Let $X$ be a topological space, covered by a collection of open sets $\{U_\alpha\}$ (or more generally a cover by subspaces whose interiors cover $X$). Consider the singular simplicial set $S(X)_\bullet$ and the simplicial subset $S^{\mathfrak{A}}(X)_\bullet$ whose $n$-simplices are those maps from the topological $n$-simplex into $X$ that factor through one of the $U_\alpha$. We know (by essentially the excision theorem) that they have the same homology. Is it true that the inclusion is, though, a homotopy equivalence of simplicial sets?

I suspect I've heard this before in a class, and it seems highly plausible, but my memory is not great, and in any case it would be nice to have a reference for this fact as a less "ad hoc" way of thinking about the construction. My guess is that the subdivision operator should give the homotopy as it gives a chain homotopy in the usual proof of excision, but I'm not very sure.


The inclusion induces isomorphism on $\pi_1$ and on (co)homology with coefficients in any local system (actually, both are isomorphic to corresponding invariants of $X$ — for $S(X)_\bullet$ this is the standard way to prove that it's weakly equivalent to $X$, AFAIK). So the inclusion is a weak homotopy equivalence — and for simplicial sets that's the same thing as homotopy equivalence.

(Re: why it is iso on $\pi_1$)

Let's first check surjectivity on $\pi_1$. A rough sketch: 0) cellular approximation allows us to compute $\pi_1$ of CW-complex; 1) any 1-cell in $S(X)$ is homotopic (in $S(X)$; rel to it's boundary) to it's subdivision; 2) any 1-cell in $S(X)$ is homotopic to it's deformation in $X$; 3) for any 1-cell $f\colon I\to X$ one can choose a finite subcover from the cover $f^{-1}(U_i)$ of $I$ and then subdivide the cell correspondingly — hence the map is indeed surjective. And injectivity is just surjectivity on the level of homotopies, so the proof is more or less analoguous.

You can, probably, see now that all proofs here are quite straightforward but somewhat tiresome, so let me stop here.

  • $\begingroup$ Thanks! Why is it an isomorphism on $\pi_1$ though? $\endgroup$ – Akhil Mathew Jun 15 '11 at 13:28
  • $\begingroup$ I see; thanks for clarifying. $\endgroup$ – Akhil Mathew Jun 16 '11 at 0:22

For a reference see Theorem 10.4.20, of the book on [Nonabelian algebraic topology}1 (EMS Tract vol 15, 2011) (pdf available) where an acyclic model argument is used in the context of crossed complexes, thus avoiding a reference to local coefficients, and indeed doing fundamental group and homology of the universal cover by the same methods).

One rather neat main idea of the proof is due to R. Schön, Acyclic models and excision, Proc AMS 59(1) (1970) 167-168. This uses a collapsing of a subdivision of an $n$-cube, and deserves publicity. Such a cubical argument also fits well with the general tenor of the above book, in which cubical arguments play a crucial role.

  • $\begingroup$ Thank you for pointing this reference out. $\endgroup$ – Akhil Mathew Jun 24 '12 at 21:57

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