$\require{AMScd}$
I'm reading Hatcher's chapter/former book on spectral sequences. I'm currently struggling to understand the last substantive step of the proof of Theorem 5.3 (the Serre spectral sequence for homology exists, basically). The proof itself is fairly long and while I have made an effort to make this question self-contained, the above link contains all details of course (pp 13-15). Please let me know if something in particular is unclear.
The most difficult bit seems to be naturality. To this end, Hatcher reduces the problem repeatedly. At the end, we are left with a homeomorphism $f : (D_i, S_i) \to (D_\beta, S_\beta)$ where both $D$s are the dimension $p-1$ disks and each $S$ is the boundary. We are also given a characteristic map attaching $(D_\beta, S_\beta)$ to the base space of our Serre fibration $\pi : E \to B$, so this homeomorphism lifts to an homeomorphism $\tilde{f} : (\tilde{D}_i, \tilde{S}_i) \to (\tilde{D}_\beta, \tilde{S}_\beta)$. It is clear that this map induces an isomorphism on the homology groups.
In each case, we have a canonical identifications $H_{p + q - 1}(\tilde{D}_i, \tilde{S}_i) \cong H_{q}(F)$ and $H_{p + q - 1}(\tilde{D}_\beta, \tilde{S}_\beta) \cong H_{q}(F)$ where $F$ is the fiber over the base point.
This identification proceeds by first identifying these homotopy groups with $H_q(F_i)$ and $H_q(F_\beta)$ respectively (the fibers over the base points of $D_i$ and $S_i$), and then using the assumption that the action of $\pi_1(B)$ on $H_*(F)$ identifying these groups with $H_q(F)$ by means of transport along any path. For the sake of being explicit, the chain of isomorphisms reducing $H_{p + q - 1}(\tilde{D}, \tilde{S})$ to $H_q(\tilde{D^0})$ is given as follows:
So, we have a (potentially non-commutative) square: $$ \begin{CD} H_{p + q - 1}(\tilde{D}_i, \tilde{S}_i) @>\tilde{f}_*>> H_{p + q - 1}(\tilde{D}_\beta, \tilde{S}_\beta)\\ @V{\epsilon}VV @VV{\epsilon}V\\ H_q(F_i) @>>> H_q(F_\beta)\\ @VVV @VVV\\ H_q(F) @= H_q(F) \end{CD} $$
We must show that if $f$ is degree 1, then this diagram commutes and if $f$ has degree $-1$, then this diagram anti-commutes. (For reference, this is stated at the bottom of page 14 in the original proof).
This is the problem I'd like to solve. I've included a bit of what Hatcher writes below, but if someone can provide a simple and direct proof of the above result then I'd happily accept that.
Hatcher claims the first case follows from naturality of $\epsilon$, and while this seems plausible enough, this naturality property is never fully stated so I find it slightly unclear. In particular, the naturality property of $\epsilon$ reduces us to considering the action of $\tilde{f}$ on the fiber of a single point, but I do not see how to conclude that this map is then the identity.
The proof of the inverse is a bit more mysterious. It again uses the naturality of $\epsilon$ to reduce to the case where $D = D^1$ is dimension 1, but he then argues that the long exact sequence of the triple $(\tilde{D}^1, \tilde{S}^0)$ "breaks up into short exact sequences": $$ 0 \to H_{q + 1}(\tilde{D}^1; \tilde{S}^0; G) \to H_q(\tilde{S}^0; G) \to H_q(\tilde{D}^1; G) \to 0 $$
I'm slightly puzzled by this because it would seem to imply that $H_{q + 1}(\tilde{D}^1) = 0$, but there is nothing special about $q$ so this would seem to imply that $H_*(\tilde{D}^1) = 0$, which is certainly false if the fiber of the fibration we're considering is non-trivial. Assuming the above makes sense, he then argues that the natural transformation $H_{q + 1}(\tilde{D}^1, \tilde{S}^1; G) \cong H_q(\tilde{D})$ when precomposed with a reflection reverses signs, but I'm afraid I don't see how this plausible sounding fact connects with our original goal. (In the original document, this is the bottom of page 14 and the very top of page 15.)
As I said above, I'm mostly curious in how to complete this last step of the proof, but I'm also very curious if I'm grossly misunderstanding what Hatcher intends here.