# 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).

• Have you thought about the case when $Y$ is just a point, say Spec $k$, so that $f_*$ is the same as computing global sections. You are then asking when the natural map $\mathcal O_X\otimes_k H^0(X,\mathcal F)\to \mathcal F$ is an isomorphism. This then becomes a useful exercise; once you solve it, you will see that the answer to your question as to when this map is an isomorphism is "not often". Oct 10 '10 at 3:39
• I'm the last person in this Grothendieck-universe which can talk about schemes, but if you write $Hom(f^* \mathcal F,\mathcal G)\cong Hom(\mathcal G,f_*\mathcal F)$ I think about some kind of adjunction. Then you're looking for its counity to be an isomorphism, am I right? (sorry for the stupid contro-question) Oct 23 '10 at 17:28
• 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. Aug 14 '12 at 1:25
• Here is a condition when that is true, but the reason is stupid. Let's assume your sheaf upstairs $\mathcal{F}$ is a pullback of a sheaf $\mathcal{G}$ downstairs, and assume moreover that the map $f: X \to Y$ has the property that $f_{*}\mathcal{O}_X=\mathcal{O}_Y$ which will be implied, for example, by requiring the map to be proper with geometrically connected fibres (as Charles Staats) said in his comment. Then your unlikely-happening-cancelation actually hold. And it's just projection formula.
– lee
Feb 4 '17 at 3:24
• In general the counit is an isomorphism if and only if the right adjoint is fully faithful. So you have to ask yourself when the pushforward is fully faithful. I cannot come up and prove any good conditions at the moment for when this is true, but on first glance it apperas to be very rare Aug 17 '19 at 13:09

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.

• This is not correct - the structure sheaf of any geometrically integral proper scheme over a field gives a counterexample to your final claim. Dec 19 '20 at 10:49
• Your answer is still wrong - either "trivial" means zero as it usually does, in which case the counterexample from the previous comment still applies, or it somewhat unusually means "isomorphic to the structure sheaf", in which case any affine scheme over $S$ that's not $S$ provides a counterexample. Furthermore, there's a reason this has sat as answered in the comments for 10 years - even if you fix this, I don't think you're really adding anything to what's contained in Emerton's hint and I don't think you're improving on his exposition either. Dec 19 '20 at 12:21

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.