Timeline for Is the counit of geometric realization a Serre fibration?
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10 events
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| Jun 25, 2013 at 15:34 | comment | added | Chris Schommer-Pries | That is a really big if though (hence probably false) since we can attach by basically any map. Also something is phishy and maybe I don't really mean the join, but only some half-join ($|L| \times I \times \Delta^k$ with only the L-end collapsed?). | |
| Jun 25, 2013 at 15:22 | comment | added | Chris Schommer-Pries | Now I am thinking this is not a serious problem, and that the cofibration problem which comes up in Neil's answer is the only serious problem. Suppose that we have an initial lift $\Delta^k \to |S(X)|$. Then this factors through some finite subcomplex $|L| \subseteq |S(X)|$. Thus we get a map from $C = |L| \cup_{\Delta^k} \Delta^k \times I$ to X. Moreover (since L is finite and hence C is compact) C is a subspace of the join $|L| * \Delta^k = |L * \Delta[k]|$. If the map $C \to |L| * \Delta^k$ is a trivial cofibration, we can extended the map and get our lift. | |
| Jun 25, 2013 at 9:16 | comment | added | Chris Schommer-Pries | Indeed, you are absolutely right. I was mistaken. In retrospect it is obvious that we only get a subdivision subordinate to an open cover, not to the CW-structure of $|S(X)|$. I don't yet have a good feeling for whether this is a serious obstacle or not. | |
| Jun 25, 2013 at 8:21 | comment | added | Neil Strickland | Think about $f:[-1,1]\to [-1,1]$ given by $f(x)=x\sin(1/x)$. Give $[-1,1]$ the obvious CW structure with $0$-cells $\\{-1,0,1\\}$. We cannot finitely subdivide the source so that each $1$-cell is mapped into a single $1$-cell of the target. | |
| Jun 25, 2013 at 7:54 | comment | added | Oscar Randal-Williams | The proof of 2C.1 in Hatcher shows that you can subdivide so that each small simplex maps into the open star of a vertex. This is not the same thing. | |
| Jun 25, 2013 at 7:30 | comment | added | Chris Schommer-Pries | In particular, I think the first paragraph of "Proof of 2C.1" on page 176 of Hatcher's Algebraic Topology does exactly this simplification. This is the first step in the proof of simplicial approximation and shows that we can subdivide $\Delta^k$ in such a way that each new simplex maps into some simplex of $|Sing(X)|$. (the rest of the proof then deforms the map to be simplicial, but we don't want to do that). | |
| Jun 24, 2013 at 20:17 | comment | added | Chris Schommer-Pries | My thinking, which might very well be faulty, was roughly as follows: Given any initial family $\Delta^k \to |Sing(X)|$, we can subdivide $\Delta^k$ into small families. We should be able to ensure that it divides into cells, each of which lands in only one cell of $|Sing(X)|$. Your argument hopefully shows that we can lift these "small" families of paths. We do this one at a time to build the global lift. | |
| Jun 24, 2013 at 14:00 | comment | added | Oscar Randal-Williams | Really? I can't see why that should be enough. | |
| Jun 24, 2013 at 13:58 | comment | added | Chris Schommer-Pries | Ah great! That basically does it since I think it suffices to check the Serre fibration property just with "small" families, where the initial lift lies inside $Sing_p(X) \times \Delta^p$ for some p. | |
| Jun 24, 2013 at 13:35 | history | answered | Oscar Randal-Williams | CC BY-SA 3.0 |