# Two definitions of homology

Let $f,g$ be arrows in an abelian category such that the composite $gf$ is defined and is given by the zero arrow. I shall try to find a definition for the quotient $\ker g /\operatorname{im} f$, which coincides with the usual notion for modules.

$gf=0$ means that $\operatorname{im}f\le\ker g$ as subobjects of $\operatorname{cod}f=\operatorname{dom}g$, so there is a (monic) arrow $u$ satisfying $\ker g\circ u=\operatorname{im}f$. I would define $\ker g/\operatorname{im}f$ as the (codomain of the) cokernel of $u$. Is that correct?

Then I have to show that $\ker g/\operatorname{im}f$ is isomorphic to the dual object $\operatorname{coim}g/\operatorname{coker}f$, i.e. the kernel of the canonical epimorphism $\operatorname{coker}f\to\operatorname{coim}g$. I have no idea how to show this. All I found out is that I have canonical arrows from $\operatorname{Ker}g$ to the object $\operatorname{coim}g/\operatorname{coker}f$ and from $\ker g/\operatorname{im}f$ to the cokernel of $f$.

As a next step (this is not an exercise) I would like to show, that also the Homology functors coincide depending on which of the two object one defines as the homology.

• Well, there is not much you can do but use the universal properties of the cokernel and kernel, along with the definition of an abelian category (i.e. the fact that the canonical morphism from the coimage to the image is an isomorphism). – user314 Jun 16 '14 at 20:39
• I know about these facts. Maybe if you could tell me, what the desired arrow $\ker g/\operatorname{im}f\to\operatorname{coim}g/\operatorname{coker}f$ looks like, I could show that it is and isomorphism. – user114885 Jun 17 '14 at 20:16

## 1 Answer

Name our arrows for now as $${\rm im\,}f\overset{i_1}\longrightarrow{\rm ker\,}g\overset{i_2}\longrightarrow B \overset{p_1}\longrightarrow{\rm coker\,}f\overset{p_2}\longrightarrow {\rm coim\,}g$$ and $\ {\rm ker\,}g\overset u\longrightarrow {\rm coker\,}i_1$, $\ {\rm ker\,}p_2\overset v\longrightarrow {\rm coker\,}f$.

We want to prove that $\ {\rm coker\,}i_1={\rm ker\,}p_2$.
(Let me write compositions from left to right.)

Note that both pairs $(i_1i_2,\,p_1)$ and $(i_2,\,p_1p_2)$ form short exact sequences.

As you already observed, by the kernel and cokernel properties, we obtain an arrow $d:{\rm coker\,}i_1\longrightarrow{\rm ker\,}p_2$.

Then the idea is to prove that $udv$ is the canonical deocmposition of $i_2p_1$, i.e. that $u={\rm coim}(i_2p_1)$ and $v={\rm im}(i_2p_1)$, thus it will follow that $d$ is an isomorphism.

Finally, e.g. $u={\rm coim}(i_2p_1)$ is equivalent, by duality, to $i_1={\rm ker}(i_2p_1)$, which can be easily verified.

• The exactness you mention is clear to me. I like to think of im,ker,etc as arrows. Then $im\, i_2i_1=im\, im\, f=ker\, coker\, ker\, coker\, f=ker\, coker\, f=ker\,p_1$. However, it is not clear to me how to obtain the arrow $d$. The fact that $(p_1i_2)i_1=0$ implies that $p_1i_2$ factors through $u=coker i_1$. But that gives me an arrow on $coker\, f$, not on $ker\, p_2$. What am I misunderstanding? – user114885 Jun 19 '14 at 11:20
• Yes, (writing this time leftwards), we get $p_1i_2=\tilde du$ with a unique $\tilde d$. Then $p_2\tilde du=p_2p_1i_2=0$, and $u$ is epi, so $p_2\tilde d=0$, hence $\tilde d=vd$ with a unique $d$, as $v={\rm ker\,}p_2$. – Berci Jun 19 '14 at 12:24