# Proof of the cellular boundary formula

I'm trying to understand the proof in Hatcher (p. 141) of the cellular boundary formula. Now there's one thing that Hatcher does several times in his book and that I don't understand very well: he states things such as "In terms of the basis for $H_{n - 1}(X^{n - 1}, X^{n - 2})$ corresponding to the cells $e^{n - 1}_{\beta}$, the map $q_{\beta *}$ is the projection of $H_{n - 1}(X^{n - 1}/X^{n - 2})$ onto its $\mathbb{Z}$ summand corresponding to $e^{n - 1}_{\beta}$". Here $X$ is a CW-complex, $e^{n- 1}_{\beta}$ is a $(n - 1)$-cell and $q_{\beta}$ is the map $X^{n -1}/X^{n-2} \to S^{n-1}_{\beta}$ that "collapses the complement of the cell $e_{\beta}^{n-1}$ to a point, the resulting quotient sphere being identified with $S_{\beta}^{n-1} = D_{\beta}^{n - 1}/\partial D_{\beta}^{n - 1}$ via the characteristic map $\Phi_{\beta}$."

Now that seems intuitively plausible and I guess I can prove it by using chains and such things, but that doesn't seem very elegant. What's the right way to understand this fact? It might be very simple but I can't see a way that seems right to me.

Thanks!

I've recently struggled with the same sort of things, so I'll try to explain how I understand it. This is all supposing you understand the technical tools such as the isomorphism $H_n(X,A) \cong H_n(X/A)$, naturality, degree theory, and the likes.
First, let me say that there are a variety of levels you can make this argument so "the right way to see it" kinda depends how thorough you want to be. At the most basic level you could air your intuitions, explaining that under the identifications of $H_{n-1}(X^{n-1}, X^{n-2})$ and $H_{n-1}(\bigvee_{\beta} S_{\beta}^{n-1})$ and $\bigoplus_{\beta} H_{n-1}(S_{\beta}^{n-1})$ and so on (indeed "and so on") the map $q_{\beta}$ is the left inverse to $i_{\beta}: S_{\beta}^{n-1} \to X^{n-1}/X^{n-2}$ which by functoriality translates into the same relation of the induced maps. Because under these identifications, $i_{\beta *}$ is the inclusion of the direct summand corresponding to $e_{\beta}^{n-1}$ it follows that under these identifications $q_{\beta *}$ is the projection onto this direct summand. Given how the book is written, I think this would pass for most purposes.
If you want to be more precise, and maybe you do because you aren't convinced by the hand-wavy proof above, then you have to (1) lay out what all these isomorphisms are, (2) make precise what is meant by the $\mathbb{Z}$ summand corresponding to $e_{\beta}^{n-1}$, and (3) prove the claim. Part of the reason it is more difficult to give a rigorous elegant proof of the statement is that in doing so we have to make the statement itself more precise with (1) and (2). The lack of infrastructure here means we are more likely to come up with clunky conventions and there goes any hope of an elegant proof. I really don't no any great way to come up with a pretty proof off the bat (but whatever you do, certainly don't go back to simplices).
I didn't feel I really understood the theorem until I wrote out all the details, so I'll suppose you are the same. The question here deals with some of the other details that you could miss. Note that in the setup for the rigorous proof $q_{\beta *}$ not only give the projection onto the $\mathbb{Z}$ summand corresponding to $e_{\beta}^{n-1}$ it will send $e_{\beta}^{n-1}$ to the chosen generator of $S_{\beta}^{n-1}$.