# Is the kernel of the product of cokernels a pullback?

I am currently reading Paolo Aluffi's textbook "Algebra: Chapter 0". I was working on exercise IX.1.18, which says:

Formulate a notion of 'intersection' of two monomorphisms with a common target $A \rightarrowtail Z$ , $B \rightarrowtail Z$ in an abelian category. Prove that the intersection of the natural monomorphisms $A \rightarrowtail A\oplus B$, $B \rightarrowtail A\oplus B$ is $0$.

If $\phi:A \rightarrowtail Z$ and $\psi:B \rightarrowtail Z$, I came up with the answer $\ker(\mathrm{coker}\: \phi,\mathrm{coker}\: \psi)$. In a slight abuse of notation, I'm using $\mathrm{coker}$ here to refer to the morphism $Z\to\mathrm{Cok}$ but I use $\ker$ to refer to the subobject of $Z$. In other words, $\ker(\mathrm{coker}\: \phi,\mathrm{coker}\: \psi)$ is the subobject of $S\subseteq Z$ given by $\{s\in Z\:|\:(\mathrm{coker}\: \phi) (s)= (\mathrm{coker}\: \psi) (s) = 0\}$.

As far as I can tell, this definition should work as expected of an intersection, but I'm not certain. In this thread, the answers instead define the intersection as the pullback of $\phi$ and $\psi$. Clearly any element in the pullback is in $\ker(\mathrm{coker}\: \phi,\mathrm{coker}\: \psi)$, but I'm not sure the converse holds. I'm curious as to whether or not it does.

## 1 Answer

Suppose you have a map $f:Y\to Z$ such that $(\operatorname{coker}\phi,\operatorname{coker}\psi)\circ f=0$. Then $\operatorname{coker}\phi\circ f=0$ and $\operatorname{coker}\psi\circ f=0$, so $f$ factors through $\ker(\operatorname{coker}\phi)=\phi$ and $\ker(\operatorname{coker}\psi)=\psi$. It follows that $f$ factors uniquely through the pullback as well. Each step of this argument is reversible: if $f$ factors through the pullback, then it is annihilated by both cokernels, so it factors through the kernel of their product. Thus $f$ factors through your $S$ iff it factors through the pullback, so they are the same subobject.