Detail in a Proof by Hatcher - Quotient of a $\Delta$-Complex 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?
 A: 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.
