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)
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$\begingroup$ 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? $\endgroup$– luwMar 18, 2017 at 2:48
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$\begingroup$ I don't understand the sign $-i$. $\endgroup$– HeinrichDMar 18, 2017 at 8:15
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$\begingroup$ Since $F$ is contravariant, the $i$-th term in $X$ becomes the $-i$-th term in $FX$. $\endgroup$– luwMar 18, 2017 at 9:43
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1$\begingroup$ Crossposted on MSE. $\endgroup$– Michael AlbaneseMar 18, 2017 at 13:22
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1 Answer
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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 $FX^{i+1}\to ker(FX^i\to FX^{i-1})$.)
answered Mar 18, 2017 at 3:52
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$\begingroup$ 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: $FX^{i+1} \to ker(FX^{i}\to FX^{i-1})$. $\endgroup$– luwMar 18, 2017 at 5:44
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$\begingroup$ Yes, and thank you for the correction. $\endgroup$ Mar 18, 2017 at 12:12