I'm trying to learn how to perform diagram-chasing in abstract Abelian categories. Instead of an approach with some elements one have to use universal properties somehow in the proof. But I reckon the need of some lemmas…
A category is Abelian if:
- it has a zero object
- it has all binary products and binary coproducts
- it has all kernels and cokernels
- every monomorphism is a kernel to some morphism
- every epimorphism is a cokernel to some morphism
The first "lemma" that coming into my mind (though I don't really know if it is true) deals with the connection between being monic and have a kernel equal to zero.
From the universal properties:
(1)$\quad f$ is a monomorphism if it given morphisms $m,n$ holds that
$\beta m=\beta n\implies m=n$
$\require{AMScd}$
\begin{CD}
X @>m>n> B@>\beta>> B'
\end{CD}
(2)$\quad k$ is a kernel to $\beta:B\to B'$ if $\beta k=0$ and for each $k'$ with
$\beta k'=0$ there is a unique morphism $\phi$ such that $k\phi=k'$
\begin{CD}
K'@>k'>>B\\
@V\exists!\phi VV\# @|\\
K @>k>> B@>\beta>> B'
\end{CD}
How to prove that $\operatorname{ker}\beta=0\implies \beta$ is mono, using (1) and (2)?
My own approach, unfortunately, consists of staring on the diagrams above.
By step 4 we have that $A \to \operatorname{Im}{f}$ and $\operatorname{Coim}{f} \to B$ are epi and mono, respectively. Therefore $\bar{f}$ is both epi and mono and we're done by step 3.
Best Answer
I assume your abelian categories are also additive.
First prove that $f$ is a monomorphism if and only if $fg = 0$ implies $g = 0$ for all $g$.
Now assume $f : X \to Y$ has kernel $(0, 0 \to X)$ and suppose $g : W \to X$ with $fg = 0$. Then by (2), there is a morphism $h : W \to 0$ such that $g = 0h = 0$.