Pullback and Pushforward Isomorphism of Sheaves

Suppose we have two schemes $X, Y$ and a map $f\colon X\to Y$. Then we know that $\operatorname{Hom}_X(f^*\mathcal{G}, \mathcal{F})\simeq \operatorname{Hom}_Y(\mathcal{G}, f_*\mathcal{F})$, where $\mathcal{F}$ is an $\mathcal{O}_X$-module and $\mathcal{G}$ an $\mathcal{O}_Y$-module (and the Homs are in the category of $\mathcal{O}_X$-modules etc). This gives a natural map $f^* f_* \mathcal{F}\to \mathcal{F}$, just by setting $\mathcal{G}=f_* \mathcal{F}$ and looking at where the identity map goes.

Are there any well-known conditions on the map or sheaves that give this is an isomorphism? For instance, I was looking through a book and saw that the map is surjective if $\mathcal{F}$ is a very ample invertible sheaf (and maybe some more hypothesis on the map and $X$ and $Y$ were assumed as well).

This question may have general interest to students: What can be said about the canonical map

$\rho: \pi^*\pi_*\mathcal{E} \rightarrow \mathcal{E}?$

where $\pi: X \rightarrow S$ any map of schemes and $\mathcal{E}$ a quasi coherent $\mathcal{O}_X$-module?

Example. Let $S:=Spec(k)$ with $k$ a field and let $X$ be a scheme of finite type over $S$. If $\mathcal{E}$ is a quasi coherent $\mathcal{O}_X$-module and $\pi: X \rightarrow S$ is the canonical map, it follows $\pi^*\pi_*\mathcal{E}\rightarrow \mathcal{E}$ is the following map:

M1. $\rho: \mathcal{O}_X\otimes \operatorname{H}^0(X, \mathcal{E}) \rightarrow \mathcal{E}.$

defined by $\rho(U)(e \otimes s_U ):= es_U \in \mathcal{E}(U)$ on the open set $U$. Here $e\in \mathcal{O}_X(U)$ and $s_U$ is the restriction of the global section $s$ to the open set $U$. Since $ \operatorname{H}^0(X,\mathcal{E})$ is a $k$-vector space it follows $\pi^*\pi_*\mathcal{E}$ is a free $\mathcal{O}_X$-module of rank $dim_k(\operatorname{H}^0(X,\mathcal{E}))$.

Example.If $\mathcal{E}$ is a rank $r$ locally trivial $\mathcal{O}_X$-module we may associate a "geometric vector bundle" $\pi: \mathbb{V}(\mathcal{E}^*)\rightarrow X$ to $\mathcal{E}$, and the fiber $\pi^{-1}(p)\cong \mathbb{A}^r_{\kappa(p)}$ is affine $r$-space over $\kappa(p)$ - the residue field of $p$. The fibration $\pi$ is by Hartshorne, Ex. II.5.18 a Zariski locally trivial fibration with affine spaces as fibers. The map $\rho$ "evaluates" a global section $s$ in the fiber $\mathcal{E}(p)$ at a point $p\in X$. We get for every $p$ an element $\rho(p)(s) \in \mathcal{E}(p)$. Here $\mathbb{A}^r_{\kappa(p)} \cong Spec(Sym_{\kappa(p)}(\mathcal{E}(p)^*))$

Hence the map $\rho$ is surjective iff $\mathcal{E}$ is generated by global sections. I believe the following claim is correct:

Lemma. Let $k=A$ be any commutative ring with $\Gamma(X,\mathcal{E})$ a free $A$-module. If the map $\rho$ is an isomorphism it follows $\mathcal{E}$ is a free (or trivial) $\mathcal{O}_X$-module.

Proof. Let $S:=Spec(k)$ and let $\pi: X \rightarrow S$ be the structure morphism. There is by definition a map

M1. $\pi^{\#}: \mathcal{O}_S \rightarrow \pi_*\mathcal{O}_X$

and this gives a canonical map

$\rho: k:=\mathcal{O}_S(S) \rightarrow \Gamma(X,\mathcal{O}_X)$. Hence $\Gamma(X,\mathcal{O}_X)$ is a $k$-algebra and left $k$-module.

It follows $\rho$ gives the abelian group $\Gamma(X, \mathcal{E})$ the structure of a $k$-module with a basis $\{e_i\}_{i\in I}$. Hence $\pi_*\mathcal{E}$ is a trivial $\mathcal{O}_S$-module on the basis $\{e_i\}_{i\in I}$. The pull back of a locally trivial $\mathcal{O}_S$-module is a locally trivial $\mathcal{O}_X$-module, and it follows there is an isomorphism

M2. $\pi^*\pi_*\mathcal{E} \cong \pi^*(\oplus_{i \in I}\mathcal{O}_S e_i) \cong \oplus_{i\in I}\mathcal{O}_X e_i$

since pull back commutes with direct sums and there is an isomorphism

$\pi^*(\mathcal{O}_S)\cong \mathcal{O}_X\otimes_{\pi^{-1}(\mathcal{O}_S)} \pi^{-1}(\mathcal{O}_S) \cong \mathcal{O}_X$

It follows $\pi^*\pi_*\mathcal{E}\cong \oplus_{i\in I} \mathcal{O}_X e_i$ is a free $\mathcal{O}_X$-module on the set $\{e_i\}_{i\in I}$. Hence if $\rho$ is an isomorphism it follows $\mathcal{E}$ is a free $\mathcal{O}_X$-module. QED

A comment on "free/trivial" modules. There seems to be confusion on what is meant by "free (or trivial) $\mathcal{O}_X$-module". I use the following definition:

Definition. Given any set $T:=\{e_i\}_{i\in I}$. We say a left $\mathcal{O}_X$-module $\mathcal{E}$ is "free (or trivial) on the set $T$" iff for any open set $U \subseteq X$ it follows

Free1. $\mathcal{E}(U) \cong \oplus_{i\in I}\mathcal{O}_X(U)e_i.$

The sum is the direct sum.

It has the following restriction maps: Given $V\subseteq U$ open subsets, it follows the restriction map

Free2. $\eta_{U,V}: \mathcal{E}(U) \rightarrow \mathcal{E}(V)$

is defined by $\eta_{U,V}(\sum_{i\in I} s_ie_i):=\oplus_{i\in I}(s_i)_Ve_i$. Here $s_i\in \mathcal{O}_X(U)$ and $(s_i)_V\in \mathcal{O}_X(V)$ is the restriction of the section $s_i$ to the open set $V$. One checks $\mathcal{E}$ is a quasi coherent $\mathcal{O}_X$-module in general.

I believe with this definition of "free/trivial", the proof of the "Lemma/observation" above is correct. I agree - it is not difficult, but some students are not used to "sheaves" and $\mathcal{O}_X$-modules.

PS: An $\mathcal{O}_X$-module $\mathcal{E}$ is called "locally trivial of rank $r$" iff there is an open cover $\{U_i\}_{i\in I}$ with $\mathcal{E}_{U_i} \cong (\mathcal{O}_X)_{U_i}^r$ for all $i$. The module $\mathcal{E}$ is "trivial of rank $r$" iff we may choose $I:=\{1\}$ and $U_1:=X$. Hence when people use the notion "$\mathcal{E}$ is a trivial $\mathcal{O}_X$-module" this means $\mathcal{E}$ is a direct sum of copies of the structure sheaf $\mathcal{O}_X$.

Example: Let $k$ be a field and let $B:=k[x], E:=\oplus_{i\in I}k[x]e_i, S:=Spec(k)$ and $C:=Spec(B)$

Let $\pi: S \rightarrow S$ be the canonical map. It follows

$\pi^*\pi_*\mathcal{E} \cong B\otimes_k (\oplus_{i\in I}Be_i)\cong \oplus_{i\in I} B\otimes_k B e_i \cong \oplus_{i\in I} k[x,y]e_i$

and the canonical map

$\phi : B\otimes_k E \rightarrow E$ defined by $b\otimes e:=be$ is not an isomorphism in this case since clearly $k[x,y] \neq k[x]$.

"If $\mathcal{F}=\mathcal{O}_X$, then the stereotypical example of when this is true is when $X=\mathbb{P}^n_Y$. More generally (but still with the $\mathcal{F}=\mathcal{O}_X$), the condition that "f is proper and all fibers are geometrically connected" may suffice, although I am not at all confident of that."

The fiber of the morphism $p: \mathbb{P}^n_Y \rightarrow Y$ over a point $s$ (with residue field $\kappa(s):=k$) is the space $\mathbb{P}^n_{k}$ and projective space is "geometrically connected" in general. Let $K$ be the algebraic closure of $k$. There is an isomorphism $Spec(K) \times_{Spec(k)}\mathbb{P}^n_{k}\cong \mathbb{P}^n_{K}$, and $\mathbb{P}^n_{K}$ is connected for any field $k$.

If $\mathcal{E}$ is a finite rank locally trivial $\mathcal{O}_X$-module we may consider $\pi:X:=\mathbb{P}(\mathcal{E}^*)\rightarrow Y$. If $Y:=Spec(k)$ with $k$ and $\mathcal{F}$ is a locally trivial $\mathcal{O}_X$-module that is not free it follows by the above Lemma that the map $\rho: \Gamma(X, \mathcal{F})\otimes \mathcal{O}_X \rightarrow \mathcal{E}$ is not an isomorphism.

The answer is: The canonical map $\rho$ is by the above argument(s) seldom an isomorphism.