Why is $f_{\ast}=f_{!}$ when $f$ is proper?

65 Views Asked by At

Let $f$ be a morphism of schemes.

I have seen that $f_{\ast}=f_{!}$ when $f$ is proper.

Why is it true (if it is not trivial where can I find a proof)?

On which generality is it true? (For example, is it true for sheaves of Abelian groups over any site on a scheme?)

1

There are 1 best solutions below

0
On BEST ANSWER

This is essentially be definition of the lower shriek functor $f_!$. Indeed, given a morphism $f\colon X\to Y$ (which is sufficiently nice), by Nagata compactification there is a factorization $X\xrightarrow{j}\overline X\xrightarrow{p}Y$ such that $p$ is a proper morphism, and $j$ is an open immersion. Then $Rf_!$ is defined to be $Rp_*\circ j_!$ where $j_!$ is the extension by zero functor. Then, one can prove that $Rf_!\colon D(Sh(X))\to D(Sh(Y))$ does not depend on the choice of compactification.

Thus, when $f$ is itself proper, we can take $p=f$ and $j=1_X$, so $Rf_!=Rf_*$.