <li><em>Monotonicity</em>: If $B \subset A$ is open and $f$ in non-negative on $A$ and integrable over $A$ then it is integrale over $B$ and $\int_{B}f\leq\int_{A}f$.</li> <li><em>Additivity</em>: If $A$ and $B$ are open and $f$ is continuous on $A\cup B$ and integrable over $A$ and $B$ then it is integrable over the union and the inetrsection and $\int_{A\cup B}f=\int_{A}f+\int_{B}f-\int_{A\cap B}f$.</li> <li>Let $A$ be open and of measure $0$. If $f$ is integrable over $A$ then $\int_{A}f=0$.</li> <li>If $f$ and $g$ agree except on a set of measure $0$ then $\int_{A}f=\int_{A}g$.</li> </ol> <p>I have been verifying these properties and seem to have a proof for each. <strong>But I would appreciate it if someone with more experience could corroborate that these properties do indeed hold for extended integrals.</strong></p> http://mathoverflow.net/questions/108167/three-half-circles-on-the-plane-may-not-meet-nicely Three half circles on the plane may not meet nicely VCF 2012-09-26T14:57:51Z 2012-09-26T19:54:41Z <p>Let $H$ denote the union of the northen hemisphere of the unit circle $S^{1}$ with the interval $[-1,1]$ on the $x$-axis. That is, $H=\{(x,\sqrt{1-x^{2}}):-1\le x\le 1\}\cup\{(x,0):-1\le x\le 1\}$</p> <p>Let us say that two copies of $H$ <em>meet nicely</em> if they intersect in exactly 6 points e.g. as the two pictures show below:</p> <p><img src="http://i50.tinypic.com/2ih8r5c.png" alt="two half circles meeting nicely"> or <img src="http://i48.tinypic.com/6rppwh.png" alt="two other circles meeting nicely"></p> <p>Note that the picture on the right shows two half-circles meeting nicely but whose centers do not lie inside their partner´s half disk.</p> <p>Now, if we have three copies of $H$, it may not seem possible to arrange them so that they <em>meet nicely</em>, i.e. both of the following two conditions hold: 1. Any two meet nicely, and 2 The intersection of the three is empty.</p> <p>Is this true? Or, if I am mistaken, I would appreciate that someome would show me, or descibe, the desired arrangement.</p> <p><strong>EDIT: I am considering ALWAYS half-circles, i.e. copies of $H$. No need to give answers related to half-disks.</strong></p> <p><strong>EDIT: It is possibel to arrange three copies of $H$ so that both 1. and 2. hold (See the seleted answer below).</strong></p> http://mathoverflow.net/questions/107285/cylinders-dividing-mathbbr3 Cylinders dividing $\mathbb{R}^{3}$ VCF 2012-09-15T20:05:55Z 2012-09-20T02:00:02Z <p>Consider $n$ affine copies of a compact cylinder, say $S^{1}\times [-3,3]$ with top and botom, sitting inside $\mathbb{R}^{3}$. </p> <p>For each $n$ we may ask ourselves how to arrange the $n$ cylinders so that they divide 3-space into the maximum number of regions possible. For example, one cylinder divides 3-space in two regions. Two cylinders, if we intersect them so as to make a cross, divide space in 6 regions, but maybe more is possible.</p> <p>If $n\ge 3$ things start to get complicated. For example, if $n=3$ we can obtain 14 regions if we start with two cylinders making a cross and then intersect the last cylinder diagonally, but I am not sure that this is the maximum. Perhaps more is possible?</p> <p>I would like to know if there is a general formula giving us the maximum number of regions into which 3-space can be divided by cylinders.</p> <p>Fundamental concepts like homology or the Euler characteristic may be of help if applied appropriately. Thus, if there is a general theory studying these kind of question I would appreciate any references on the matter.</p> http://mathoverflow.net/questions/94830/homotopy-equivalence-of-certain-kinds-of-adjunction-spaces Homotopy equivalence of certain kinds of adjunction spaces VCF 2012-04-22T10:02:01Z 2012-08-22T16:17:55Z <p>Suppose that $X$, $Y$ and $Z$ are topological spaces, with $A\subset X$, a map $f:A\rightarrow Y$, and a homotopy equivalence $\phi:Y\rightarrow Z$. It seems fair to think that the adjunction spaces $Y\cup_{f}X$ and $Z\cup_{\phi\circ f}X$ will be homotopy equivalent provided that $(X,A)$ has the homotopy extension property.</p> <p>A reasonable candidate for a homotopy equivalence seems to arise from the map $(Id,\phi):X+Y\rightarrow X+Z$ after passing to the quotient ($X+Y$ denotes disjoint union, and $(Id,\phi)$ is the map defined to be the identity on $X$ and $\phi$ on Y).</p> <p>Any suggestions will be appreciated.</p> <p>Thanks!</p> http://mathoverflow.net/questions/97747/topological-version-of-two-results-in-smooth-morse-theory Topological version of two results in smooth Morse theory VCF 2012-05-23T11:33:49Z 2012-05-28T14:04:56Z <p>Morse theory is generally presented in the DIFF category. However, there is a version of Morse theory in TOP (see the post <a href="http://mathoverflow.net/questions/96883/morse-theory-in-top-and-pl-categories" rel="nofollow"><em>Morse theory in TOP and PL categories?</em></a> for references).</p> <p>It is well known that a TOP (resp. DIFF) closed $n$-manifold that admits a TOP (resp. DIFF) Morse function having exactly two non-degenerate critical points must be homeomorphic to the $n$-sphere. </p> <p>In DIFF this is due to G. Reeb, and it remains true even if the critical points are degenerate as proven by Milnor and Rosen (Theorem 1, p.124 of [Mil2007]).</p> <p>In TOP it is due to Kuiper, [Kui1961]; and my first question is: <strong>Does the result remain true even if the points are allowed to be degenerate?</strong></p> <p>Working in DIFF, let $M$ be an $n$-manifold and $f$ a Morse function on $M$. Then if $[a,b]$ is an interval of regular values of $f$, it is not difficult to show that $f^{-1}(a)$ is homeomorphic to $f^{-1}(b)$ (e.g. Milnor's <em>Morse Theory</em>, p. 12). However, <strong>is this true in the topological case?</strong></p> <p>[Kui1961] N.H. Kuiper, A continuous function with two critical points, Bull. Amer. Math. Soc., 67(1961), 281-285.</p> <p>[Mil2007] J. Milnor, Collected Papers of John Milnor: Vol. III, Differential Topology, AMS, 2007. </p> http://mathoverflow.net/questions/96883/morse-theory-in-top-and-pl-categories Morse theory in TOP and PL categories? VCF 2012-05-14T10:53:40Z 2012-05-24T18:37:35Z <p>Apparently there are topological and piecewise linear versions of Morse theory. I would like to know of references that treat these topics.</p> <p>How is a Morse function defined for compact manifolds (with boundary) in the TOP and PL categories?</p> <p>It is well known that smooth Morse functions always exits for compact smooth manifolds. Are there similar results in the TOP and PL categories?</p> <p>It is possible to classify closed smooth surfaces via smooth Morse theory. Is there a classification theorem for closed TOP (respestively PL) surfaces via topological (respectively PL) Morse theory?</p> <p>Thanks</p> http://mathoverflow.net/questions/96883/morse-theory-in-top-and-pl-categories/97724#97724 Answer by VCF for Morse theory in TOP and PL categories? VCF 2012-05-23T07:08:22Z 2012-05-24T18:37:35Z <p>I think that Daniele and Sergey have, between the two pretty much answered my question. However, I would like to add the following:</p> <ol> <li><p>Regarding TOP Morse theory, in [Mor1959] M. Morse laid the foundations of the theory of topological non-degenerate functions, and proved the TOP Morse inequalities. </p> <p>Also, in [Kui1961] the topological version of the Reeb-Milnor theorem for DIFF manifolds is proven. That is, a TOP n-dimensional closed manifold admiting a TOP Morse function having exactly two non-degenerate critical points is homeomorphic to the n-sphere.</p> <p>Another source containing further results for topological manifolds via TOP Morse theory that parallel those obtained for differentiable manifolds is J. Cantwell's paper [Can1967].</p> <p>Kirby and Siebenmann's "Foundational essays on topological manifolds, smoothings, and triangulations is freely available <a href="http://www.maths.ed.ac.uk/~aar/papers/ks.pdf" rel="nofollow">here</a>. In section 3 of Essay III (p. 80) they define Morse functions in the DIFF and TOP categories for manifolds possibly with boundary.</p> <p>Finally, simple examples of non-differentiable TOP Morse function are easily found. For example, the absolute value function on $\mathbb{R}$, $x\rightarrow |x|$. The origin is a non-degenerate critical point. Also the height function restricted to the double cone (i.e., the space formed by the cones $x^{2}+y^{2}=(z\pm1)^{2}$) has exactly two non-degenerate critical points (the tips of the cones).</p></li> <li><p>Regarding PL Morse theory, J. Harer's <a href="http://www.ima.umn.edu/2005-2006/PISG7.10-28.06/activities/harer/PL-Morse_Theory.pdf" rel="nofollow">slides</a> contain an interesting approach using homology. In particular, a PL Morse function is defined using Betti numbers.</p></li> </ol> <p>REFERENCES:<br> [Cant1967] J. Cantwel, Topological non-degenerate functions, Tohoku Math. Journ., 20 (1968), 120-125.</p> <p>[Kui1961] N.H. Kuiper, A continuous function with two critical points, <em>Bull. Amer. Math. Soc.</em>, 67(1961), 281-285.</p> <p>[Mor1959] M. Morse, Topological non-degenerate functions on a compact manifold $M$, <em>Journal d'Analyse Math.</em>, 7 (1959), 189-208.</p> http://mathoverflow.net/questions/96670/classification-of-surfaces-and-the-top-diff-and-pl-categories-for-manifolds/97019#97019 Answer by VCF for Classification of surfaces and the TOP, DIFF and PL categories for manifolds VCF 2012-05-15T15:53:04Z 2012-05-15T16:33:22Z <p>I have come to believe that answering the questions I posted would be more enlightening if I try to provide an overview of the larger context that they are part of.</p> <p>The literature treating and generalizing the topics mentioned in the post for surfaces is as extensive as it is interesting. The 1960's and 70's were times of very active research in this part of topology, and it still is today. Three wonderul resources are Kirby &amp; Siebenmann's book <a href="http://www.maths.ed.ac.uk/~aar/papers/ks.pdf" rel="nofollow">Foundational Essays on Topological Manifolds, Smoothings and Triangulations</a>, Milnor's paper <a href="http://www.ams.org/notices/201106/rtx110600804p.pdf" rel="nofollow">Differential Topology Forty-six Years Later</a>, and A.Ranicki's <a href="http://www.maths.ed.ac.uk/~aar/slides/orsay.pdf" rel="nofollow">slides</a>. I will refer to these as [KS], [Mil2011] and [Ran], respectively. Also, a word on notation: <em>uniqueness</em> will mean up to PL, DIFF or TOP homeomorphism, depending on the category at hand. Unless otherwise stated, words like <em>manifold</em> or <em>surface</em> will have general meaning (i.e. possibly with boundary and possibly non-compact). Finally, a list of references is included at the bottom. </p> <ol> <li><p>[<strong>DIFF &amp; PL</strong>] (Strictly speaking PL and DIFF are not comparable. One uses the category <a href="http://en.wikipedia.org/wiki/PDIFF" rel="nofollow">PDIFF</a>, which is equivalent to PL. However, this distinction is not normally made unless technicalities may require so.) Differentiable manifolds admit canonical PL structures. A differentiable manifold can be triangulated uniquely up to PL equivalence. S.S. Cairns first proved this result for compact $C^{1}$ manifolds, including those having a finite number of boundary components (See [Cai1934], [Cai1936]), although he generalized these results later (see [Cai1961]). J.H.C. Whitehead proved it for $C^{1}$ manifolds without boundary (see [Whi1940]), and J. Munkres finally included $C^{r}$ manifolds with boundary, $1\le r\le\infty$ (see [Mun1966] or Theorem 3.10.2 in [TL]). </p> <p>A given PL structure on a topological manifold may have compatible differentiable structures that are inequivalent. That is, $$\mathrm{DIFF}\rightarrow \mathrm{PL}$$ is <strong>not injective</strong>. In [Mil1956] J. Milnor gave an example of a manifold PL-homeomorphic to the usual 7-dimensional sphere $S^{7}$, but not diffeomorphic to it. In fact, it is known that for $n\neq 4$ a topological $n$-sphere admits a unique PL structure (For $n\le 3$ see [Moi1977] or [TL], for $n\ge 5$ is due to Smale and can be found in [Sma1962]. The case $n=4$ is an open question). Therefore, the inequivalent differentiable structures that Milnor constructed in [Mil1956] are all compatible with the usual PL structure on $S^{7}$. </p> <p>Even More, on the topological manifold $\mathbb{R}^{4}$ it is possible to define uncountably many inequivalent PL or differentiable structures. An excellent account of this exotic $\mathbb{R}^{4}$'s can be found in Chapter XIV of [Kir1989].</p> <p>The functor above is also <strong>not surjective</strong>. That is, there are PL manifolds that do not admit a compatible differentiable structure. <a href="http://en.wikipedia.org/wiki/Kervaire_manifold" rel="nofollow">M. Kervaire</a> gave such an example in [Ker1960]. Later, J. Ells and N.H. Kuiper (see [EK1961]), and I. Tamura (see [Tam1961]) gave examples in dimension 8, the lowest possible. </p> <p>In dimensions 7 or less, PL manifolds always admit compatible differentiable structure, and in dimensions 6 or less this happens in an a unique way (See Theorem 2 in [Mil2011] and Theorem 3.10.8 and Problems 3.10.19-20 in [TL] for dimension up to three). In this sense, DIFF=PL for manifolds of dimension $n\leq 6$, which means that the number of inequivalent differentiable structures on a topological 4-sphere is also unknown. </p> <p>The obstruction to finding a differentiable structure on a given PL manifold is called the <em>Munkres-Hirsch-Mazur obstruction</em> (see the last paragraph on [Mil2011]).</p></li> <li><p>[<strong>TOP &amp; PL</strong>] $$\mathrm{PL}\rightarrow \mathrm{TOP}$$ is <strong>neither surjective nor injective</strong>. Indeed, there are topological manifolds, such as Freedman's <a href="http://en.wikipedia.org/wiki/E8_manifold" rel="nofollow">E8 manifold</a>, that do not admit a PL structure, or are even triangulable even if we allow non-PL <a href="http://en.wikipedia.org/wiki/Triangulation_%28topology%29" rel="nofollow">triangulations</a>. (A proof of this now follows from the proof of the 3-dimensional Poincaré conjecture, which implies that any triangulation of a 4-dimensional manifold is necessarily a PL-triangulation). </p> <p>The exotic $\mathbb{R}^{4}$'s mentioned above provide an example of a topological manifold having uncountably many inequivalent PL structures. This disproves the manifold version of the Hupvermutung. The non-manifold version of the Haupvermutung was disproven by J. Milnor ([Mil1961]), who found two homeomorphic compact simplicial complexes that are not PL homeomorphic. </p> <p>In dimension $3$ or less, the Hupvermutung is true (see Chapters 35 &amp; 36 in [Moi1977] or Thurston/Levy's book). In this sense PL=TOP for manifolds of dimension $n\le 3$. Moreover, as mentioned earlier, except possibly for $n=4$ there is only one $n$-dimensional PL sphere. </p> <p>The obstruction to finding a PL structure on a given topological manifold culminated with the resuts of Kirby and Siebenmann (the <a href="http://en.wikipedia.org/wiki/Kirby-Siebenmann_class" rel="nofollow"><em>Kirby–Siebenmann class</em></a>). (see [KS] and Theorem 1 in [Mil2011]).</p></li> <li><p>[<strong>TOP &amp; DIFF</strong>] As John Klein points out in the comments, smoothing a topological manifold is in general a question formulated by first putting a combinatorial structure on the manifold (normally a handlebody structure or a PL structure). The examples of Kervaire, Ells &amp; Kuiper and Tamura mentioned above yield topological manifolds having no differentiable structure. However, these are still PL manifolds. </p> <p>More striking is the E8 manifold which, not being triangulable, cannot have a differentiable structure. It provides and example of a topolgical manifold of dimension four that admits neither PL nor differentiable structures. </p> <p>The exotic $\mathbb{R}^{4}$'s mentioned earlier give an example of a topological manifold having uncountably many inequivalent differentiable structures. </p> <p>In dimension 3 or less the results above yield DIFF=PL=TOP. </p></li> </ol> <p>Coming back to surfaces I want to point out that Theorem 8.3 in [Moi1997] shows that <em>Every surface is triangulable</em>. At the begining of the proof it is shown that triangulations and PL structures are equivalent notions on a surface. Moreover, Theorem 8.5 is the Hauptvermutung for surfaces. Therefore, a complete classification of non-compact surfaces (with boundary) seems to have been achieved by the results contained and mentioned in Prishlyak and Mischenko's paper.</p> <p>Finally, I want to point out that the result that two smooth surfaces are diffeomorphic iff they are homeomorphic is due to J. Munkre's and can be found in his dissertation "Some Applications of Triangulation Theorems", U. of Michigan, 1955. The proof uses the triangulation theorems proven by E.E. Moise, who was Munkres' advisor.</p> <p>REFERENCES:</p> <p>[Cai1934] S.S. Cairns, On the triangulation of regular loci, Ann. of Math. 35 (1934), 579–587.</p> <p>[Cai1936] S.S. Cairns, Polyhedral approximation to regular loci, Ann. of Math. 37 (1936), 409–419.</p> <p>[Cai1961] S.S. Cairns, A simple triangulation method for smooth manifolds, Bull. Amer. Math. Soc. 67 (1961), 380–390.</p> <p>[Ker1960] M. Kervaire, A manifold which does not admit any differentiable structure, Comm. Math. Helv. 34 (1960), 257–270.</p> <p>[Kir1989] R. Kirby, The Topology of 4-Manifolds", Lecture Notes in Mathematics no. 1374, 1989.</p> <p>[EK1961] J. Eells, and N.H. Kuiper, Manifolds which are like projective planes, Publ. Math. IHES 14 (1961), 5-46.</p> <p>[Mil1961] J. Milnor, Two complexes which are homeomorphic but combinatorially distinct. Annals of Mathematics, 74-2 (1961), 575–590</p> <p>[Moi1977] E.E. Moise, Geometric Topology in Dimensions 2 and 3. New York, Springer-Verlag, 1977.</p> <p>[Mun1960] J.R. Munkres, Obstructions to smoothing piecewise differential homeomorphisms, Annals of Mathematics, 72 (1966), 521-544.</p> <p>[Mun1966] J.R. Munkres, Elementary Differential Topology, (rev. ed.), Princeton University Press, Princeton, N.J., 1966.</p> <p>[Sma1962] S. Smale, On the structure of manifolds. Amer. J. Math., 84 (1962), 387--399</p> <p>[Tam1961] I. Tamura, 8-manifolds admitting no differentiable structure, J. Math. Soc. Japan 13 (1961), 377-382.</p> <p>[TL] W.P. Thurston and S. Levy (ed.), Three-Dimensional Geometry and Topology, Princeton University Press, Princeton, NJ, 1997.</p> <p>[Whi1940] J.H.C. Whitehead, On $C^{1}$-complexes, Ann. of Math. 41 (1940), 809–824.</p> http://mathoverflow.net/questions/96670/classification-of-surfaces-and-the-top-diff-and-pl-categories-for-manifolds Classification of surfaces and the TOP, DIFF and PL categories for manifolds VCF 2012-05-11T13:40:38Z 2012-05-15T16:33:22Z <p>A surface is simply a 2-manifold. The classification theorem for compact connected surfaces (with boundary) is commonly regarded in the categories <strong>TOP</strong>, <strong>DIFF</strong> and <strong>PL</strong>. Well known proofs (e.g. via triangulations, or Morse theory) yield the <em>same</em> classification because of results that <em>connect</em> these categories for surfaces. Informally speaking, here is what I know to be true for compact connected surfaces </p> <p>1) [<strong>TOP</strong> &amp; <strong>PL</strong>]. <em>Topological surfaces always admit a triangulation, and any two triangulations of a surface are piecewise-linear equivalent</em> (<a href="http://en.wikipedia.org/wiki/Hauptvermutung" rel="nofollow">Hauptvermutung</a> for surfaces)</p> <p>2) [<strong>DIFF</strong> &amp; <strong>PL</strong> (without using 1.)]. <em>Every smooth surface admits a PL-structure, as every smooth manifold does</em> (See the paper "<em>On $C^{1}$ Complexes</em>", by J.H.C. Whitehead).</p> <p>Next is where I seek to be enlightened:</p> <p>3) [<strong>TOP</strong> &amp; <strong>DIFF</strong>, (without using either 1. or 2.)]. <em>Two smooth surfaces are diffeomorphic iff they are homeomorphic, and a topological surface always admits a smoothing.</em></p> <p><strong>Where can I find a formal statement, and a complete proof of 3.?</strong></p> <p>Finally, consider non-compact connceted surfaces (with boundary). There seems to be a complete classification of non-compact connected triangulable surfaces with boundary (See the paper <a href="http://www.imath.kiev.ua/~mfat/html/papers/2007/1/pri_mis/art.pdf" rel="nofollow">"<em>Classification of Noncompact Surfaces with Boundary</em>"</a>, by A.O. Prishlyak and K.I. Mischenko).<strong>What about the TOP and DIFF categories? That is, do the results 1-3 above hold for non-compact surfaces?</strong> </p> <p>NOTE: I want to mention the post <a href="http://mathoverflow.net/questions/4155/classification-problem-for-non-compact-manifolds" rel="nofollow">Classification problem for non-compact manifolds</a> for a related, yet different discussion. The paper: "<em>On the Classification of Noncompact Surfaces</em>", by Ian Richards is mentioned there in a comment. This paper considers the case of non-compact triangulable surfaces without boundary.</p> <p>Thank you!</p> http://mathoverflow.net/questions/96071/general-gluing-theorem-for-adjunction-spaces General gluing theorem for adjunction spaces VCF 2012-05-05T15:57:59Z 2012-05-06T17:49:32Z <p>Consider the following interesting theorem:(7.5.7, p.294 in <em>Topology and Groupoids</em> by Ronald Brown)</p> <p><strong>Gluing theorem for adjunction spaces:</strong> <em>Suppose that we have the following commutative diagram of topological spaces and continuous maps:</em></p> <p><img src="http://i48.tinypic.com/sdixrc.jpg" alt="alt text"></p> <p><em>where</em> $\varphi_{A}$, $\varphi_{X}$ <em>and</em> $\varphi_{Y}$ <em>are homotopy equivalences, and the inclusions</em> $i$ <em>and</em> $i'$ <em>are closed cofibrations. Then the map</em></p> <p>$$\varphi:X\cup_{f}Y\rightarrow X'\cup_{f\phantom{l}'}Y'$$</p> <p><em>induced by</em> $\varphi_{A}$, $\varphi_{X}$ <em>and</em> $\varphi_{Y}$ <em>is a homotopy equivalence</em>. </p> <p><a href="http://mathoverflow.net/questions/94830" rel="nofollow">In this post</a> I asked a question that would become a corollary if the gluing theorem were true for arbitrary cofibrations $i$ and $i'$. <strong>I would like to know if this is the case. Otherwise, does anybody know of a counterexample?</strong></p> <p>Perhaps, it is possible that although the map $\varphi$ may not in general be a homotopy equivalence when the cofibrations $i$ and $i'$ are arbitrary, the adjunction spaces $Y\cup_{f}X$ and $Y'\cup_{f'}X'$ will nonetheless be homotopy equivalent. A possible approach to this would be to try to find a third space containing both as deformation retracts. </p> <p><strong>Note on notation:</strong> That an inclusion map $j:B\rightarrow Z$ is a closed cofibration means that it is a cofibration (i.e., the pair $(B,Z)$ has the homotopy extension property) and the set $B$ is closed in $X'$.</p> <p>Thank you</p> http://mathoverflow.net/questions/94830/homotopy-equivalence-of-certain-kinds-of-adjunction-spaces/95132#95132 Answer by VCF for Homotopy equivalence of certain kinds of adjunction spaces VCF 2012-04-25T08:52:14Z 2012-05-03T02:20:59Z <p>Thanks to Tom Goodwillie and Ronnie Brown for their answers. Also thanks to Theo for pointing out that I posted the problem on math.stackexchange.com/q/135173/5363.</p> <p>After some thought it became clear that the pair $(X,A)$ needs to have the homotopy extension property (HEP). A simple counterexample is the following: </p> <p>Let $X=S^{1}$ (unit circle in the complex plane), $A=S^{1}-\{(1,0)\}=Y$, and $Z=\{(-1,0)\}$. The pair $(X,A)$ does not have the HEP, for if it did we would have a retraction $r$ of $X\times I$ onto $X\times\{0\}\cup A\times I$, which is not compact. The point $Z$ is a strong deformation retract of $Y$, and the deformation retraction is given by $(e^{2\pi i t},s)\mapsto e^{\pi i +(1-s)(2\pi i t-\pi i)}$.</p> <p>Now, letting $f:A\rightarrow Y$ be the identity, we can easily see that $Y\cup_{f}X$ is a circle, while $Z\cup_{\phi\circ f}X$ is a 2-point space $\{a,b\}$ with the topology $\{\emptyset,\{a,b\},{a}\}$. This two spaces are not homotopy equivalent since the latter is contractible.</p> <p>With the extra hypothesis that the pair $(X,A)$ has the HEP I believe that we can answer the original question I posted in the affirmative.</p> <p><strong>Proposition:</strong> Let $X,Y,Z$ be topological spaces. Suppose that $(X,A)$ has the HEP and that we have a map $f:A\rightarrow Y$ and a homotopy equivalence $\phi:Y\rightarrow Z$. Then the adjunction spaces $Y\cup_{f}X$ and $Z\cup_{\phi\circ f}X$ are homotopy equivalent.</p> <p><strong>Sketch of proof:</strong> The idea is to construct a topological space $W$ having both $Y\cup_{f}X$ and $Z\cup_{\phi\circ f}X$ as strong deformation retracts. For this we begin with the mapping cylinder $M_{\phi}$ of the homotopy equivalence $\phi$, and the map $F:A\times I\rightarrow M_{\phi}$ induced by the map $i\circ(f,Id):A\times I\rightarrow Y\times I + Z$ after passing to the quotient (Here $i$ is simply the inclusion of $Y\times I$ in the disjoint union $Y\times I + Z$, and $(f,Id)$ is the map defined to be $f$ on $A$ and the identity on $I$). Then, we define $W$ to be the adjunction space $M_{\phi}\cup_{F}(X\times I)$. </p> <p>In order to show that $W$ will be the right space for the job we need to combine two things:</p> <p>First, For any $u\in I$, let $R_{u}=X\times u\cup A\times I$. The HEP of $(X,A)$ guarantees that $X\times I$ strong deformation retracts onto $R_{u}$. Indeed, the HEP gives us a retraction $r^{u}$ of $X\times I$ onto $R_{u}$. Note that $r^{u}$ has two components $(r_{1}^{u},r_{2}^{u})$, which we can use to define the strong deformation retraction by </p> <p>$$H_{u}(x,t,s)=(r_{1}^{u}(x,st),(1-s)t+sr_{2}^{u}(x,t)).$$</p> <p>Now define </p> <p>$$(H_{u},\rho):X\times I\times I + M_{\phi}\times I\rightarrow X\times I + M_{\phi},$$</p> <p>where $\rho(m,s)=m$ for all $s\in I$. Then, after passing to the quotient, this map induces a strong deformation retraction of $W$ onto $M_{\phi}\cup_{F}R_{u}$. In particular, $H_{0}$ and $H_{1}$ will induce strong deformation retractions of $W$ onto $M_{\phi}\cup_{F}R_{0}$ and $M_{\phi}\cup_{F}R_{1}$, respectively.</p> <p>Second, Since $\phi$ is a homotopy equivalence, the mapping cylinder $M_{\phi}$ has its top $Y$ and bottom $Z$ as strong deformation retracts.</p> <p>Now, combine the strong deformation retraction of $W$ onto $M_{\phi}\cup_{F}R_{0}$ with the strong deformation retraction of $M_{\phi}$ onto its bottom to yield a strong deformation retraction of $W$ onto $Z\cup_{\phi\circ f}X$. Similarly, combine the strong deformation retraction of $W$ onto $M_{\phi}\cup_{F}R_{1}$ with the strong deformation retraction of $M_{\phi}$ onto its top to yield a strong deformation retraction of $W$ onto $Y\cup_{\phi\circ f}X$.</p> <p>Overall the idea seems right. Any insights will be appreciated.</p> <p>Thanks!</p> http://mathoverflow.net/questions/115795/extended-integral-in-spivaks-calculus-on-manifolds Comment by VCF VCF 2012-12-08T15:18:39Z 2012-12-08T15:18:39Z No need for more details, thank you for the corroboration. http://mathoverflow.net/questions/108167/three-half-circles-on-the-plane-may-not-meet-nicely/108182#108182 Comment by VCF VCF 2012-09-26T18:49:48Z 2012-09-26T18:49:48Z What do you mean by &quot;supporting half-planes determined by the diameters&quot;? My understanding is that you mean planes orthogonal to the diameters passing through the centers of the half-circles. http://mathoverflow.net/questions/108167/three-half-circles-on-the-plane-may-not-meet-nicely/108182#108182 Comment by VCF VCF 2012-09-26T17:49:08Z 2012-09-26T17:49:08Z About the paragraph right after the picture. Are you saying that it is impossible to arrange four copies of $H$ so that both 1. Any two meet nicely AND 2. Any three of more have empty intersection? http://mathoverflow.net/questions/108167/three-half-circles-on-the-plane-may-not-meet-nicely Comment by VCF VCF 2012-09-26T17:18:10Z 2012-09-26T17:18:10Z It means exactly what it says. That is, that the intersection of the three haf-circles is empty. http://mathoverflow.net/questions/108167/three-half-circles-on-the-plane-may-not-meet-nicely/108178#108178 Comment by VCF VCF 2012-09-26T16:59:59Z 2012-09-26T16:59:59Z Even if you are thinking about half-disks your assumption is incorrect. See the picture on the right bove. http://mathoverflow.net/questions/108167/three-half-circles-on-the-plane-may-not-meet-nicely/108178#108178 Comment by VCF VCF 2012-09-26T16:34:13Z 2012-09-26T16:34:13Z I think that you are thinking about half-disks, not half-circles. http://mathoverflow.net/questions/107285/cylinders-dividing-mathbbr3/107640#107640 Comment by VCF VCF 2012-09-21T08:49:32Z 2012-09-21T08:49:32Z Ah, the pleasures of disagreeing agreefully with the agreeable Gerhard. http://mathoverflow.net/questions/107285/cylinders-dividing-mathbbr3/107640#107640 Comment by VCF VCF 2012-09-20T16:45:15Z 2012-09-20T16:45:15Z @Gerhard: It&#180;s fine you disagree. Joseph has certainly shown good effort. I voted up his first answer and acepted it. But I stand by my opinion that it is good practice to write answers that address the question somehow, not simply the comments. It has nothing to do with manners. http://mathoverflow.net/questions/107285/cylinders-dividing-mathbbr3/107640#107640 Comment by VCF VCF 2012-09-20T12:06:07Z 2012-09-20T12:06:07Z Yes, I was wrong. Certainly 6 is the most with two. I had gotten the same picture with Maple. $n$ ellipsoids of any size will divide space into the maximum number of regions if any two intersect in two ellipses as the two in your configuration (i.e. in two points), any three in 8 points and any four or more have empty intersection. Roughly speaking, you may think of this conditions as happening when the defining equations are &quot;linearly independent&quot;. The max is $n(4n^2-9n+11)/3$. I give you -1 as your answer is not related to the original question and because of the ambiguous final remark http://mathoverflow.net/questions/107285/cylinders-dividing-mathbbr3/107295#107295 Comment by VCF VCF 2012-09-19T11:39:36Z 2012-09-19T11:39:36Z @Gerhard: This is just my mind wondering, but if we have two ellipsoids of the same size I think we can arrange arrange them so that we get 8 regions. First make a perfect cross (this yiels 6 regions) and then slide the vertical ellipsoid to the right so that two mnew patches appear thus yielding 2 more regions. What do you think? http://mathoverflow.net/questions/107285/cylinders-dividing-mathbbr3/107297#107297 Comment by VCF VCF 2012-09-16T12:35:22Z 2012-09-16T12:35:22Z @Joseph: It would be nice to have the kind of closed formula I am looking for. But the asymtotic complexity point of view makes me think that perhaps it isn&#180;t necessary to have one. http://mathoverflow.net/questions/107285/cylinders-dividing-mathbbr3 Comment by VCF VCF 2012-09-16T12:06:54Z 2012-09-16T12:06:54Z Yes, the cylinders have top and bottom. That is, they are compact. More explicitely, they are copies of $D\times\{0\}\cup S^{1}\times [-3,3]\cup D\times\{1\}$, where $D$ denotes the unit 2-disk. http://mathoverflow.net/questions/107285/cylinders-dividing-mathbbr3 Comment by VCF VCF 2012-09-15T20:49:47Z 2012-09-15T20:49:47Z Thank you, I didn&#180;t kow that. It makes sense to me since, for example, I know that if we consider $n$ planes instead of cylinders we get at most $(n^3+5n+6)/6$ regions. http://mathoverflow.net/questions/97747/topological-version-of-two-results-in-smooth-morse-theory Comment by VCF VCF 2012-05-30T05:30:16Z 2012-05-30T05:30:16Z It will be worth it! Milnor's works are examples of great mathematics, and of great mathematical writing http://mathoverflow.net/questions/97747/topological-version-of-two-results-in-smooth-morse-theory Comment by VCF VCF 2012-05-24T16:11:59Z 2012-05-24T16:11:59Z Sorry, I don't have a good example. The height function, when restricted to e.g., the unit 2-sphere in $\mathbb{R}^{3}$ has exactly two non-degenerate critical points (both in the DIFF and TOP senses). In DIFF, it is possible to smoothly deform the sphere slightly in such a way that the height function on the new manifold still has exactly two critical points, but now the Hessians will vanish, thus giving us degenerate ones (there is an illustrative picture in p.127 of Milnor's work cited above). In TOP this will not work, I will think about it more.