1
$\begingroup$

Let $\mathcal{A}$ be an Abelian category, $X$ be a complex, $F$ be a contravariant exact functor. I am wondering whether F preserves the homology of X, that means whether $H^{i}(FX)=F(H^{-i}(X)),\ \forall i$? (Obviously, this is true for covariant functors)

$\endgroup$
4
  • $\begingroup$ I know that a contravariant functor is a covariant functor in the opposite category, but what is the relationship between the homology in the opposite category and the original homology? $\endgroup$
    – luw
    Mar 18 '17 at 2:48
  • $\begingroup$ I don't understand the sign $-i$. $\endgroup$
    – HeinrichD
    Mar 18 '17 at 8:15
  • $\begingroup$ Since $F$ is contravariant, the $i$-th term in $X$ becomes the $-i$-th term in $FX$. $\endgroup$
    – luw
    Mar 18 '17 at 9:43
  • 1
    $\begingroup$ Crossposted on MSE. $\endgroup$ Mar 18 '17 at 13:22
4
$\begingroup$

Yes. There are two ways to describe $H^iX$: it is the cokernel of $X^{i-1}\to ker(X^i\to X^{i+1})$, and it is also the kernel of $coker(X^{i-1}\to X^i)\to X^{i+1}$. From the first of these it is clear that $F(H^iX)$ is the kernel of $coker(FX^{i+1}\to FX^i)\to FX^{i-1}$. (And from the second it is clear that $F(H^iX)$ is the cokernel of $FX^{i+1}\to ker(FX^i\to FX^{i-1})$.)

$\endgroup$
2
  • $\begingroup$ Thank you very much! Your answer is quite clear and helpful. One thing I want to make sure is that were you using the fact: if $F$ is contravariant exact, then $Fcok(f)=ker(Ff)$ and $F(kerf)=cok(Ff)$ ? By the way, I think there is a minor mistake in your last expression, it should be: $FX^{i+1} \to ker(FX^{i}\to FX^{i-1})$. $\endgroup$
    – luw
    Mar 18 '17 at 5:44
  • $\begingroup$ Yes, and thank you for the correction. $\endgroup$ Mar 18 '17 at 12:12

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.