How to show the commutativity of diagrams involving homology in abelian categories?

Question

In MacLane's book, it is shown that diagram chases can be made in any abelian category using "members" instead of elements (page 204-208). But I have two problems (concerns) about the technology of chasing members:

1. For any two arrows $h_1,h_2:A\to B$, $h_1x\equiv h_2x$ for any $x\in_m A$ does not seem to imply $h_1=h_2$. Therefore, it may be not able to show the commutativity of some diagrams by chasing members.

2. When homology is involved, the effect of some morphisms is hard to be described by "members", such as the connecting morphisms in homology long exact sequences. In this case, need we resort to the complex construction of connecting morphisms by Snake lemma?

For examples, to avoid using Mitchell's embedding theorem when learning homological algebra: 1) I cannot prove Lemma 1.5.7 in Weibel's book; 2) I can finally prove Lemma 14.6 in homology in the Stacks project, but the complex complete construction of connecting morphisms in homolgy long exact sequence have to be fully refered.

Answer 1

For 1: Observe that $\operatorname{id}_A \in_m A$.

For 2: Suppose we have a short exact sequence $0 \to A_\cdot \to B_\cdot \to C_\cdot \to 0$ of complexes, and we want to construct the connecting morphism $H_{n+1}(C_\cdot) \to H_n(A_\cdot)$. Now suppose we have some element $c \in \operatorname{Hom}(U, H_{n+1}(C_\cdot))$ for some test object $U$. Then there exists some epimorphism $\pi : V \to U$ and some $\tilde c \in \operatorname{Hom}(V, C_{n+1})$ with image $c\circ \pi$ in $\operatorname{Hom}(V, H_{n+1}(C_\cdot))$. Similarly, there exists some epimorphism $\pi' : W \to V$ and some $b \in \operatorname{Hom}(W, B_{n+1})$ whose image in $C_{n+1}$ is $\tilde c \circ \pi'$. Now, continue the usual diagram chasing to end up with some $a \in \operatorname{Hom}(W, H_n(A_\cdot))$.

Now, the equivalent I tend to use to the argument of independence of choices is:

Lemma: Suppose $\pi : W \to U$ is an epimorphism in an abelian category. Then $\pi$ is the coequalizer of the two projection morphisms $W \times_U W \to W$.

(The proof is easy to do by another member-chasing type of argument.)

Using this lemma, what we need to show is that $a \circ \pi_1 = a \circ \pi_2$ in $\operatorname{Hom}(W \times_U W, H_n(A_\cdot))$. From here, the argument that that is true looks very much like the argument by independence of choices, using the fact that $a\circ \pi_1$ and $a\circ \pi_2$ act as if they came from two different choices of preimages in $B_{n+1}$. Therefore, since $a\circ \pi_1 = a\circ \pi_2$, we can conclude that $a$ induces a morphism $\tilde a \in \operatorname{Hom}(U, H_n(A_\cdot))$.

Now all that is left is to verify that this construction is natural, so that it gives a morphism of functors $\operatorname{Hom}({-}, H_{n+1}(C_\cdot)) \to \operatorname{Hom}({-}, H_n(A_\cdot))$. By Yoneda's lemma, we then get a corresponding morphism $H_{n+1}(C_{\cdot}) \to H_n(A_{\cdot})$.

(It might also be useful to think of what the lemma means in the case of a category of sheaves of abelian groups on a topological space, or a category sheaves of modules over a ring object. In that case, passing to $V$ and then $W$ can be viewed as corresponding to passing to open covers of an open set $U$. And then, applying the lemma corresponds to verifying a gluing condition holds to get from sections on the open cover to a section over the original open set $U$.)

(And on second thought, maybe all that's really needed is to observe that every epimorphism in an abelian category is regular, i.e. it is the coequalizer of some two morphisms. The two morphisms probably don't necessarily need to be the two projection morphisms I used here. I just tend to like this pair because of the connection I mentioned to how the construction works in the sheaf case.)