## The Question :

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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).

*The Question Comments :*

## The Answer 1

*0 people think this answer is useful*

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.

## The Answer 2

*-1 people think this answer is useful*

Students might want to get more details since many books in algebraic geometry “leave the details to the reader”.

Let $f:X\rightarrow S$ be a morphism of schemes with $V:=Spec(A)\subseteq S$ and $U:=f^{-1}(V):=Spec(B)$ and $\mathcal{E}$ a locally trivial $\mathcal{O}_X$-module of rank $r$ with $\mathcal{E}(U):=E$. Let

$\pi:Y:=\mathbb{V}(\mathcal{E}^*):=Spec(Sym_{\mathcal{O}_X}(\mathcal{E}^*))\rightarrow X$.

be the geometric vector bundle of $\mathcal{E}$ as defined in Hartshornes book Exercise II.5.18. (Note: Hartshorne does not dualize).

Let us understand the map $\rho$ in terms of this exercise.

It follows for any point $s\in U$ there is an isomorphism $\pi^{-1}(s)\cong \mathbb{A}^r_{\kappa(s)}$ where $\kappa(s)$ is the residue field of $s$. This follows from the following “functoriality argument”:

$Sym_A(E^*)\otimes_A \kappa(s) \cong Sym_{\kappa(s)}(E^*\otimes_A \kappa(s)):=Sym_{\kappa(s)}(E^*(s))$.

Hence for any vector $v\in \mathcal{E}(s)$ it follows we get a canonical $\kappa(s)$-linear map

$u_v:\kappa(s)^*\rightarrow \mathcal{E}(s)$.

Taking duals we get a canonical map

$u_v: \mathcal{E}(s)^* \rightarrow \kappa(s).$

We get an element

$u_v\in Hom_{\kappa(s)}(\mathcal{E}(s)^*, \kappa(s))$ and by functoriality an element

$u_v\in Hom_{\kappa(s)-alg}(Sym_{\kappa(s)}(\mathcal{E}(s)^*), \kappa(s))$.

Hence the map $u_v$ gives a canonical $\kappa(s)$-rational point

$u_v \in \mathbb{V}(\mathcal{E}^*(s))(\kappa(s)) \cong \mathbb{A}^r_{\kappa(s)}(\kappa(s))$

since the map $u_v$ is a map of schemes

$u_v:Spec(\kappa(s)) \rightarrow Spec(Sym_{\kappa(s)}(\mathcal{E}^*(s))$

over $\kappa(s)$.

Lemma. The above construction gives a one-to-one correspondence between vectors $v\in \mathcal{E}(s)$ and $\kappa(s)$-rational points $u_v$ in $\pi^{-1}(s)\cong \mathbb{V}(\mathcal{E}^*(s))$.

Proof: This follows from the construction above QED

Hence if we view a global section $t$ of $\mathcal{E}$ as a section of the projection morphism $\pi$, it follows the “value” of $t$ at the point $s$ is the element $u_{t(s)}\in \mathbb{V}(\mathcal{E}^*(s))(\kappa(s))$ from the Lemma corresponding to the value $t(s)\in \mathcal{E}(s)$.

The canonical map $\rho:f^*f_*\mathcal{E}\rightarrow \mathcal{E}$ gives a map on the fiber

$\rho(s): \Gamma(X, \mathcal{E})\otimes_{\mathcal{O}_{X,s}} \kappa(s) \rightarrow \mathcal{E}(s)$

with $v:=\rho(s)(t)$ for a global section $t$. The corresponding $\kappa(s)$-rational points $u_{v}\in \pi^{-1}(s)\cong \mathbb{A}^r_{\kappa(s)}$ is the evaluation of the global section $t$ at the point $s$.

Example. Let $C:=\mathbb{P}^1_k$ with $k$ a field and $\mathcal{L}_i:=\mathcal{O}(l_i)$ for $i=1,..,n$ and let $\mathcal{L}_d:=\mathcal{O}(d)$. Let $\mathcal{E}:=\oplus_i \mathcal{L}_i$ with $l_i\geq 1$ for all $i$. Consider the map $\rho$ in this case:

$\rho: \Gamma(C, \mathcal{E}\otimes \mathcal{L}_d)\otimes \mathcal{O}_C \rightarrow \mathcal{E}\otimes \mathcal{L}_d \rightarrow 0$. In this case the rank of $\mathcal{E}\otimes \mathcal{L}_d$ is $n$ for all choices of $d$.

There is an isomorphism

F1. $\Gamma(C, \mathcal{E}\otimes \mathcal{L}_d)\cong \oplus_n \Gamma(C, \mathcal{O}(l_i+d))$

hence the map $\rho$ is surjective for all $d\geq 0$, but it is not injective. The dimension in F1 can be arbitrary large.

Example. If $A$ is any commutative unital $k$-algebra with $k$ a field, and $I\subseteq A\otimes_k A$ the kernel of the multiplication map $m:A\otimes_k A\rightarrow A$ defined by $m(a\otimes b):= ab$, we may form the left and right $A$-module $J(l):=A\otimes_k A/I^{l+1}$. There are two maps $p,q:A\rightarrow J(l)$ defined by $p(a):=1\otimes a$ and $q(a):=a\otimes 1$. Let $E$ be a left $J(l)$-module. It follows $p^*p_*(E)\cong J(l)\otimes_A E$ and there is a short exact sequence

$0 \rightarrow J \rightarrow J(l)\otimes_A E \rightarrow^{\rho} E \rightarrow 0$

where $\rho:p^*p_*(E)\rightarrow E$ is the canonical map. Hence in this situation the map $\rho$ is always surjective but seldom injective. For affine schemes the map $\rho$ is always surjective.