User jeremy brazas - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T10:04:26Z http://mathoverflow.net/feeds/user/5801 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115004/the-integers-as-a-sequential-but-non-first-countable-topological-group The integers as a sequential but non-first countable topological group Jeremy Brazas 2012-11-30T17:08:25Z 2013-05-12T11:34:00Z <p>Completely unaware of the Bohr topology, I recently <a href="http://mathoverflow.net/questions/114816/hausdorff-group-topologies-on-finitely-generated-groups" rel="nofollow">asked</a> whether or not there was a Hausdorff group topology on the integers $\mathbb{Z}$ which made the group fail to be first countable. For me, this topological group is a bit extreme since there are no non-trivial convergent sequences. I'm very interested to know if there is a sequential example.</p> <p>If $\mathbb{Z}$ is given a Hausdorff group topology which makes it a sequential space, must it be first countable?</p> http://mathoverflow.net/questions/99712/refining-open-covers-in-locally-path-connected-spaces Refining open covers in locally path connected spaces Jeremy Brazas 2012-06-15T15:05:10Z 2013-04-02T03:52:03Z <p>Suppose $X$ is a locally path connected topological space and $\mathcal{U}$ is an open cover of $X$ (consisting of path connected sets if we want). </p> <p>One often wants the intersection $A\cap B$ of pairs of elements $A,B\in \mathcal{U}$ to be path connected, or perhaps stronger, that the intersection of finitely many elements of $\mathcal{U}$ be path connected. This, for instance, is the case in some groupoid versions of the van Kampen theorem (like the one in Peter May's <em>A Concise Course in Algebraic Topology</em>). Having covers with this property also simplifies life a bit in the study of shape invariants constructed via <a href="http://en.wikipedia.org/wiki/Nerve_of_a_covering" rel="nofollow">nerves of covers</a>. If we don't start with a cover this nice perhaps we can at least get to one by refinement.</p> <p><strong>Question:</strong> Is it always possible to find an open cover $\mathcal{V}$ of $X$ refining $\mathcal{U}$ such that the intersection of every pair of elements in $\mathcal{V}$ is path connected (or empty)? Can we do even better and find $\mathcal{V}$ such that the intersection of finitely many elements of $\mathcal{V}$ is path connected?</p> <p>I'm less confident such refinement is possible for general locally path connected $X$. I'd be perfectly content to assume $X$ is paracompact Hausdorff.</p> http://mathoverflow.net/questions/118833/making-cw-complexes-metrizable Making CW-complexes metrizable Jeremy Brazas 2013-01-13T17:24:50Z 2013-03-09T20:04:16Z <p>It is a basic topological fact that CW-complexes aren't typically metrizable (they must satisfy a certain local finiteness condition) and the quotient topology is to blame.</p> <p>Suppose $X$ is a finite dimensional CW-complex with countably many cells in each dimension. Is it possible weaken the topology of $X$, without changing homotopy type, so the resulting space is metrizable?</p> http://mathoverflow.net/questions/122932/a-free-topological-group-as-a-topological-module/123008#123008 Answer by Jeremy Brazas for a free topological group as a topological module Jeremy Brazas 2013-02-26T17:57:31Z 2013-02-26T17:57:31Z <p>I don't know the answer to the general question but the answer seems to be "yes" when $G$ is compact. Here is a straightforward argument.</p> <p>The corresponding action of $G$ on the free topological monoid $M=\coprod_{n\geq 0}(G\sqcup G^{-1})^n$ is certainly continuous. When $G$ is compact $F(G)$ is the topological quotient of $M$ via word reduction $r:M\to F(G)$. When the action of $G$ on itself is conjugation, we have $^{g}(r(g_{1}^{\epsilon_1}...g_{n}^{\epsilon_n}))=r(^{g}(g_{1}^{\epsilon_1}...g_{n}^{\epsilon_n}))$ for any $g_{1}^{\epsilon_1}...g_{n}^{\epsilon_n}\in M$. The action is continuous since $id\times r:G\times M\to G\times F(G)$ is a topological quotient map (compactness of $G$ helps out here too).</p> <p>This argument should also work when $G$ is a $k_{\omega}$-space (inductive limit of nested compact subspaces) but doesn't extend to the general case since $r:M\to F(G)$ is not quotient when $G=\mathbb{Q}$ is the rationals.</p> http://mathoverflow.net/questions/116980/is-the-wedge-sum-of-two-cones-over-the-hawaiian-earring-contractible/116999#116999 Answer by Jeremy Brazas for Is the wedge sum of two cones over the hawaiian earring contractible? Jeremy Brazas 2012-12-22T00:56:48Z 2012-12-22T00:56:48Z <p>The space you are talking about is sometimes called the Griffiths space. As Henry Horton suggests, to prove it is not contractible you can show the fundamental group is non-trivial (even though its Cech homotopy groups are trivial). The reference already given is a good one though I'll add that Griffith was the first to show this in</p> <p>H. B. Griffiths, The fundamental group of two spaces with a common point, Quart. J. Math. 5 (1954), 175-190.</p> <p>Another discussion on non-contractible one-point unions of simply connected spaces is in</p> <p>K. Eda, A locally simply connected space and fundamental groups of one point unions of cones, Proc. Amer. Math. Soc. 116 No. 1 (1992) 239-249.</p> http://mathoverflow.net/questions/114816/hausdorff-group-topologies-on-finitely-generated-groups Hausdorff group topologies on finitely generated groups Jeremy Brazas 2012-11-28T22:03:00Z 2012-11-28T23:08:45Z <p>Suppose $G$ is a finitely generated Hausdorff topological group. Must $G$ be first countable (or perhaps a sequential space)? What if we restrict to the abelian case?</p> <p>I wonder if this is even true for the additive group of integers $\mathbb{Z}$. There certainly are non-discrete, Hausdorff group topologies on $\mathbb{Z}$ where a basis at $0$ consists of subgroups (such as that used in Furstenberg's proof of the infinitude of primes). On the other hand, determining if there is a Hausdorff group topology that makes a given sequence converge to $0$ is non-trivial. For instance, it is known that the sequence of squares $n^2$ can't converge to $0$ in any Hausdorff group topology and that if there is a Hausdorff group topology on $\mathbb{Z}$ such that the sequence of primes $2,3,5,...,p,..$ converges to $0$, then the twin prime conjecture is false.</p> http://mathoverflow.net/questions/111310/universal-covering-space-for-non-semilocally-simply-connected-spaces/111388#111388 Answer by Jeremy Brazas for Universal covering space for non-semilocally simply connected spaces Jeremy Brazas 2012-11-03T16:15:34Z 2012-11-03T16:15:34Z <p>My answer to <a href="http://mathoverflow.net/questions/58341/is-a-space-with-no-covering-spaces-simply-connected" rel="nofollow">this question</a> gives an example of a locally path connected (but non-semilocally simply connected) space $HA\subset\mathbb{R}^3$ called the Harmonic archipelago: draw the Hawaiian earring on a disk and between each hoop push the surface straight up to make a hill of height 1. </p> <p><img src="http://www.freeimagehosting.net/newuploads/f8n4t.png" alt="Harmonic archipelago"></p> <p>The fundamental group is uncountable but the only covering of $HA$ is itself so it is the unique object in its category of coverings. This weird covering behavior happens because you can deform any loop as close as you want to the basepoint (over finitely many hills) but cannot actually contract it all the way <strong>to</strong> the basepoint because the homotopy would have to take the loop over infinitely many of hills (contradicting compactness).</p> <p>Topologies on the fundamental group detect this behavior. For instance the quotient topology from the loop space (the <em>quasitopological fundamental group</em>, $\pi_{1}^{qtop}(X)$) satisfies: If $p:Y\to X$ is a covering map, then $p_{\ast}:\pi_{1}^{qtop}(X)\to \pi_{1}^{qtop}(Y)$ is an open embedding of quasitopological groups. So if $\pi_{1}^{qtop}(X)$ is an indiscrete group there are no proper open subgroups so the only covering of $X$ is itself.</p> <p>This works for all spaces including non-locally path connected ones. Really, I'd say it is easier to produce non-locally path connected examples. Here is a compact space $X$ with $\pi_{1}^{qtop}(X)\cong \mathbb{Z}_n$ indiscrete.</p> <p>Let $T\subset \mathbb{R}^2$ be the <a href="http://en.wikipedia.org/wiki/Topologist%27s_sine_curve" rel="nofollow">closed topologists sine curve</a>, $a\in T$ be the endpoint in the open path component and $b$ be a point in the other path component - say the origin. Now let $$Y=\frac{T\times S^1}{T\times \{1\}\cup \{a\}\times S^1}$$</p> <p>In other words, if $a$ is the basepoint of $T$, $Y$ is the reduced suspension $\Sigma T$.</p> <p>You have $\pi_1(Y)\cong \mathbb{Z}$ where the loop $L:S^1\to Y$, $t\mapsto (b,t)$ represents a generator, but every neighborhood of $L$ contains a trivial loop so the quotient topology on $\pi_1(Y)$ is the indiscrete topology. If you attach a 2-cell to $Y$ using $L^n$ as an attaching map, you get a space $X$ where $\pi_{1}^{qtop}(X)\cong \mathbb{Z}_n$ is indiscrete. Both $Y$ and $X$ spaces have no nontrivial coverings.</p> <p>Constructing locally path connected examples with countable indiscrete fundamental groups is much harder. In fact, I think it is an open question whether or not there exists a Peano continuum with finite, indiscrete fundamental group. For such spaces it is more efficient to talking about coverings via the "shape topology" and Spanier groups (the last section of <a href="http://arxiv.org/abs/1207.1310" rel="nofollow">this paper</a> shows the shape topology consists precisely of the data of the category of coverings). Applied to your question: locally path connected $X$ has a categorical universal covering iff there is a minimal open subgroup in $\pi_1(X)$ with the shape topology. If the minimal open subgroup is the trivial subgroup you get back a classical universal covering and a discrete group.</p> http://mathoverflow.net/questions/110805/fundamental-groups-and-homology-groups-of-closed-subsets-of-the-plane/110830#110830 Answer by Jeremy Brazas for Fundamental groups and homology groups of closed subsets of the plane Jeremy Brazas 2012-10-27T13:38:36Z 2012-10-27T13:38:36Z <p>The fundamental group of a closed planar set naturally injects into the first Cech homotopy group, which is an inverse limit of finitely generated free groups. In particular, the algebraic restrictions gained are: the fundamental group must be locally free, fully residually free, and residually finite.</p> <p><a href="http://arxiv.org/abs/math/0512343" rel="nofollow">Fischer, H., Zastrow, A., The fundamental groups of subsets of closed surfaces inject into their first shape groups, Algebraic and Geometric Topology 5 (2005) 1655-1676.</a></p> <p>The first homology pro-group then consists of finitely generated free groups and the first Cech homology group is the inverse limit of these.</p> <p>When $X\subset \mathbb{R}^2$ is compact and locally path connected, the canonical map $H_1(X)\to \check{H}_1(X)$ is surjective but the kernel can be difficult to understand even for the <a href="http://jlms.oxfordjournals.org/content/62/1/305.short" rel="nofollow">Hawaiian earring.</a></p> <p><a href="http://130.44.194.100/proc/2000-128-05/S0002-9939-99-05670-1/S0002-9939-99-05670-1.pdf" rel="nofollow">Eda, K., Kawamura, K. The surjectivity of the canonical homomorphism from singular homology to Cech homology Proc. Amer. Math. Soc. 128 No. 5 (1999) 1487-1495</a></p> http://mathoverflow.net/questions/109364/when-is-a-locally-homeo-a-covering-map/109418#109418 Answer by Jeremy Brazas for when is a locally homeo a covering map? Jeremy Brazas 2012-10-11T21:52:52Z 2012-10-11T21:52:52Z <p>This answer takes a more general viewpoint than Alexandre's. The generality is in response to the small number of assumptions on the spaces involved.</p> <p>First, you should assume that $Y$ is locally path connected. Only on special occasions does anything good come from looking at coverings of non-locally path connected spaces.</p> <p>Covering maps are highly structured, so promoting a local homeomorphism (even with the locally compact assumption) to a covering map requires strong conditions. Even if $f$ has the following (rather strong) lifting property, $f$ is not always a covering map: For each map $g:W\to Y$ from a locally path connected space $W$ such that $g_{\ast}(\pi_1(W,w))\subseteq f_{\ast}(\pi_1(X,x))$, there is a unique lift $\tilde{g}:W\to X$, $\tilde{g}(w)=x$ such that $f\tilde{g}=g$. </p> <p>Here are (necessary) conditions you must have for $f:X\to Y$ to be a covering map.</p> <p>1) $f$ has <strong>unique path lifting</strong> (equivalently, if $P(X,x)$ is the set of paths in $X$ starting at $x$, then the induced function $Pf:P(X,x)\to P(Y,p(y))$ is a bijection for each $x\in X$.</p> <p>2) $f$ has unique lifting of homotopies of paths (this doesn't follow from 1. and you can formulate it in the same way as a bijection on path spaces)</p> <p>Properties 1) &amp; 2) still aren't enough to promote $f$ to be a covering map. You need to strengthen unique path lifting.</p> <p>3) $f$ has <strong>continuous</strong> unique path lifting if $P(X,x)$ has the compact-open topology and the induced map $Pf:P(X,x)\to P(Y,p(y))$ is a <em>homeomorphism</em> for each $x\in X$.</p> <p>A surjective local homeomorphism with 2) and 3) and $Y$ locally path connected now has the general lifting property listed above but this is still not enough. Such a map is a semicovering (shameless plug - but it's open-access so its ok right?): <a href="http://www.intlpress.com/HHA/v14/n1/a3/v14n1a3.pdf" rel="nofollow">Semicoverings: a generalization of covering space theory</a>, Homology, Homotopy and Appl. 14 (2012) pp.33-63 ). There are semicoverings of the Hawaiian earring - which is locally compact - that are not coverings (example 3.8 in that paper). On the other hand, a covering map always has properties 2) and 3).</p> <p>This suggests that in the end you will need to assume $Y$ is semilocally simply connected.</p> <p>If you have $Y$ locally path connected, semilocally simply connected and $f$ has properties 2) and 3), then $f$ is a covering map. With these new assumptions on $Y$, I suspect you can weaken 3) to 1).</p> http://mathoverflow.net/questions/41616/what-is-1bg-for-an-arbitrary-topological-group-g What is π_1(BG) for an arbitrary topological group $G$? Jeremy Brazas 2010-10-09T17:29:24Z 2011-11-25T17:28:55Z <p>The classifying space $BG=|Nerve(G)|$ of an arbitrary topological group $G$ does not necessarily have the homotopy type of a CW-complex but the fundamental group should still be accessible. What is $\pi_{1}(BG)$? A reference on this would be great. My initial guess: $\pi_{1}(BG)$ is the quotient group $\pi_{0}(G)$ for arbitrary $G$</p> <p>Motivation: There is a natural way to make $\pi_1$ a functor to topological groups. I am interested in relating the topologies of $G$ and $\pi_{1}(BG)$ but the topology on $\pi_{1}(X)$ is boring (discrete) when $X$ is a CW-complex.</p> http://mathoverflow.net/questions/64617/mathematical-ideas-named-after-places/64735#64735 Answer by Jeremy Brazas for Mathematical ideas named after places Jeremy Brazas 2011-05-12T03:37:03Z 2011-05-12T03:37:03Z <p>The <a href="http://ncatlab.org/nlab/show/Warsaw+circle" rel="nofollow">Warsaw circle</a> is a motivating example in shape theory.</p> http://mathoverflow.net/questions/59371/looking-for-general-approaches-to-show-connectedness-of-topological-groups/59482#59482 Answer by Jeremy Brazas for Looking for general approaches to show connectedness of topological groups Jeremy Brazas 2011-03-24T21:08:00Z 2011-03-24T21:08:00Z <p>Perhaps this is helpful for the locally compact case.</p> <p>Corollary 3.1.12 in <a href="http://books.google.com/books?id=B785AETmFKEC&amp;printsec=frontcover&amp;dq=topological+groups+and+related&amp;hl=en&amp;ei=Ka6LTanVG-GG0QHh_rnzDQ&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=1&amp;ved=0CCgQ6AEwAA#v=onepage&amp;q&amp;f=false" rel="nofollow">"Topological groups and related structures"</a> by Arhangel'skii and Tkachenko gives a nice characterization of connectedness in the locally compact case: $G$ is connected $\Leftrightarrow$ $G$ has no proper open subgroups $\Leftrightarrow$ Every neighborhood of the identity algebraically generates $G$. </p> <p>I don't know how practical this is for the groups you have in mind but the generality is nice.</p> http://mathoverflow.net/questions/58811/normal-subgroups-of-braid-groups/58900#58900 Answer by Jeremy Brazas for Normal subgroups of braid groups Jeremy Brazas 2011-03-19T03:11:54Z 2011-03-19T03:11:54Z <p>If you're willing to consider "any classification" you might consider the covering space theory of configuration spaces. </p> <p>Equivalence classes of regular (finite) connected coverings of the configuration space $C_{n}(\mathbb{R}^{2})$ of n unordered points in $\mathbb{R}^{2}$ completely classify (finite index) normal subgroups of $B_n=B_n(\mathbb{R}^{2})=\pi_{1}(C_{n}(\mathbb{R}^{2}))$ via the usual Galois correspondence.</p> <p>This may not be very enlightening but may offer a practical approach to studying the lattice of normal subgroups of $B_n$. <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.pjm/1102785282" rel="nofollow">Hansen's polynomial covering theory</a> might also help to identify interesting subgroups of $B_n$.</p> http://mathoverflow.net/questions/58341/is-a-space-with-no-covering-spaces-simply-connected/58344#58344 Answer by Jeremy Brazas for Is a space with no covering spaces simply connected? Jeremy Brazas 2011-03-13T16:54:35Z 2011-03-13T16:54:35Z <p>No, the <a href="http://arxiv.org/abs/math/0501426" rel="nofollow">harmonic archipelago</a> (An illustration is on pg 7 of <a href="http://people.oregonstate.edu/~bogleyw/research/ups.pdf" rel="nofollow">W. A. Bogley and A. J. Sieradski, Universal path spaces, preprint</a>) is a locally path connected subspace of $\mathbb{R}^{3}$ and has uncountable fundamental group but every connected cover is trivial.</p> <p>This is part of a more general phenomenon. Let $\pi_{1}^{top}(X)$ be the fundamental group of a space $X$ with the quotient topology of the loop space $\Omega(X)$ with the compact-open topology (sometimes called the "topological fundamental group"). This is a quasitopological group (in that inversion is continuous and multiplication is continuous in each variable) but is not always a topological group. If $p:X\rightarrow Y$ is a covering map the induced homomorphism $p_{\ast}:\pi_{1}^{top}(X)\rightarrow \pi_{1}^{top}(Y)$ is an open embedding of quasitopological groups (i.e. $\pi_{1}^{top}(X)$ embeds as an open subgroup). One consequence of this is that if $\pi_{1}^{top}(X)$ has the indiscrete topology, then either $\pi_{1}(X)=1$ or every connected covering of $X$ is trivial. There are lots of examples of spaces where this occurs other than the harmonic archipelago.</p> http://mathoverflow.net/questions/54391/topological-vs-pro-fundamental-groups/54477#54477 Answer by Jeremy Brazas for Topological vs pro fundamental groups Jeremy Brazas 2011-02-06T00:23:34Z 2011-02-09T16:56:58Z <p>I don't really have a great feeling for the information captured by the "toposophic" fundamental group so this answer is a bit one-sided towards what I can say about the quotient topology on the fundamental group and what it has to do with covering maps. Maybe you can form something useful from this.</p> <p>First, I would say quotient $\pi_{1}(X)$ is "rarely" a topological group but is always "quasi"topological group in that you know multiplication is separately continuous. It comes with some serious topological baggage. The requirement that $X$ be locally simply connected (in the sense I define in the comments) does rule out a good number of "wild" spaces of interest but there are some very interesting locally simply connected, non-locally path connected examples.</p> <p>The connection to coverings is: If $p:Y\rightarrow X$ is a covering map, the induced homomorphism $p_{\ast}:\pi_{1}(Y)\rightarrow \pi_{1}(X)$ is an open embedding of quasitopological groups. </p> <p>This also implies that if $X$ is nice enough (for instance, locally simply connectivity in the sense of Wikipedia) to have a universal cover, then $\pi_{1}(X)$ is a discrete group and the topology gives no new information.</p> <p>Constructing covers of non-locally path connected spaces is tricky so in general there will not be a complete Galois correspondence between open subgroups and covers. There should be a correspondence in the locally path connected case.</p> <p>Added:</p> <p>An enlightening <a href="http://arxiv.org/abs/math/0501482" rel="nofollow">difference</a> between quotient and inverse limit topology is in the case of the Hawaiian earring. </p> <p>It has been known for less than two years that quotient $\pi_1$ is not always a topological group. Besides that, its theory is not very developed. I suppose it is "wrong" if you demand a functor that takes values in the category topological groups. But as long as we are discussing other potential topologies...</p> <p>There is a natural "fix" to the quotient topology using a reflection from topological algebra. It is defined in <em>The fundamental group as topological group</em> on my personal <a href="http://euclid.unh.edu/~jbrazas/" rel="nofollow">page</a>. The resulting topologized fundamental group is defined for all spaces, takes values in topological groups, and is universal with respect to continuous homomorphisms from quotient $\pi_1$ to topological groups. It is well-behaved in that it admits a number of topological analogues: On "non-discrete wedges of circles" $X_+\wedge S^1$ it is free topological, every topological group is realized as a fundamental group with this topology by attaching 2-cells to one of these "wedges," van Kampen theorems involving pushouts of topological groups are possible, etc.</p> http://mathoverflow.net/questions/23940/why-free-topological-groups-on-tychonoff-spaces Why free topological groups on Tychonoff spaces? Jeremy Brazas 2010-05-08T17:03:20Z 2010-11-27T12:47:52Z <p>This is a question of the motivation for a common assumption found in the literature.</p> <p>The free topological group $F(X)$ on a space $X$ exists for all spaces $X$ (It seems this was first shown by Katutani and Samuel). I mean "free topological group" in the sense that $F:Top\rightarrow TG$ is left adjoint to the forgetful functor $U:TG\rightarrow Top$ from the category of topological groups to the category of topological spaces.</p> <p>$F(X)$ is well studied when $X$ is a Tychonoff space. This permits the application of pseudometrics which seems to be a powerful tool for describing the complicated topological structure of $F(X)$. Also, it seems to be a useful fact that the canonical map $\sigma:X\rightarrow F(X)$ is an embedding when $X$ is Tychonoff. </p> <p>These two conveniences do seem to make it convenient to study $F(X)$ when $X$ is Tychonoff but it seems almost no one is interested in $F(X)$ when $X$ is not Tychonoff. Why is this? Are these uninteresting for some reason?</p> http://mathoverflow.net/questions/47118/quotients-of-topological-groupoids Quotients of topological groupoids Jeremy Brazas 2010-11-23T16:57:17Z 2010-11-24T05:32:10Z <p>The issues that arise when moving from topological groups to topological groupoids are (at least to me) both subtle and interesting. Recently, I was reading a <a href="http://www.bangor.ac.uk/~mas010/pdffiles/b-hardy1.pdf" rel="nofollow">paper</a> of R. Brown and J.P.L. Hardy from 1974 which lists the following as an open problem:</p> <p><strong>Problem:</strong> Find a topological groupoid $G$ and a totally disconnected normal subgroupoid $N$ such that $G\rightarrow G/N$ (the map of morphism spaces) is not an open map.</p> <p>Of course, this is not an issue with topological groups and normal subgroups.</p> <p>Has this problem been settled? It has been a while and seems pretty accessible so I would imagine some progress has been made. If such $G$ and $N$ do exist, are there reasonable conditions we can place on them that do make $G\rightarrow G/N$ an open map?</p> <p>I will make a couple of clarifications based on David's comment.</p> <p><strong>Clarification:</strong> Totally disconnected in this context is not topological. It means that $N(x,y)=\emptyset$ when $x\neq y$ so the components of $N$ (as a groupoid, not a space) are groups. R. Brown talks about these and quotient groupoids (but not topological groupoids) in "Topology and Groupoids."</p> <p>As David points out, the problem refers to whether or not the morphism component of the canonical functor $G\rightarrow G/N$ is an open map of topological spaces.</p> http://mathoverflow.net/questions/46771/fundamental-group-of-r2-minus-the-irrationals/46776#46776 Answer by Jeremy Brazas for Fundamental group of R^2 minus the (ir)rationals Jeremy Brazas 2010-11-20T21:38:02Z 2010-11-20T22:44:47Z <p>When you want to compute the fundamental group of a wild space very often the thing to do is identify it as the subgroup of an inverse limit of simpler fundamental groups (often the first shape group). A result of Fischer and Zastrow says that if $X\subseteq \mathbb{R}^{2}$, then the canonical homomorphism of $\pi_1(X,x)$ into the shape group $\check{\pi}_{1}(X,x)$ is injective for any $x\in X$. Of course, this homomorphism is not always injective (even for 2-dimensional compacta) but for a subset of the plane like you have this approach should work. This is, for instance, how you compute the fundamental group of the Hawaiian earring. I would begin by looking for some simple approximating spaces (probably with free fundamental groups) with projection maps and figuring out which elements of the inverse limit of the fundamental groups of these spaces are represented by loops.</p> <p>Here is the paper I mentiond:</p> <p>Fischer, Zastrow, <a href="http://arxiv.org/abs/math/0512343" rel="nofollow">The fundamental groups of subsets of closed surfaces inject into their first shape groups</a>. Algebraic and Geometric Topology. Volume 5 (2005) 1655–1676. </p> http://mathoverflow.net/questions/39780/when-does-a-closed-inclusion-induce-a-closed-inclusion-on-free-topological-groups When does a closed inclusion induce a closed inclusion on free topological groups? Jeremy Brazas 2010-09-23T17:53:10Z 2010-11-09T18:06:53Z <p>Suppose $A$ is a closed subspace of a Tychonoff space $X$. Does the inclusion $i:A\hookrightarrow X$ induce a closed embedding of topological groups on the free (Markov -the unbased version) topological groups $F_{M}(i):F_{M}(A)\rightarrow F_{M}(X)$?</p> <p>Since the underlying algebraic groups are just the usual free groups the injection is immediate. It is not too hard to see that the subgroup $\langle A\rangle$ generated by $A$ in $F_{M}(X)$ is closed (this shows up in the paper "The free topological group over the rationals"). But what about the embedding part? Does this always hold? If not, what if $X$ is a metric space? If it can be helped, I would like to avoid placing compactness conditions on $A$ and $X$.</p> <p>Update: I have recently come upon a result which makes this easier. A theorem of O.V. Sipacheva in <a href="http://www.sciencedirect.com/science?_ob=ArticleURL&amp;_udi=B6V1K-3Y6Y7XC-1&amp;_user=1967573&amp;_coverDate=03%252F03%252F2000&amp;_rdoc=1&amp;_fmt=high&amp;_orig=search&amp;_origin=search&amp;_sort=d&amp;_docanchor=&amp;view=c&amp;_searchStrId=1534002766&amp;_rerunOrigin=google&amp;_acct=C000053403&amp;_version=1&amp;_urlVersion=0&amp;_userid=1967573&amp;md5=114a60ecca2eb6643fe08f1cd2faf99a&amp;searchtype=a" rel="nofollow">Free topological groups of spaces and their subspaces</a> says that for Tychonoff $X$ and arbitrary $A\subset X$, $i:F_{M}(A)\rightarrow F_{M}(X)$ is a topological embedding if and only if $A$ is P-embedded in $X$. This is also equivalent to saying that $A$ with its universal uniformity is a uniform subspace of $X$ with its universal uniformity.</p> <p>So I guess the updated question is: If $A$ is closed in Tychonoff $X$, then does $A$ with its universal uniformity become a uniform subspace of $X$ with its universal uniformity. If not, what are some sufficient conditions (on $A$) for this to occur?</p> http://mathoverflow.net/questions/45351/does-pi-1-have-a-right-adjoint/45375#45375 Answer by Jeremy Brazas for Does $\pi_1$ have a right adjoint? Jeremy Brazas 2010-11-08T22:30:38Z 2010-11-08T22:30:38Z <p>In the more general setting the answer is no. Left adjoints preserve colimits and it is not true that $\pi_{1}(X\vee Y)\cong \pi_{1}(X)\ast\pi_{1}(Y)$ for all spaces $X,Y$ (even compact metric spaces). For instance, if $(\mathbb{HE},x)$ is the usual Hawaiian earring, let $X=Y=C\mathbb{HE}=\mathbb{HE}\times I/\mathbb{HE}\times{1}$ be the cone on the Hawaiian earring with basepoint the image of $(x,0)$ in the quotient. It is a theorem of Griffiths that $\pi_{1}(C\mathbb{HE}\vee C\mathbb{HE})$ is uncountable and not the free product of trivial groups.</p> http://mathoverflow.net/questions/30769/example-of-a-quasitopological-group-with-discontinuous-power-map Example of a quasitopological group with discontinuous power map Jeremy Brazas 2010-07-06T12:55:47Z 2010-07-12T10:21:06Z <p>A quasitopological group is a group $G$ with topology such that multiplication $G\times G\rightarrow G$ is continuous in each variable (i.e. all translations are continuous) and inversion $G\rightarrow G$ is continuous. Sometimes these are called semitopological or semicontinuous groups. What (if it exists) is an example of a quasitopological group such that at least one of the $n$-th power maps $g\mapsto g^{n}$ (for $n\geq 2$) is discontinuous?</p> <p>I am pretty sure such an example exists but I am having a hard time finding one in the literature.</p> http://mathoverflow.net/questions/27660/applications-of-compactness/27748#27748 Answer by Jeremy Brazas for Applications of compactness Jeremy Brazas 2010-06-10T23:35:17Z 2010-06-11T16:16:34Z <p>Compactifications were mentioned in general above... but I think this might be worth mentioning.</p> <p>The Stone-Cech compactification $\beta$ is used all the time since it produces a compact Hausdorff space from an arbitrary space in the "most efficient way." Just looking at applications of $\beta$ might be a more pointed question than asking about the wide world of compactness.</p> <p>$\beta$ is used frequently in topological algebra since the topological structure of universal algebras on non-compact spaces is often highly complicated. Specifically, it is used in <a href="http://www.jstor.org/pss/2041864" rel="nofollow">Applications of the Stone-Cech compactification to free topological groups</a> to give extremely short proofs of some important results in the study of free topological groups. The original proof of Joiner's Fundamental Lemma is rather long and complicated. The paper by Hardy, Morris, and Thomas-Smith I have linked here (sorry if you don't have free access to this) gives a two page proof. One can also prove some nice embedding theorems for topological groups in just a few lines using $\beta$. To see some more applications of $\beta$ to topological groups see Arhangel'skii and Tkachenko's book.</p> http://mathoverflow.net/questions/26911/how-do-you-know-when-a-reflective-subcategory-of-top-is-quotient-reflective How do you know when a reflective subcategory of Top is quotient-reflective? Jeremy Brazas 2010-06-03T11:19:30Z 2010-06-05T18:30:19Z <p>A subcategory $\mathcal{C}$ of the category $Top$ of topological spaces is a <a href="http://en.wikipedia.org/wiki/Reflective_subcategory" rel="nofollow">reflective subcategory</a> if the inclusion functor $i:\mathcal{C}\hookrightarrow Top$ has a left adjoint $R:Top\rightarrow \mathcal{C}$. In other words, for every space $X$, there is a space $RX\in \mathcal{C}$ and a map $r:X\rightarrow RX$ such that for each map $f:X \rightarrow Y$ with $Y\in \mathcal{C}$, there is a unique map $\tilde{f}:RX\rightarrow Y$ such that $\tilde{f}\circ r=f$.</p> <p>If the reflection map $r$ is always a quotient map, $\mathcal{C}$ is said to be a quotient-reflective subcategory. For example, it seems that the subcategories of $T_0$ spaces and $T_2$ spaces are quotient-reflective but the subcategories of completely regular spaces and compact Hausdorff spaces are reflective but not quotient-reflective. I'd like to hear of other examples of quotient-reflective subcategories as well.</p> <p>There seem to be conditions fully characterizing when a subcategory is reflective. For example, some necessary conditions are mentioned in the answer to this MO <a href="http://mathoverflow.net/questions/9504/why-is-top-4-a-reflective-subcategory-of-top-3" rel="nofollow">question</a>. Are there also conditions characterizing subcategories which are quotient-reflective?</p> http://mathoverflow.net/questions/26680/fundamental-group-as-topological-group/26712#26712 Answer by Jeremy Brazas for Fundamental group as topological group Jeremy Brazas 2010-06-01T13:22:13Z 2010-06-02T00:48:35Z <p>Andrew's answer is right, but I'll just throw in a few comments since "topological" homotopy invariants are of great interest to me. <a href="http://arxiv.org/abs/0909.3086" rel="nofollow">Here</a> Paul Fabel has shown that $\pi_{1}^{top}$ on the Hawaiian earring is not a topological group. It turns out that multiplication can fail to be continuous even for some reasonably nice spaces (like locally simply connected planar continua). This and a nice connection with free topological groups appears in a manuscript I just posted on arXiv (perhaps shameless self-promotion...but I'll post the link when it becomes available).</p> <p>This quotient map mistake has appeared many places in the literature, even in an appendix by Peter May from the 70's who described the topological fundamental groupoid (giving each hom-set the quotient topology of fixed endpoint Moore path spaces). The same false assertion that products of quotient maps are again quotient maps was also used to "show" the higher topological homotopy groups are topological groups.</p> <p>With this quotient topology, the topological fundamental group(oid) is discrete on spaces that have universal covers (are path connected, locally path connected, and semi-locally simple connected). When it is non-discrete, this topology is often difficult to deal with. After all...there are going to be many homotopic loops that we identify in the quotient but which "look nothing alike" in the space.</p> <p>While we don't always have a topological group, we're not completely out of luck. $\pi_{1}^{top}(X)$ is always a quasitopological group in the following sense:</p> <p><strong>definition:</strong> A quasitopological group is a group $G$ with topology such that inversion is continuous and multiplication $G\times G\rightarrow G$ is continuous in each variable.</p> <p>A basic theory of these objects can be found in "Topological Groups and Related Structures" some of which you can get on google books. </p> <p><strong>Edit:</strong> <a href="http://arxiv.org/abs/1006.0119" rel="nofollow">The topological fundamental group and free topological groups</a> is the new paper I mentioned. It is a bit denser than the one from Andrew's answer but if anyone is interested enough to read it, I'd greatly appreciate any comments, suggestions, or corrections.</p> http://mathoverflow.net/questions/6775/what-is-known-about-module-categories-over-general-monoidal-categories/25560#25560 Answer by Jeremy Brazas for What is known about module categories over general monoidal categories? Jeremy Brazas 2010-05-22T07:07:22Z 2010-05-22T07:07:22Z <p>In <a href="http://arxiv.org/abs/0911.4979" rel="nofollow">Monoidal 2-structure of Bimodule Categories</a> Greenough discusses some of what you seem to be asking here for arbitrary abelian categories. He describes the tensor product $\mathcal{M}\boxtimes_{\mathcal{C}}\mathcal{N}$ of $\mathcal{C}$-bimodule categories $\mathcal{M},\mathcal{N}$ over a tensor category $\mathcal{C}$ as universal with respect to "$\mathcal{C}$-balanced" functors on the Deligne product $\mathcal{M}\boxtimes\mathcal{N}$.</p> http://mathoverflow.net/questions/23430/what-is-an-example-of-a-non-regular-totally-path-disconnected-hausdorff-space/23461#23461 Answer by Jeremy Brazas for What is an example of a non-regular, totally path-disconnected Hausdorff space? Jeremy Brazas 2010-05-04T17:38:14Z 2010-05-04T17:38:14Z <p>One of the easiest examples is the rational numbers with the subspace topology of the real line with the K-topology. Total path disconnectedness is not entirely necessary for multiplication of $\pi_{1}(\Sigma X_{+})$ to fail to be continuous. It just makes the path component space of $X$ equal to $X$, greatly simplifying complications.</p> http://mathoverflow.net/questions/23369/powers-of-quotient-maps Powers of quotient maps Jeremy Brazas 2010-05-03T19:40:32Z 2010-05-04T14:55:37Z <p>Suppose $q:X \rightarrow Y$ is a quotient map of topological spaces such that the product map $q^2:X^2\rightarrow Y^2$ is also a quotient map. Are the maps $q^n:X^n\rightarrow Y^n$ quotient maps for all $n\geq 3$? If not, are there sufficient conditions that make this the case?</p> http://mathoverflow.net/questions/99712/refining-open-covers-in-locally-path-connected-spaces/126090#126090 Comment by Jeremy Brazas Jeremy Brazas 2013-04-01T12:29:38Z 2013-04-01T12:29:38Z The cone over any space is simply connected and moreover is contractible. It has nothing to do with path connectivity. http://mathoverflow.net/questions/99712/refining-open-covers-in-locally-path-connected-spaces/126090#126090 Comment by Jeremy Brazas Jeremy Brazas 2013-03-31T13:59:22Z 2013-03-31T13:59:22Z I am doubtful that restating my question in terms of 0-th homology is helpful. http://mathoverflow.net/questions/122932/a-free-topological-group-as-a-topological-module Comment by Jeremy Brazas Jeremy Brazas 2013-03-01T12:57:30Z 2013-03-01T12:57:30Z The last section of Pestov's paper <a href="http://topology.auburn.edu/tp/reprints/v24/tp24221.pdf" rel="nofollow">topology.auburn.edu/tp/reprints/v24/tp24221.pdf</a> may be helpful. http://mathoverflow.net/questions/122932/a-free-topological-group-as-a-topological-module/123008#123008 Comment by Jeremy Brazas Jeremy Brazas 2013-03-01T12:42:04Z 2013-03-01T12:42:04Z Oh, of course you are right! That was ridiculous to suggest. I suppose what I was going for was a non-locally compact group. http://mathoverflow.net/questions/122932/a-free-topological-group-as-a-topological-module/123008#123008 Comment by Jeremy Brazas Jeremy Brazas 2013-03-01T00:11:29Z 2013-03-01T00:11:29Z I am interested to know the answer to your general question but remain skeptical. I would be surprised if $\mathbb{Q}$ were not a counterexample. You might try to contact someone who works with free topological groups regularly to see if it is known. http://mathoverflow.net/questions/118833/making-cw-complexes-metrizable Comment by Jeremy Brazas Jeremy Brazas 2013-01-15T14:30:48Z 2013-01-15T14:30:48Z I did not mean to stress the &quot;non-explicit homotopy&quot; as much as &quot;preferred CW-structure&quot; but I will see if I can make the construction work for me. Regardless, I still hope someone might know the answer to my question. Thanks again! http://mathoverflow.net/questions/118833/making-cw-complexes-metrizable Comment by Jeremy Brazas Jeremy Brazas 2013-01-14T15:21:20Z 2013-01-14T15:21:20Z Thank you for these comments @Igor and @Misha. I am aware of Whitehead's result, however, I don't want to lose my preferred CW-structure with a non-explicit homotopy equivalence. What I am asking seems plausible when you consider the 1-dimensional case. For instance, a countably infinite wedge of circles is not first countable but you can weaken the topology at the basepoint so that it embeds in $\mathbb{R}^2$. The homotopy inverse of the continuous (but non-open) identity map comes from collapsing a small closed ball about the basepoint. http://mathoverflow.net/questions/115004/the-integers-as-a-sequential-but-non-first-countable-topological-group/116385#116385 Comment by Jeremy Brazas Jeremy Brazas 2012-12-14T19:22:01Z 2012-12-14T19:22:01Z Thanks very much Ramiro. http://mathoverflow.net/questions/114816/hausdorff-group-topologies-on-finitely-generated-groups/114821#114821 Comment by Jeremy Brazas Jeremy Brazas 2012-11-30T14:19:03Z 2012-11-30T14:19:03Z Thanks, this is an excellent reference. http://mathoverflow.net/questions/114816/hausdorff-group-topologies-on-finitely-generated-groups/114821#114821 Comment by Jeremy Brazas Jeremy Brazas 2012-11-28T23:38:15Z 2012-11-28T23:38:15Z Ok, I now know what the Bohr topology is. Is there a standard text for reading up on this (that includes the failure to be sequential)? http://mathoverflow.net/questions/109779/is-the-fundamental-group-functor-a-left-adjoint Comment by Jeremy Brazas Jeremy Brazas 2012-10-16T11:45:01Z 2012-10-16T11:45:01Z Related: <a href="http://mathoverflow.net/questions/45351/does-pi-1-have-a-right-adjoint/45375#45375" rel="nofollow" title="does pi 1 have a right adjoint">mathoverflow.net/questions/45351/&hellip;</a> http://mathoverflow.net/questions/59371/looking-for-general-approaches-to-show-connectedness-of-topological-groups/59482#59482 Comment by Jeremy Brazas Jeremy Brazas 2011-03-25T02:59:15Z 2011-03-25T02:59:15Z <i>proper open subgroups</i> http://mathoverflow.net/questions/59371/looking-for-general-approaches-to-show-connectedness-of-topological-groups/59482#59482 Comment by Jeremy Brazas Jeremy Brazas 2011-03-25T02:58:21Z 2011-03-25T02:58:21Z Yes, this seems to be the simplest example. Every neighborhood of $0$ algebraically generates all of $\mathbb{Q}$ so there are no open subgroups. http://mathoverflow.net/questions/59371/looking-for-general-approaches-to-show-connectedness-of-topological-groups/59482#59482 Comment by Jeremy Brazas Jeremy Brazas 2011-03-25T00:58:20Z 2011-03-25T00:58:20Z Hugo, definitely connectedness implies the other two (which are equivalent for arbitrary $G$) for arbitrary $G$ but this does not seem to be what you are interested in. Local compactness (or maybe something slightly weaker) should be needed for the other direction. http://mathoverflow.net/questions/56700/the-fundamental-group-of-space-which-has-both-an-h-and-a-co-h-structure/56837#56837 Comment by Jeremy Brazas Jeremy Brazas 2011-02-27T22:20:02Z 2011-02-27T22:20:02Z Fair enough. I did not read this into &quot;space.&quot;