# algebraic geometry – Pullback and Pushforward Isomorphism of Sheaves

## 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).

• 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”.
• 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)
• 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.
• 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.
• 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

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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.

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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.