# $\operatorname{Im}f\cong A/\operatorname{Ker}f$ in abelian categories

Let $f:A \rightarrow B$ be an arrow in some abelian category. There is the usual epi-mono factorization of any such arrow, but can we go further and prove isomorphism of the objects: $\operatorname{Im}f\cong A/\operatorname{Ker}f$? How can one prove this?

Here, $A/\operatorname{Ker}f$ is the codomain of the quotient object $A\twoheadrightarrow A/\operatorname{Ker}f$ defined to be the cokernel of any monic representing the subobject $\operatorname{Ker}f\rightarrowtail A$.

• If you use \operatorname{Ker} instead of \mathrm{Ker} then you automoatically get proper spacing before and after "Ker". But I wonder if you object to writing \ker f, which gives you $\ker f$? ${}\qquad{}$ – Michael Hardy Jan 17 '15 at 17:46
• @MichaelHardy I just try to capitalize the first letter when referring to objects. – user153312 Jan 17 '15 at 19:59

$A / \operatorname{Ker} f$ is called the coimage of $f$. There is always a canonical map $\operatorname{CoIm} f \to \operatorname{Im} f$, which can be defined in two dual ways (exercise!). An abelian category is often defined to be a preabelian category in which this canonical map is an isomorphism for every $f$.
If instead you define an abelian category to be a preabelian category in which every monomorphism is a kernel and every epimorphism is a cokernel, then it takes a bit of work to show that the coimage and image coincide. But it follows easily from this definition that an abelian category is balanced -- monic epis are isos. So it suffices to show that $\operatorname{CoIm} f \to \operatorname{Im} f$ is a monic epic. This follows from the fact that $\operatorname{Im}f \to B$ is monic, and dually $A \to \operatorname{CoIm} f$ is epic.