Intuitive explanation of Four Lemma

Question

In the Short Five Lemma

where the rows are exact, it is a fact that $$\alpha \text{ and }\gamma \text{ injective (surjective) }\implies \beta \text{ injective (surjective)}.$$ I've heard this fact summarized as "if $\beta\left.\right|_A$ is injective (surjective) and $\tilde{\beta}:B/A \to B'/\beta(A)$ is injective (surjective), then $\beta$ is injective (surjective)".

This explanation makes a good deal of sense to me, and I'm looking for a similar explanation for the four lemma, and also for questions like this:

In these cases, we could talk about $B$ containing $A / \ker \psi$, and induced maps on this. Is this the way to go?

Would it be advantageous to consider this from a categorical perspective?

I'm trying to have exercises like the above be not a collection of random facts, but a picture of what's really going on, and explanations like the one above are very helpful.

Answer 1

Hre's one way of looking at (a) of your posted question:

• $\beta$ injective means that you can replace $B'$ by $B$ (we've identified $B$ with $\beta(B)$), as long as you now interpret $\psi'$ as $\psi'|_{{\rm image}(\alpha)}$. Now there's no point in separating the two occurrences of $B$ nor keeping the map $\beta$; once we identify the two occurrences of $B$, the diagram becomes a bowtie.
• $\alpha$ surjective means you can replace $A$ by $A/\ker\alpha$ (commutativity of the bowtie's left triangle forces $\psi|_{\ker\alpha}=0$). But $A/\ker\alpha \cong A'$, so now there's no point in separating the two occurrences of $A'$ nor keeping the map $\alpha$. Now we basically have the same map twice: $\psi=\psi'$ (technically the maps induced by them via this process, but we'll use the same names) both go from $A'$ to $B$.
• We can replace $B$ by $B/\ker\varphi \cong C$; by exactness; this forces $\psi$ (hence $\psi'$) to be the zero map. Again we identify $B/\ker\varphi$ with $C$; now the diagram is just $A'\longrightarrow C\longrightarrow C'$, where the first map is the zero map.
• By exactness, the second map $\varphi'$ is injective. But by now $\gamma=\varphi'$, so $\gamma$ is injective.