Let $\mathscr C$ be an Abelian category and $W,X,Y,Z$ objects in $\mathscr C$.
How to prove the following lemma?
Lemma: The square $$ \require{AMScd}\begin{CD} W @>u >> X \\ @VVV @VVV \\ Y @>v >> Z \end{CD} $$ is a pullback square if and only if both of the following conditions are true:
- The induced map $\ker(u) \to \ker(v)$ is an isomorphism
- The induced map $\operatorname{coker}(u) \to \mathrm{coker}(v)$ is monic
On
In the above answer given by Sofie Verbeek, there is a paragraph as follows.
We find that $u$ is the kernel of the map $X \to \operatorname{coker}(u)$ as well as the map $X \to \operatorname{coker}(u) \to \operatorname{coker}(v)$. These observations imply that the map $\operatorname{coker}(u) \to \operatorname{coker}(v)$ must be a monomorphism. Certainly you'll find that this is the case by embedding your category into a category of $R$-modules, but I can't yet see a direct abstract-nonsense proof for this.
Now I try to give a direct abstract-nonsense proof for this.
Suppose $u:W\to X$ is monic, and the kernel of $X\to X/W\to L$ is also $u:W\to X$.
We have the following commutative diagram:
$$ \require{AMScd}\begin{CD} W @>u>> X @>f>> X/W @>>> 0 \\ @| @| @Vh VV @VVV \\ W @>u>> X @>g>> L @>>> \operatorname{coker}(g) \end{CD} $$
Because $\operatorname{id}_W: W\to W$ is an isomorphism and $0\to \operatorname{coker}(g)$ is monic, according to the sufficient condition of the lemma we want to prove, the square in the middle is a pullback square.
$\forall \alpha,\beta: V\to X/W$ such that $h\alpha=h\beta$, thus $h(\alpha-\beta)=0$, and we let $V\to X$ be the zero map, then we have an unique morphism $\varphi: V\to X$ such that $\operatorname{id}_X\varphi=0$ and $f\varphi=\alpha-\beta$. Because $\varphi=\operatorname{id}_X\varphi=0$, therefore $\alpha-\beta=0$, hence $\alpha=\beta$, so $h$ is monic.
Here's a sketch of one of the directions. Sorry for not including diagrams --- it's a pain without tikzcd.
Suppose that your diagram is a pullback square.
The natural map $\operatorname{ker}(v) \to Y$ together with the zero map $\operatorname{ker}(v) \to X$ induce a map $\operatorname{ker}(v) \to W$. The universal property of $\operatorname{ker}(u)$ forces this map to factor through $\operatorname{ker}(u)$. You should try to verify that the map $\operatorname{ker}(v) \to \operatorname{ker}(u)$ that you find in this way, is an inverse of the natural map $\operatorname{ker}(u) \to \operatorname{ker}(v)$.
Now consider the map $\operatorname{coker}(u) \to \operatorname{coker}(v)$. Suppose first that $v$ is a monomorphism. Then recall (or try to prove) that pullbacks always preserve monomorphisms, so that $u$ is also a monomorphism. Thus $u$ is the kernel of the map $X \to \operatorname{coker}(u)$ and similarly for $v$.
On the other hand, one can also show that $u$ is the kernel of $X \to Z \to \operatorname{coker}(v)$ (hence to $X \to \operatorname{coker}(u) \to \operatorname{coker}(v)$) as follows. Suppose you have a map $V \to X$ such that $V \to X \to Z \to \operatorname{coker}(v)$ yields the zero map. Then the part $V \to X \to Z$ factors through $Y$ (since we assumed that $v$ is the kernel of $Z \to \operatorname{coker}(v)$). Apply the universal property of the pullback to induce a unique map $V \to W$. This is the map that we wanted to find.
We find that $u$ is the kernel of the map $X \to \operatorname{coker}(u)$ as well as the map $X \to \operatorname{coker}(u) \to \operatorname{coker}(v)$. These observations imply that the map $\operatorname{coker}(u) \to \operatorname{coker}(v)$ must be a monomorphism. Certainly you'll find that this is the case by embedding your category into a category of $R$-modules, but I can't yet see a direct abstract-nonsense proof for this.
In an abelian category, you can verify that every morphism can be written as the composition of a (regular) epimorphism followed by a monomorphism; indeed, a map $f : c\to d$ can be written as $c \to \operatorname{Im}(f) \to d$. We decompose the map $v$ in your diagram into an epimorphism $v_1$ followed by a monomorphism $v_2$. In the diagram $$\begin{CD} W @>{u_1}>> X' @>{u_2}>> X \\ @V{u'}VV @VVV @VV{v'}V \\ Y @>>{v_1}> Y' @>>{v_2}> Z \end{CD}$$ we can either consider the pullback of $v_2 \circ v_1$, or the pullback $X'$ of $v_2$ followed by the pullback of $v_1$ over $X' \to Y'$; the results are the same (both being $W$). So we need not worry about that.
By the above discussion, the map between cokernels is a monomorphism in the case that $v$ is a monomorphism, so we have a monomorphism when looking at the cokernels of $v_2$ and of its lift. Now try and verify that composing with $v_1$ and its lift does not change the cokernel. (This is because $v_1$ and its lift are both epimorpisms.)
Edit: I have my doubts but here's an idea for the converse under a slightly weaker hypothesis, namely that in the diagram $$\begin{CD} \operatorname{ker} @>>> W @>{u}>> X @>>> \operatorname{coker}(u) \\ @VVV @V{u'}VV @VV{v'}V @VVV \\ \operatorname{ker} @>>> Y @>>{v}> Z @>>> \operatorname{coker}(v) \\ & @VVV @VVV \\ & & \operatorname{coker}(u') @>>> \operatorname{coker}(v') \end{CD}$$ both maps between cokernels are monomorphisms. (Note: Some parts in the diagram shouldn't be there. I don't understand how these diagrams work.) By symmetry this is really not a far-fetched assumption.
Take maps $V \to X$ and $V \to Y$. Try and use commutativity to show that the compositions $V \to X \to \operatorname{coker}(u) \to \operatorname{coker}(v)$ and $V \to Y \to \operatorname{coker}(u') \to \operatorname{coker}(v')$ are the zero maps. Then use the fact that the map of cokernels are monomorphisms to conclude that the maps $V \to X$ and $V \to Y$ factor through $\operatorname{Im}(u)$ and $\operatorname{Im}(v)$.
From this point, we do a diagram chase. Take an element $v$ of $V$, send it to $x$ and to $y$ in $X$ and in $Y$. As they lie in the image of $u$ and $u'$, we may assume $x = u(w_1)$ and $y = u'(w_2)$. Notice that the difference $w_1 - w_2$ gets sent to $0$ within $Z$, so $u'(w_1-w_2)$ lies in the kernel of $v$. The isomorphism of the kernels of $u$ and $v$ show that $u'(w_1-w_2)$ lies in the kernel of $v$, and by the isomorphism of kernels, it has a pre-image in the kernel of $u$. One of these pre-images is $w_1 - w_2$, from which we find $u(w_1 - w_2) = u(w_1) - u(w_2) = 0$. We find that when choosing pre-images in $W$, we need not worry about well-definedness issues. Use this to construct an explicit map that fits our needs $V \to W$.