Timeline for Multisimplicial geometric realization
Current License: CC BY-SA 3.0
7 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Mar 19, 2013 at 1:56 | comment | added | Peter May | I agree completely: I am lazy about point-set topology, and of course some (mild?) condition on a pair $(X,A)$, A closed, is needed to ensure the existence of a map u from X to I with A = u^{0}. It would be nice if some expert on point-set topology weighed in, but that is not the kind of thing I'm willing to spend time on for recreation. | |
| Mar 19, 2013 at 0:19 | comment | added | Ricardo Andrade | I know the above example is not Hausdorff, so certainly not compactly generated Hausdorff, but it is the best I could come up with. However, it does not seem to me that assuming that X is compactly generated and Hausdorff would be sufficient to establish the NDR property either. | |
| Mar 19, 2013 at 0:13 | comment | added | Ricardo Andrade |
(continuation) Further, let $X=[0,1]^\delta$ be the unit interval (or any other set with at least two points, really) with the indiscrete topology: it has only two open subsets, so it is certainly not Hausdorff. Then $X$ deformation retracts to $\{0\}$, and the homotopy underlying one such deformation retraction gives a map $H:X\to\operatorname{Map}(I,X)$ such that $H(x)\in s(X)$ iff $x=0$. In conclusion, the composite $f\circ H: X\to I$ is a non-constant continuous map. This is not possible because $X$ is indiscrete and has more than two points. Please let me know if I made a mistake.
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| Mar 19, 2013 at 0:10 | comment | added | Ricardo Andrade | I absolutely apologize for prolonging this discussion into point-set topology. However, it seems to me that one requires non-trivial conditions on the space $X$ for the NDR condition that Peter states to hold ($X$ metrizable would certainly suffice). Perhaps that is what Peter means, but I acknowledge I am a bit dense. Let us consider the simplest case: the single degeneracy map $\newcommand{\Map}{\operatorname{Map}} s:X=\Map(\Delta^0,X)\to\Map(\Delta^1,X)=\Map(I,X)$. Assume there exists a map $f:\Map(I,X)\to I$ such that $f^{-1}(0)=s(X)$ is the image of the degeneracy. (to be continued) | |
| Mar 18, 2013 at 21:47 | history | edited | Peter May | CC BY-SA 3.0 |
added 795 characters in body
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| Mar 18, 2013 at 2:46 | history | answered | Peter May | CC BY-SA 3.0 |