Kernel of a composition of morphisms Let $\mathcal{A}$ be an abelian category. I am trying to understand what the kernel of the composition of two morphisms $f,g$ would be like.
Take $W$ to be the pullback of $f$ and $i$. Hence, $j$ is a monomorphism. And if there is any object $V$ such that $g \circ f \circ a=0$; then we have a unique morphism $c$ by the universal property of kernel. And consequently, by the universal property of pullback, we would have a unique morphism $b$. Therefore, $W$ satisfies the universal property of $Ker(g \circ f)$

What I find surprising is that there is no role of $Ker(f)$ in the proof, though I can see that $Ker(f)$ is a subobject of $W$. Is the solution correct?
 A: Your argument is definitely correct. Moreover, this result shouldn't be surprising:

*

*The kernel of a morphism $f:X\to Y$ can be thought of as "those $x\in X$ for which $f(x)=0$".

*The pullback of morphisms $X\xrightarrow fY\xleftarrow iK$ can be thought of as "those $(x,k)\in X\times K$ for which $f(x)=i(k)$".

Therefore, the pullback of $\ker g\to Y$ along $f:X\to Y$ should be thought of as "those $(x,k)\in X\times(\ker g)$ for which $f(x)=k$" or in other words, "those $(x,y)\in X\times Y$ for which $f(x)=y$ and $g(y)=0$". In this form, perhaps it's more obvious why the pullback of $\ker g\to Y$ along $f$ is just $\ker(g\circ f)$:
\begin{align}
X\times_Y(\ker g) &\approx \{(x,k)\in X\times(\ker g) \mid f(x)=k\} \\
 &\approx \{(x,y) \in X\times Y \mid f(x)=y, g(y) = 0\} \\
 &\approx \{(x,f(x)) \in X\times Y \mid g(f(x)) = 0\} \\
 &\approx \{x\in X \mid g(f(x)) = 0 \} \\
 &\approx \ker(g\circ f)
\end{align}
(this argument would be precise in your favourite abelian category; e.g., $R\mathbf{Mod}$.)
For an even more intuitive explanation why $\ker f$ doesn't show up in the construction of $\ker(g\circ f)$: it doesn't really matter what $f$ does to the elements of $X$ (in particular, it doesn't really matter if $f$ kills the element of $X$); rather, it only matters if $f$ sends the element of $X$ into the kernel of $g$. This means finding $\ker(f\circ g)$ should boil down to seeing what the preimage of $\ker g$ is in $X$ under $f$: that is, $\ker(g\circ f) = f^{-1}(\ker g)$. How, then, do we compute the preimage? By pulling back the inclusion $\ker g\hookrightarrow Y$ along $f$!
