Let $\mathscr C$ be a Abelian category, how to show a morphism $\varphi^•: A^•\to B^•$ of complexes induce the morphism of cohomology objects $H^i(A^•)\to H^i(B^•)$?

$A^•:\qquad \cdots\to A^{i-1}\stackrel{f^{i-1}}{\to}A^i\stackrel{f^i}{\to}A^{i+1}\stackrel{f^{i+1}}\to\cdots$

$B^•:\qquad \cdots\to B^{i-1}\stackrel{g^{i-1}}{\to}B^i\stackrel{g^i}{\to}B^{i+1}\stackrel{g^{i+1}}\to\cdots$

However, I can only give the morphism $\operatorname{ker}f^i\to \operatorname{ker}g^i$.

  • 3
    $\begingroup$ Have you tried diagram chasing? $\endgroup$ – Ashar Tafhim Sep 3 '18 at 7:24
  • 1
    $\begingroup$ $\DeclareMathOperator\im{im}$Show that your map $\ker f^i \to \ker g^i$ sends $\im f^{i-1}$ into $\im g^{i-1}$ so you get a well defined map in homology. $\endgroup$ – Christoph Sep 3 '18 at 9:38

Let $\varphi^\bullet : (A^\bullet, d_A^\bullet)\to (B^\bullet,d_B^\bullet)$ be a morphism of cochain complexes. Concretely, recall that this means that for any $i,$ we have $d_B^{i}\circ\varphi^i = \varphi^{i+1}\circ d_A^{i}$ as maps $A^i\to B^{i+1}.$

  1. First, observe that $\varphi^i$ induces a morphism $\varphi^i : \ker d_A^i\to\ker d_B^i$ by restriction. Given $\alpha\in\ker d_A^i\subseteq A^i,$ we have $d_B^i\circ\varphi^i(\alpha) = \varphi^{i+1}\circ d_A^i(\alpha) = \varphi^{i+1}(0) = 0.$ This means that $\varphi^i(\alpha)\in\ker d_B^i,$ which is what we wanted.
  2. In order to show that we obtain a well-defined map at the level of cohomology, we must show that the composition $$ \ker d_A^i\xrightarrow{\varphi^i}\ker d_B^i\xrightarrow{\pi}\ker d_B^i/\operatorname{im}d_B^{i-1} = \mathrm{H}^i(B^\bullet) $$ factors through $H^i(A^\bullet);$ i.e., that any $\alpha\in\operatorname{im}d_A^{i-1}$ is sent to $0$ in the quotient. Precisely, this means that we must check that such an $\alpha$ is mapped into $\operatorname{im}d_B^{i-1}$ under $\varphi^i.$ To that end, let $\alpha\in\operatorname{im}d_A^{i-1},$ and write $\alpha = d_A^{i-1}(\beta)$ for some $\beta\in A^{i-1}.$ Then it follows that $$ \varphi^i(\alpha) = \varphi^i(d_A^{i-1}(\beta)) = d_B^{i-1}(\varphi^{i-1}(\beta))\in\operatorname{im}d_B^{i-1}. $$ This shows that the map $\pi\circ\varphi^i : \ker d_A^i\to \mathrm{H}^i(B^\bullet)$ factors through a unique map $(\varphi^i)^\ast : \mathrm{H}^i(A^\bullet)\to\mathrm{H}^i(B^\bullet).$ This is the desired map.

As a remark addressing a comment: though diagram chases do not work in an arbitrary Abelian category a priori, in situations like this they are fine. In general, the Freyd-Mitchell embedding theorem tells you that a small Abelian category is equivalent to a full subcategory of $R$-$\mathsf{Mod}$ for some commutative ring $R.$ While your original Abelian category might not be small, this allows you to diagram chase using elements nonetheless: you can consider the smallest Abelian subcategory of your original category generated by all the objects and morphisms in the relevant diagram, which will again be small. Then the embedding theorem will apply, and diagram chasing may commence!

  • $\begingroup$ Your answer only works for some particular categories (e.g. category of $R$-modules), not for general abelian categories. $\endgroup$ – Aolong Li Oct 23 '18 at 16:54
  • 1
    $\begingroup$ @AolongLi Is your complaint that the Freyd-Mitchell embedding theorem only applies to small abelian categories? I recall reading that you can rig things to work in a situation like this where the abelian category is potentially unwieldy by considering the smallest abelian subcategory containing everything in the diagram (which can be forced to be small), and then use Freyd-Mitchell to get the result. $\endgroup$ – Stahl Oct 23 '18 at 21:33
  • $\begingroup$ Thanks for your comments. I learned a lot from it! Alternatively, we can also just prove this directly using the categorical definition of kernel and image. $\endgroup$ – Aolong Li Oct 23 '18 at 22:14
  • $\begingroup$ Please add something to your original answer so that I can cancell my downvote. Because my vote is locked unless the answer is editted... $\endgroup$ – Aolong Li Oct 23 '18 at 22:19
  • 1
    $\begingroup$ @AolongLi Sure :) And I agree - you can perform everything categorically, but categorical diagram chases can become rather unwieldy, and I find it much easier to use the embedding theorem if it works. $\endgroup$ – Stahl Oct 24 '18 at 1:16

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.