(I apologize in advance if there's a way simpler way to formulate things, or an obvious result that I'm just missing.)
Let $\require{AMScd}\newcommand{\C}{\mathscr C}\C$ be an abelian category.
By the universal property of the kernel, a commutative diagram
\begin{align}
\begin{CD}
X_1 @>\sigma_X>> X_2\\
@Vf_1VV @Vf_2VV \\
Y_1 @>\sigma_Y>> Y_2
\end{CD}
\end{align}
gives rise to a morphism $\DeclareMathOperator{\Ker}{Ker} \Ker f_1 \to \Ker f_2$.
This is done as follows:
If $\kappa_i\colon \Ker f_i\to X_i$ is the respective kernel injection and $;$ denotes covariant composition, we can consider the composition
$\kappa_1 ; \sigma_X ; f_2 = \kappa_1;f_1;\sigma_Y = 0;\sigma_Y = 0$.
Hence $\kappa_1;\sigma_X$ factors through $\kappa_2$ by means of a (unique) morphism
$\Ker(\sigma)\colon \Ker f_1 \to \Ker f_2$.
In fact, this assignment ist functorial, so $\Ker\in \C^{\ast \to \ast}$, where the superscript denotes the small category containing two objects and one non-identity morphism.
To verify that this turns into a functor, consider the diagram
\begin{align}
\begin{CD}
\Ker f_1 @>\Ker\sigma>> \Ker f_2 @>\Ker \tau>> \Ker f_3 \\
@V\kappa_1VV @V\kappa_2VV @V\kappa_3VV\\
X_1 @>\sigma_X>> X_2 @>\tau_X>> X_3\\
@Vf_1VV @Vf_2VV @Vf_3VV\\
Y_1 @>\sigma_Y>> Y_2 @>\tau_Y>> Y_3,
\end{CD}
\end{align}
whose upper squares commute by the definition of $\Ker \sigma$.
We now see that $\Ker \sigma;\Ker \tau$ factors $\kappa_1;(\sigma_X\tau_X) = \kappa_1;(\sigma;\tau)_X$, but this is the property that uniquely defines $\Ker (\sigma;\tau)$.
Question
In the former diagram, under which condition is the top row exact?
The reason I ask is that chaining epis $\alpha;\beta\colon A\to B\to C$ gives rise to natural transformations of morphisms $\sigma\colon\alpha \to \alpha;\beta$
and $\tau\colon \alpha;\beta \to \beta$ as depicted here:
\begin{align}
\begin{CD}
\Ker \alpha @>\Ker\sigma>> \Ker \alpha;\beta @>\Ker \tau>> \Ker \beta \\
@V\kappa_1VV @V\kappa_2VV @V\kappa_3VV\\
A @= A @>\alpha>> B\\
@V\alpha VV @V\alpha;\beta VV @V\beta VV\\
B @>\beta >> C @= C.
\end{CD}
\end{align}
Interpreting the top row e.g. in the case of $R$-modules (and treating $\Ker$ as a literal zero set) enlightens us as follows:
- $\Ker \sigma$ is just „subspace inclusion“ since every element in the kernel of $\alpha$ is already in the kernel of $\alpha;\beta$
- $\Ker \tau$ maps an $x\in \Ker\alpha;\beta$ to $\alpha(x)\in \Ker\beta$, so we just have restriction of $\alpha$ to $\Ker\alpha;\beta = \alpha^{-1}(\Ker \beta)$.
Therefore $\Ker (\Ker \tau) = \Ker \alpha\vert_{\alpha^{-1}(\Ker \beta)} = \Ker \alpha$, so we indeed have an exact sequence.
The corresponding dual statement is very natural however:
Taking instead cokernels and monos (interpreting the monos $\beta',\alpha'$ as subspace inclusions $C\subset B \subset A$) yields an exact sequence $\DeclareMathOperator{\coker}{Coker}\coker\beta' \to \coker\beta';\alpha' \to \coker\alpha'$, which is precisely the statement that $(A/C)/(B/C)\simeq A/B$.
This suggests to me that the top row being exact gives us some sort of generalized isomorphism theorem, but I am unsure whether one can generalize the demonstrated diagram chase. I've tried to look for something like a snake lemma, but failed to find something helpful.
Best Answer
The snake lemma tells you exactly what you want to know. The exactness of the kernel sequence on top tells you that kernel is left exact as a functor from the arrow category. The existence of the connecting morphism tells that kernel is not right exact in general.