User cary - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T20:53:07Z http://mathoverflow.net/feeds/user/1874 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88448/mapping-spaces-of-diagrams mapping spaces of diagrams Cary 2012-02-14T18:53:50Z 2012-02-15T01:55:12Z <p>I'm studying a small category $A$ and diagrams of based spaces or spectra indexed by $A$ (so let's say diagrams in a category $C$ that's closed symmetric monoidal, has a compatible model structure, etc.). I'm told that in this setting there's a projective model structure on diagrams $A \rightarrow C$, where the fibrations and weak equivalences are defined objectwise; and an injective model structure, where the cofibrations and weak equivalences are defined objectwise.</p> <p>I want to take two diagrams $X,Y$ and form a mapping space $\text{Map}(X,Y)$ in a homotopy-invariant way. (The mapping space should be a subspace of the product of the mapping spaces for each level. In a more formal setting, I can pass to the twisted arrow category on $A$ and take an equalizer.) I should be able to accomplish this by taking a cofibrant replacement of $X$ and a fibrant replacement of $Y$, but there are two model structures available for doing this. This gives me two different mapping spaces; I want to show that they are equivalent. Any suggestions for approaches/tools/references?</p> <p>(An intermediate step might be to show that they are both equivalent to the space I get by taking a projective-cofibrant replacement of $X$ and an injective-fibrant replacement of $Y$.)</p> http://mathoverflow.net/questions/84458/do-spectra-have-diagonal-maps do spectra have diagonal maps? Cary 2011-12-28T18:00:17Z 2012-01-02T00:34:59Z <p>Topological spaces have diagonal maps $X \rightarrow X \times X$ and $X \rightarrow X \wedge X$, and suspension spectra also have diagonal maps $\Sigma^\infty X \rightarrow \Sigma^\infty(X \wedge X) \cong (\Sigma^\infty X) \wedge (\Sigma^\infty X)$. What about general spectra? (i.e. symmetric spectra, S-modules, or any other convenient definition.) I always assumed you could, but I haven't thought through it carefully. And if not, can we still get a cup product on $E^*(X)$ when $E$ and $X$ are spectra?</p> http://mathoverflow.net/questions/83015/what-does-bg-classify-i-e-what-is-a-principal-fibration what does BG classify? i.e. what is a principal fibration? Cary 2011-12-09T00:28:39Z 2011-12-09T22:45:33Z <p>I'm looking for cold hard facts about just what $BG$ classifies, if $G$ is any grouplike topological monoid. I have some vague idea that $[X,BG]$ is in bijection with equivalence classes of "principal fibrations" over $X$. What exactly is a principal fibration?</p> <p>May's <em>Classifying Spaces and Fibrations</em> looks like a good source, but I'm having trouble teasing out whether his notion of $G\mathcal U$-fibration (which he proves is classified by $BG$) is equivalent to the notion of a fibration $E \rightarrow B$ with a fiberwise right $G$-action giving weak equivalences $G \rightarrow E_b$, $g \mapsto xg$ for each point $x$ in the fiber $E_b$. (I can show the $\Rightarrow$ but not the $\Leftarrow$. Maybe there needs to be another condition in my naive description.)</p> <p>Also, any grouplike monoid $G$ is weakly equivalent to $\Omega BG$, so "principal fibrations," whatever they are, should correspond (in some sense I want to make precise) to pullbacks of the path-loop fibration over $BG$.</p> http://mathoverflow.net/questions/44859/cw-complexes-and-paracompactness CW complexes and paracompactness Cary 2010-11-04T18:46:36Z 2011-06-25T09:12:20Z <p>It seems like when we assume "niceness" in homotopy theory we assume that $X$ has the homotopy type of a CW complex, and in fiber bundle theory we assume that $X$ is paracompact. How do these two interact? Is any space with the homotopy type of a CW complex paracompact? (In particular, is $I^I$ paracompact?)</p> <p>(CW complexes are always paracompact and Hausdorff. According to Milnor (http://www.jstor.org/stable/1993204) a paracompact space that is "equi locally convex" will have the homotopy type of a CW complex. Also according to that paper, if $X$ has the homotopy type of a CW complex and $K$ is actually a finite complex then $X^K$ has the homotopy type of a CW complex.)</p> http://mathoverflow.net/questions/40857/is-the-mapping-cylinder-of-a-serre-fibration-also-a-serre-fibration Is the mapping cylinder of a Serre fibration also a Serre fibration? Cary 2010-10-02T17:49:58Z 2010-10-02T20:52:05Z <p>If we have a Serre fibration $p: E \rightarrow B$ with fiber of homotopy type $S^{k-1}$, then we can create a fibration with contractible fiber by first taking the mapping cylinder $M_p$ of $p$ to get a map $M_p \rightarrow B$ (not necessarily a fibration) with contractible "fiber" and then applying the "space of paths" construction to get a fibration $M_p \times_B B^I \rightarrow B$. My question is, is this last part of the construction necessary, or is the mapping cylinder $M_p$ already a Serre fibration?</p> <p>I tried lifting a homotopy $f_t: X \times I \rightarrow B$ with starting point $\tilde f_0: X \rightarrow M_p$ by cutting $X$ into the closed preimage $C$ of $B \subset M_p$ and the open preimage $U$ of $E \times [0,1) \subset M_p$. On $C \times I$ we set $\tilde f_t(x) = f_t(x) \in B \subset M_p$. On $U \times I$ we lift $f_t|U: U \times I \rightarrow B$ to $g_t: U \times I \rightarrow E$ and then set $\tilde f_t(x) = (g_t(x),(1-t)(t$ coordinate of $\tilde f_0(x)) + t(1))$. This defines a continuous lift on $C$ and on $U$ separately. If the continuous lift on $U$ extends to the closure of $U$ then we're done. The map $U \rightarrow E$ could be nasty though near the boundary of $U$. Perhaps a better approach is to first construct a map from $X \times I$ that is only "close to" a lift, then use obstruction theory (I'm not an expert on this) to show that it is homotopic to some lift.</p> http://mathoverflow.net/questions/15830/confusion-over-a-point-in-basic-category-theory Confusion over a point in basic category theory Cary 2010-02-19T18:43:38Z 2010-02-21T18:51:19Z <p>"Let Top be the category of topological spaces." If I see a definition like this, in which homeomorphic (isomorphic in the category) spaces are not identified together, then for each given topological space, how many times does it show up as an object in Top? Once? Countably many times? Uncountably many times? Is there a semantic crisis if we don't identify all homeomorphic topological spaces to the same object?</p> http://mathoverflow.net/questions/88448/mapping-spaces-of-diagrams/88461#88461 Comment by Cary Cary 2012-02-14T22:46:42Z 2012-02-14T22:46:42Z If we make the degree of $\{1,\ldots,n\}_+$ equal to $n$, then $R_+$ can be all injective order-preserving maps, and $R_-$ can be all surjective maps. Then every map factors uniquely into a map in $R_-$ followed by a map in $R_+$. Unfortunately not every map in $R_-$ lowers degree, and not every map in $R_+$ raises degree. http://mathoverflow.net/questions/88448/mapping-spaces-of-diagrams/88461#88461 Comment by Cary Cary 2012-02-14T21:46:05Z 2012-02-14T21:46:05Z I need a weak homotopy equivalence between my two constructions of mapping spaces, and I think that's stronger than saying that the homotopy categories are equivalent. Also, my A is the category of based sets $\emptyset_+$, $\{1\}_+$, $\ldots$, $\{1,\ldots,n\}_+$ and based maps between them. I believe this is not a Reedy category but I could be mistaken. http://mathoverflow.net/questions/84458/do-spectra-have-diagonal-maps/84462#84462 Comment by Cary Cary 2012-01-07T16:21:35Z 2012-01-07T16:21:35Z Thank you, this was very helpful. http://mathoverflow.net/questions/84458/do-spectra-have-diagonal-maps/84462#84462 Comment by Cary Cary 2012-01-01T15:28:58Z 2012-01-01T15:28:58Z So if I had a natural &quot;diagonal&quot; $X \rightarrow X \wedge X$ then it would give a bilinear pairing $[X,A] \times [X,B] \rightarrow [X,A \wedge B]$. By your argument, such a pairing must be zero. So the only candidate for $X \rightarrow X \wedge X$ is the zero map. Did I understand that correctly? http://mathoverflow.net/questions/83015/what-does-bg-classify-i-e-what-is-a-principal-fibration/83066#83066 Comment by Cary Cary 2011-12-10T20:49:30Z 2011-12-10T20:49:30Z OK, so in your setup, the answers are (1): all principal homogeneous spaces; (2): quasifibrations, Serre fibrations, Hurewicz fibrations, OR G-fibrations; (3): fiberwise maps commuting with the G-action. Using May's answer, anything in (2) with a fiberwise right G-action is (3)-equivalent to May's notion of G-fibration, which is classified up to (3)-equivalence by BG. We get the equivalence by applying Γ, essentially the usual trick for replacing maps by fibrations. So now we have four distinct classification problems that are all solved by BG. http://mathoverflow.net/questions/83015/what-does-bg-classify-i-e-what-is-a-principal-fibration/83082#83082 Comment by Cary Cary 2011-12-10T18:18:49Z 2011-12-10T18:18:49Z Thank you! This notion is really rather nice; thanks for helping me understand it. http://mathoverflow.net/questions/83015/what-does-bg-classify-i-e-what-is-a-principal-fibration/83066#83066 Comment by Cary Cary 2011-12-09T18:20:35Z 2011-12-09T18:20:35Z Thank you! This is exactly what I'm looking for: some type of fibration/quasifibration with a fiberwise G-action, up to fiberwise maps commuting with this action. As I said, there's something similar to this in May's book. And you're also correct in guessing that I want to preserve G, or at least work with some monoid mapping into or out of G, rather than resorting to a zig-zag. http://mathoverflow.net/questions/83015/what-does-bg-classify-i-e-what-is-a-principal-fibration Comment by Cary Cary 2011-12-09T06:53:21Z 2011-12-09T06:53:21Z Thanks for the tip David. Could you give me a reference for this notion of concordance class? http://mathoverflow.net/questions/83015/what-does-bg-classify-i-e-what-is-a-principal-fibration Comment by Cary Cary 2011-12-09T00:47:04Z 2011-12-09T00:47:04Z The other question is similar, but I'm looking for a different answer. In particular, I want to describe the objects being classified as fibrations with extra structure, or be told that this is not possible. http://mathoverflow.net/questions/44859/cw-complexes-and-paracompactness Comment by Cary Cary 2010-11-04T20:36:22Z 2010-11-04T20:36:22Z since the usual proof that two bundles are isomorphic uses the bundle homotopy theorem, and this theorem needs paracompactness. http://mathoverflow.net/questions/44859/cw-complexes-and-paracompactness Comment by Cary Cary 2010-11-04T20:35:50Z 2010-11-04T20:35:50Z In the homotopy category of CW-complexes, every object is actually a CW complex, right? Because then principal G-bundles are represented by maps into BG. If we enlarge the class of objects to spaces with the homotopy type of a CW-complex, then each space X has a homotopy equivalence from a CW complex K, pulling back the bundle gives a bundle over K, and then we classify this with a map $K \rightarrow BG$. Composing maps, we get a &quot;classifying map&quot; $X \rightarrow BG$. The problem, though, is that pulling back the bundle over $BG$ to a bundle over X doesn't necessarily give an isomorphic bundle, http://mathoverflow.net/questions/40857/is-the-mapping-cylinder-of-a-serre-fibration-also-a-serre-fibration/40862#40862 Comment by Cary Cary 2010-10-02T21:45:43Z 2010-10-02T21:45:43Z I think I've managed to convince myself that this works by thinking about paths in $U$ that end on $C$. It would be nice if I could finish this proof with more precision though. http://mathoverflow.net/questions/15830/confusion-over-a-point-in-basic-category-theory/15833#15833 Comment by Cary Cary 2010-02-19T19:14:05Z 2010-02-19T19:14:05Z That's true, but I can see the usefulness of formulating things this precisely. Part of the problem is that my intuition is a little loose.