"PushPull" of a locally free sheaf

Question

Let $F \to Y$ be a locally free sheaf over $X$, and $\pi:X\to Y$. Is it true that

$$ \pi_*\pi^{-1}F = F \quad ? $$

This seems obvious to me, but I'm asking here because I'm often wrong with these things ;-)

My way to prove the above claim is simply to consider an affine open $V\subset Y$ and compute

$$ \pi_*\pi^{-1}F(V) = \pi^{-1}F(\pi^{-1}V) = \lim_{W\supset V}F(W) = F(V) $$ This really looks correct to me. In case it is wrong, could you point out which is the step that goes wrong?

Bonus question: What about the case of a quasichoerent sheaf on a scheme? Is it true that $$\pi_* \pi^* F = F \quad ?$$

Answer 1

Neither is true. In the first your mistake is in this step: $$(\pi^{-1}F)(\pi^{-1}V) = \lim_{W \supset V}F(W)$$ It should be $$(\pi^{-1}F)(\pi^{-1}V) = \lim_{W \supset \pi\pi^{-1}V}F(W)$$ because in general we don't have $\pi(\pi^{-1}(V)) = V$. Take, for example, $\pi$ to be the inclusion map of a point.

For the second assume $X$ and $Y$ are affine varieties over a field $k$. Then quasicoherent sheaves correspond to modules and $\pi^*$ and $\pi_\ast$ correspond to tensor and restriction respectively. If $A \to B$ is a ring map and $M$ an $A$-module then $B \otimes_A M$ is not necessarily isomorphic to $M$ as an $A$-module. Again, choosing the inclusion of a point provides a counterexample. This corresponds to a ring map $A \to k$ so $A/\mathfrak m \simeq k$ where $\mathfrak m$ is the kernel of the inclusion. Then $M \otimes k \simeq M/\mathfrak mM$ and if $\mathfrak mM \neq 0$ then this is not isomorphic to $M$ as an $A$-module.

Answer 2

The second formula you quote is true in the following case:

$Z \overset{\pi}\to X$ is a closed immersion with $Z\,$ equal to the support of a coherent sheaf $\mathcal{F}$ on $X$.

Here is a proof of this fact. We may assume that $X$ is affine (equal to $\operatorname{Spec} A$) with $\mathcal{F} = \widetilde{M}$ for some finite type $A$-module $M$. Hence $\pi$ is the closed immersion $\operatorname{Spec}(A/\operatorname{Ann}(M)) \to \operatorname{Spec} A$.

Now we have $\pi_{\ast} \pi^\ast \widetilde{M} = \hspace{1mm}_A(M \otimes_A A/\operatorname{Ann}(M))$ which is canonically isomorphic to $M$ as an $A$-module because $M$ is canonically an $A/\operatorname{Ann}(M)$-module. This concludes the proof.