2

I am having trouble with the general definition of Chain homology. Given an Abelian category $\textbf{A}$, I have defined the category of Chain Complexes $\text{Ch}_\bullet(\textbf{A})$ as usual. Moreover, I have defined subobjects and quotient objects as equivalence classes of monomorphisms and epimorphisms. In the special case where $f:X\to Y$ is a monomorphism, it gives rise to a quotient object $Y/X$ of $Y$ as the equivalence class of $\text{coker}(f)$.

In the case of the Chain Homology, it is supposedly defined as : $$H_n(C)=\ker(\partial_n)/\text{Im}(\partial_{n+1}).$$

Now, if I consider the following diagram, I cannot put things all together :

The map $\ker(\partial_n)\to\text{Im}(\partial_{n+1})$ is the wrong way around to consider its cokernel. In fact, it has no reason to be either monic or epic...

What am I doing wrong ? Could someone explain to me the correct definition ? I cannot find anything else than the nLab article about it...

Anthony
  • 1,379
  • 1
    Your $\mathrm{Im},\partial_{n+1}$ is,kernel of the cokernel of $\partial_{n+1},$ and so it should have an arrow $c$ to, and not from $C_n$. It is also the cokernel of the kernel of $\partial_{n+1},$ so there is a canonical arrow from $C_{n+1}$ to it, making the triangle (with $\partial_{n+1}$) commutative. Since the map $C_{n+1}\to \mathrm{im},\partial_{n+1}$ is an epimorphism, from $\partial_n \circ \partial_{n+1}=0$ one obtains that $\partial_n \circ c=0$ also, and thus, by the universal property of $\ker\partial_n$, one obtains a map $\mathrm{im},\partial_{n+1} \to \ker \partial_n $. – Pavel Čoupek Jul 03 '20 at 18:04
  • Thanks for your answer ! It helps me a lot ! One last thing : when you say that $\text{Im}(\partial_{n+1})$ is the cokernel of $\ker(\partial_{n+1})$, you mean that the map $C_{n+1}\to\text{Im}(\partial_{n+1})$ is given to as $\text{coIm}(\partial_{n+1})$ post-composed with the iso from the image to the coimage ? – Anthony Jul 03 '20 at 18:51
  • 1
    Yes, I mean that in any abelian category, image and coimage are isomorphic via the canonical morphism between them (I don't remember which is which, so I just call image both; since they are canonically isomorphic, hopefully this terminology is not too abusive.) – Pavel Čoupek Jul 04 '20 at 02:43

0 Answers0