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The proof in Hatcher's Algebraic Topology that simplicial and singular homology are equivalent has a topological detail that's throwing me off. The setup is this: We have a $\Delta$-complex $X$ which is assumed finite dimensional with skeletons $X_k$ defined as the union of all $k$-simplicies in $X$. Fixing some $k$, we have a map $\Phi : \bigsqcup_\alpha(\Delta_\alpha^k,\partial \Delta_\alpha^k) \to (X^k, X^{k-1})$ formed by the characteristic maps of each $k$-simplex $\Delta_\alpha^k \to X$. He then claims that this map induces a homeomorphism of quotient spaces $(\bigsqcup_\alpha\Delta_\alpha^k)/(\bigsqcup_{\alpha}\partial \Delta_\alpha^k) \to X^k/X^{k-1}$.

I'm having trouble proving this is a homeomorphism. $\Phi$ definitely induces a continuous map of quotient spaces, and it's not hard to see that it's bijective, but I can't make that last step from bijective continuous map to homeomorphism. If the source were compact, this wouldn't be a problem because the target is Hausdorff (I think--I know $\Delta$-complexes are Hausdorff but I'm not totally sure about the quotient of a $\Delta$-complex by a skeleton). But the source could contain infinitely many simplices, so it won't always be compact. How do we get around this? Do we have to construct a continuous inverse for the induced map by hand?

Nick A.
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    Do you believe $\Phi$ is an open map? If so, https://math.stackexchange.com/questions/1339399/a-bijective-continuous-map-is-a-homeomorphism-iff-it-is-open-or-equivalently-i You might benefit from the paragraph in Hatcher immediately prior to the heading "Simplicial Homology", p. 104 in the online copy (PDF page 8 of the linked chapter). – Eric Towers Sep 24 '20 at 23:07
  • @EricTowers $\Phi$ might not be an open map, but the induced map must be - not totally sure how to show this though. If an open set in the source quotient is in the interior of the simplices, then its image is definitely open, but how would we show that an open neighborhood of $\bigsqcup \partial \Delta_\alpha^k$ maps to an open set in $X^k$ under $\Phi$? – Nick A. Sep 24 '20 at 23:15
  • Oops. You're right. Should have been talking about the induced map. – Eric Towers Sep 24 '20 at 23:19
  • Since all spaces involved here are compactly generated Hausdorff, it suffices to observe that a compact subset of $X^k/X^{k-1}$ lies in the image of finitely many cells, and the restriction to finitely many cells is a homeomorphism for the reason you described. –  Sep 25 '20 at 16:18

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To save on some notation, let $Y=(\bigsqcup \Delta_{a}^{k})/(\bigsqcup \partial \Delta^{k}_{a})$, and let $\phi_{a}$ denote the image $\phi_{a}(\Delta_{a}^{k})$.

The characteristic maps for each $\Delta^{k}$ are nice in that for $\phi_{a}:\Delta^{k}_{a}\rightarrow X^{k}$, we have $\phi_{a} \cap \phi_{b}\neq \varnothing$ only on $X^{k-1}$.

Let $A\subset Y$ be a closed set and let $\tilde{A}$ be the preimage of the quotient map. If $\tilde{A}\cap \partial \Delta^{k}_{a}\neq\varnothing$ for any $a$, then $\sqcup\partial \Delta^{k}_{a}\subset \tilde{A}$. So $Y-A$ is a collection of open sets completely in the interior of each $\Delta^{k}_{a}$. On the interior of the $\Delta^{k}$'s, $\Phi$ is a homeomorphism. Hence $\Phi(A^{c})=(\Phi(A))^{c}$ is open. So $\Phi(A)$ is closed.

Now suppose that $\tilde{A}$ doesn't intersect any boundary of a $\Delta^{k}$, The it is contained in the interior of these simplices and again $\Phi$ is a homeomorphism when when restricted to them. Hence $\Phi(A)$ is closed. So $\Phi$ is a closed map.

Daniel H. Hartman
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