# Show that in pre-abelian categories, $0 \to A \to B$ is cokernel-exact $\iff$ $A \to B$ is monic

I am working on Chapter 7: Abstract Homological Algebra of M.Scott Osborne's Basic Homological Algebra and have trouble with the following exercise, which seems easy:

Suppose $$\mathscr A$$ is a pre-Abelian category. $$A,B\in obj\mathscr A, f \in Hom(A,B)$$. Show that:

$$0 \to A \to B$$ is kernel-exact $$\iff$$ $$0 \to A \to B$$ is cokernel-exact $$\iff$$ $$A \to B$$ is monic.

In this book, the definitions of kernel-exact and cokernel-exact are as follows:

Suppose $$\mathscr A$$ is pre-Abelian, and suppose $$\require{AMScd}$$ $$\begin{CD} A @>f>> B @>g>> C \end{CD}$$ is a diagram in $$\mathscr A$$ with $$gf=0$$

1. Let $$j:K \to B$$ be a kernel of $$g$$ and suppose $$f$$ factor through $$K$$ with $$\bar f:A\to K$$. The diagram above is called kernel-exact if $$\bar f$$ is epic.
2. Let $$p:B \to D$$ be a cokernel of $$f$$ and suppose $$g$$ factor through $$K$$ with $$\bar g:D\to C$$. The diagram above is called cokernel-exact if $$\bar g$$ is monic.

I have already proved that "$$0 \to A \to B$$ is kernel-exact" is equivalent to "$$A \to B$$ is monic", but failed to show that $$0 \to A \to B$$ is cokernel-exact $$\iff$$ $$A \to B$$ is monic.

Here are my efforts:

For one direction: if $$A \to B$$ is monic (denote it as $$h$$), then $$0 \to A$$(denoted as $$i$$) is its kernel. Suppose the cokernel of $$i$$ to be $$l: A \to D$$ and the induced map to be $$j:D \to B$$ (by the definition of cokernel). Since $$jl=h$$ is monic, then $$l$$ must be monic, so $$l$$ is a bimorphism. However, $$\mathscr A$$ is not necessarily balanced since it is just pre-abelian, so I do not know how to go on.

I know there are two ways to prove j is monic: one is to prove the kernel of $$j$$ is $$0$$, and the other is to use the definition of monic directly(i.e. $$j$$ is monic $$\iff$$ $$\forall M \in obj \mathscr A$$, $$s \in Hom(M,D), js=0$$ implies $$s=0$$). I tried them both but do not know how to move on. I got stuck in the reverse direction,too.

Can anyone give me some hints?

• The cokernel of $0\to A$ is always an iso, in any pointed category. – Arnaud D. Feb 15 at 11:10

as Arnaud D. already said, cokernel $$\varphi: A \to C$$ of $$0 \to A$$ is always an iso, hence we get that the property of $$\bar{g}$$ being monic immediately gives $$g= \bar{g} \circ \varphi$$ is a monic.
Now assume that $$g$$ is a monic, then we again have that $$\varphi: A \to C$$ is an iso. Now $$\bar{g} = \varphi^{-1} g$$ has to be a monic as well.
it suffices to proof that $$id: A \to A$$ is a cokernel. Hence consider a morphism $$f: A \to B$$ such that $$f \circ 0 = 0$$, which is an empty statement, hence the defining property holds. Furthermore A is universal with this property, since those morphisms factors uniquely over $$A$$ (it is in the end a morphism from $$A$$ to somewhere).