Let $\mathscr{A}$ be an abelian category i want to show that $\mathcal{H^i}$ ( the i-th homology group) is a functor from the category of complexes of $\mathscr{A}$ to $\mathscr{A}$. I showed this for modules and when i did that proof I only had to show that maps of complexes respects the quotient.

I know that there's an embedding theorem for abelian categories but if i wanted to show this only with category theory I don't know how to show that the map of complexes behaves well under the quotient.

I tried expressing $\mathcal{H^i}(A)$ as the cokernel of the inclusion : $Im(f_{i-1}) \rightarrow Ker(f_{i})$ but after that i don't know what to do. I guess since $\mathcal{H^i}(B)$ is also another cokernel playing with the universal property I could get the desired map but I don't know.

Thanks as always!

  • 4
    $\begingroup$ Cycles and boundaries are both functors, and the natural map from boundaries to cycles is a natural transformation. Homology is the cokernel, as a functor, of this natural transformation. $\endgroup$ – Qiaochu Yuan Sep 11 '15 at 23:05
  • $\begingroup$ Sounds very clear thanks! It's very late here so I'll give it a try tomorrow and I'll ask if I have further questions. $\endgroup$ – Abellan Sep 11 '15 at 23:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.