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

abelian-categoriescategory-theory

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?

Best Answer

I will only do the second equivalence (on request I can do the first too), which is in my opinion a lot nicer:

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.

Here is also a proof about Arnaud D.s statement:

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).

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