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As suggested by Daniel Rust I'll pose this as a separate question.

Let $C$ be a category. Denote by $Ar(C)$ the following category: an object in $Ar(C)$ is a morphism $X_1 \rightarrow X_2$ in $C$. A morphism in $Ar(C)$ from $X_1 \rightarrow X_2$ to $Y_1 \rightarrow Y_2$ is a commutative diagram.

Show that $F:Ar(Ab) \rightarrow Ar(Ab)$ with $F(X_1 \rightarrow X_2) = (\mathrm{ker}(f) \rightarrow X_1)$ defines a functor.

I'm not sure here how the functor works on morphisms. Let $(\varphi,\gamma) \in \mathrm{Hom}(f,g)$. We must have $F(\varphi,\gamma):F(f) \rightarrow F(g)$, or in words, $(\varphi,\gamma)$ must be mapped to a pair making $\mathrm{ker}(f) \rightarrow X_1$ and $\mathrm{ker}(g) \rightarrow Y_1$ commute. I see two ways here; either introduce some map $\mathrm{ker}(f) \rightarrow \mathrm{ker}(g)$ and restrict $\varphi$ somehow, neither of which I'd have a clue on how to do, or look at how $F$ works on $\varphi,\gamma$ separately. In the latter case, $F(\varphi):\mathrm{ker}(\varphi) \rightarrow X_1$ and $F(\gamma):\mathrm{ker}(\gamma) \rightarrow X_2$. The first might make sense, but $F(\gamma)$ makes no sense to me in trying to make the new square diagram commute. Hints on how to show that $F$ is a functor?

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We have a commutative diagram $\require{AMScd}$ $\newcommand{\cd}[8]{ \begin{CD} #1 @>#5>> #2\\ @V #6 V V\# @VV #7 V\\ #3 @>>#8> #4 \end{CD}}$

$$\cd{X_1}{Y_1}{X_2}{Y_2}{}{f}{g}{}$$

and want to find a morphism $\ker(f) \to \ker(g)$. Well, if we have an element of $\ker(f)$, it lies in $X_1$, and gets mapped to some element in $Y_1$. A simple diagram chase shows, that this element lies in the kernel of $g$. This defines a map $\ker(f) \to \ker(g)$. It is just the restriction of $X_1 \to Y_1$. Hence, it is a group homomorphism, and the diagram

$$\cd{\operatorname{Ker}f}{\operatorname{Ker}g}{X_1}{Y_1}{}{}{}{}$$

commutes, as desired. That this defines a functor, is very easy to check.

The same holds more generally in every linear category with kernels.

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