Pushforward of pullback of a sheaf

Are there any reasonable hypotheses on a map $f: X \to Y$ and a sheaf $E$ on $Y$ so that $f_* f^* E \cong E$?

Solution 1:

Have you looked at this question? -This is a sort of reciprocal of yours.

Both functors $f^*$ and $f_*$ are adjoints ($f^*$ left, $f_*$ right). In the previous question, it was asked when the counit of the adjunction $f^*f_*{\cal F} \longrightarrow {\cal F}$ was an isomorphism. The case $Y=*$ a point gives a nice example of what you can expect then.

As for your case, you are asking when the unit of the adjunction ${\cal F} \longrightarrow f_*f^*{\cal F}$ is an isomorphism. An interesting case for you could be the following: take $X = Y_\mathrm{dis}$; that is, $Y$ with the discrete topology, in which every point is an open set, and $f= \mathrm{id} :Y_\mathrm{dis} \longrightarrow Y $ to be the identity map. Then, $f_*f^*{\cal F} = \prod_{y\in Y} y_*({\cal F}_y) $, where ${\cal F}_y $ is the stalk of ${\cal F}$ at the point $y$ and $y_*$ is the direct image functor induced by the inclusion of a one-point space ${*} \hookrightarrow Y$ at the point $y$. That is, $y_*({\cal F}_y)$ is a skyscraper sheaf.

Or, if you want it simpler, take $X = {*}$ a one-point space and you can delete that product $\prod_{y\in Y}$, getting just ${\cal F} \longrightarrow y_*({\cal F}_y)$.

Anyway, for that $f = \mathrm{id}$, $f_*f^*{\cal F}$ is the first stage of the Godement cosimplicial resolution and gives you always a flasque sheaf out of an arbitrary one.

Solution 2:

This question may have general interest to students in algebraic geometry.

If $f:X:=Spec(B)\rightarrow S:=Spec(A)$ is a map of affine schemes and if $E$ is an $A$-module with $\mathcal{E}$ the corresponding $\mathcal{O}_X$-module, it follows $f_*f^*(\mathcal{E})$ is the sheafification of $B\otimes_A E$ as left $A$-module. There are canonical $A$-linear map $i:E\rightarrow B\otimes_A E$ and $\rho:B\otimes_A E \rightarrow E$ defined by $i(e):=1\otimes e$ and $\rho(b\otimes e):=be$. The maps $i,\rho$ are seldom isomorphisms.

Question. "Are there any reasonable hypotheses on a map $f:X\rightarrow Y$ and a sheaf $\mathcal{E}$ on $Y$ so that $f_*f^*\mathcal{E} \cong \mathcal{E}$?"

Answer: In the affine situation it follows the sheaf $f_*f^*(\mathcal{E})$ is seldom isomorphic to $\mathcal{E}$. In the trivial case when $X=S$ it is true.