# What conditions on a map of schemes guarantee that pullback of global sections is injective?

Consider a morphism of schemes $f: X \rightarrow Y$. What conditions on $f, X, Y$ are sufficient to guarantee that $H^0(Y, \mathscr{F}) \rightarrow H^0(X, f^* \mathscr{F})$ is injective for

• All $\mathscr{O}_Y$ modules $\mathscr{F}$?
• All (quasi-)coherent $\mathscr{O}_Y$ modules $\mathscr{F}$?

A necessary condition for the first and second questions is that $f$ is surjective on closed points, since if $y \in Y \setminus f(X)$ is closed, we can take $\mathscr{F} = i_* \mathscr{O}_{k(y)}$. Then for any $x \in X$, $(f^* \mathscr{F})_x \simeq \mathscr{F}_{f(x)} = 0$, so $f^* \mathscr{F} = 0$. But $H^0(Y, \mathscr{F}) = k(y) \neq 0$.

In the case that $\mathscr{F}$ is a locally free sheaf, $f$ is quasi-compact and $Y$ is reduced, this condition is also sufficient (and in fact, we only need $f$ to be dominant). To see this, the map $H^0(Y, \mathscr{F}) \rightarrow H^0(X, f^* \mathscr{F})$ is the global part of the canonical map of sheaves $\mathscr{F} \rightarrow f_* f^* \mathscr{F}$. Let $U$ be an affine open set such that $\mathscr{F}|_U \simeq \mathscr{O}_U$. Then on $U$, the canonical map is identified with the structure map $\mathscr{O}_U \rightarrow f_* \mathscr{O}_{f^{-1}(U)}$, which is injective since $\mathscr{O}_Y \rightarrow f_* \mathscr{O}_X$ is injective under these hypotheses.

Is this condition also sufficient for the first two questions?

The affine version of the question is:

• Which extensions of rings $A \subseteq B$ have the property that for any (resp. any finitely presented) $A$-module $M$, the map $M \rightarrow M \otimes_A B$ is injective?

Since this always holds when $M$ is flat (which implies locally free in the finitely presented case), it seems natural to guess that we should require $B$ to be faithfully flat over $A$. However, I cannot find a way to use this property or a counterexample when $\mathrm{Spec} \ B \rightarrow \mathrm{Spec} \ A$ is surjective but $B$ is not flat over $A$.

EDIT I found a partial answer to this question in EGA IV-2 2.2.8: if $X, Y$ are arbitrary and $f$ is faithfully flat, then the canonical morphism is injective for all sheaves of quasi-coherent modules. This goes by identifying the global sections of $\mathscr{F}$ with morphisms $u: \mathscr{O}_Y \rightarrow \mathscr{F}$ and noting that the canonical map agrees with the map $u \rightarrow f^*(u)$. Faithful flatness says that if $f^*(u) = 0$ then $u = 0$. Actually, Remark 2.2.9 says that the $\mathscr{O}_Y$-module $\mathscr{F}$ does not need to be quasi-coherent for this proof to go through, although the proof is unclear to me.

Also, Alex provided a counterexample in the case that $f$ is surjective but not flat. So now, the remaining question is: If $X$ and $Y$ are "nice enough", does $f$ really have to be flat?

EDIT 2 Here is an easy counterexample when $X, Y$ are both smooth but not integral. Let $Y = \mathbf{A}^1$, $X = (\mathbf{A}^1 \setminus \{0\}) \sqcup \{0\}$, and $f$ the map given on rings by $k[x] \rightarrow k \times k[x,x^{-1}]$, $x \mapsto (0, x)$. Then let $\mathscr{F}$ be the coherent sheaf corresponding to the module $k[x]/x^2$. $f^* \mathscr{F} \simeq k$, since $(0,1) = x^2 * x^{-2}$ is killed. Then, $x$ maps to $0$ in the global section map.

• If $f$ is faithfully flat and locally of finite presentation this follows from fppf descent for quasi-coherents which shows that the functor $\mathcal{F}_\text{fl}:\mathsf{Sch}/Y\to\mathsf{Set}$ given by $(g:T\to Y)\mapsto (g^\ast\mathcal{F})(T)$ is a sheaf on $\mathsf{Sch}/Y$ with the fppf topology. Indeed, since then $\{f:X\to Y\}$ is an fppf cover and thus you have injectivity of the map by unicity of gluing. Thus, you should just look at the proof of this result--the key technical thing is the 'Amitsur complex trick' which I leave to you to google. For the case of Commented Nov 28, 2016 at 10:20
• Do you know an example that works if $X, Y$ are both integral? (Preferably smooth over $\mathbb{C}$). I'm trying to convince myself that we really can't avoid talking about flatness by assuming that $X$ and $Y$ are sufficiently nice. My original thought was to try examples when $X$ is the blowup of a point in $Y$, but I was not able to make this work. Commented Nov 28, 2016 at 10:35
• Hmm, smooth I don't know. But, here's a possible integral example. Hopefully I haven't made a mistake. Let's think about your affine criterion for a second. We know that we have the SES $0\to A\to B\to B/A\to 0$ (I'm assuming everything is integral so surjectivity implies injectivity of functions). Then, using this sequence I think you can see from the long exact sequence on $\text{Tor}$'s that the injectivity of $M\to M\otimes_A B$ is equivalent to the fact that the map $\text{Tor}^1(M,B)\to\text{Tor}^1(M,B/A)$ is an isomorphism. In particular, suppose that $b\in B-A$ but $ab\in A$ for some Commented Nov 28, 2016 at 11:35
• some $a$. Then, $B/A$ has $A$-torsion so that $\text{Tor}^1(B/A,A/(a))\ne 0$. But, if $B$ is a domain then $B$ has no $A$-torsion so $\text{Tor}^1(B,A/(a))=0$. So, for example, think about the normalization map of the cuspidal cubic $\text{Spec}(k[x,y]/(y^2-x^3))$ and the element $b=\frac{y}{x}$. Doesn't that give you an integral example? Of course, this is surjective but not flat. It's also clear why this geometrically doesn't work. Commented Nov 28, 2016 at 11:36
• PS, you may want to look at my comment regardless since it's relevant to the question beyond tracing's nice answer. Commented Nov 30, 2016 at 1:40

## 1 Answer

If $Y$ is Noetherian, so that any quasi-coherent sheaf can be written as a direct limit of coherent sheaves, then we see that the questions for coherent and quasi-coherent sheaves are equivalent.

Furthermore, the pushforward of a quasi-coherent sheaf on any open $U\subset Y$ is quasi-coherent on $Y$, and so the question for $f:X \to Y$ is equivalent to the analogous question for each $f^{-1}(U) \to U$, as $U$ ranges over the members of an open cover of $Y$, say an affine open cover.

If $f$ is quasi-compact (e.g. if $X$ is also Noetherian), then $f^{-1}(U)$ is a union of finitely many affines, and so we may find a surjective map $V \to f^{-1}(U)$ with affine source which is locally on the domain an open immersion, and so the question reduces to the case of $V \to U$, with both $U$ and $V$ affine.

In this case, as has basically been observed already, a sufficient condition is that the corresponding ring map $A \to B$ admit a section (as a morphism of $A$-modules), or even admits a section after a faithfully flat base-change. (Thus if $X\to Y$ is faithfully flat, we get a positive answer, because if $A \to B$ is faithfully flat, then the base-changed morphism $B \to B\otimes_A B$ admits a section, even as a map of rings, namely the multiplication map $B\otimes_A B \to B$.)

There are other cases where one necessarily has sections: e.g. Hochster's direct summand conjecture, whose proof was recently completed by Andre (see here) shows that if $R \subset S$ is a finite extension with $R$ a Noetherian regular ring splits as a map of $R$-modules. So if $X\to Y$ is a finite surjective morphism of Noetherian schemes, with $Y$ furthermore regular, then the question has a positive answer (but you should check that my reductions to the affine case are correct!).

• Hey tracing--nice answer. I didn't think to consider the direct summand conjecture (also, for those reading Bhatt gave a shorter proof soon after)--hopefully answering this question doesn't require anything that hard! Specifically, in the case that the OP is interested in (I think the case of smooth over characteristic zero fields) then the direct summand conjecture is 'easy'. Namely, if $A\to B$ is the corresponding map of coordinate rings, let's say which is finite of degree $n$, then one can consider the trace map $\text{tr}_{B/A}:B\to A$ divided by $n$ Commented Nov 30, 2016 at 1:36
• (see this mathoverflow.net/questions/7655/… for a definition of the trace map when both are normal) to obtain the desired section. In fact, this shows that if our map is finite and both $X$ and $Y$ are normal (and we're in characteristic $0$) then we have an affirmative answer--this explains why (as I show above in my comments) intuitively this fails for normalization maps. I wonder if, in general, assuming that $X$ and $Y$ are normal is good enough or, more generally, whether this is a regular in codimension $1$ phenomenon. Commented Nov 30, 2016 at 1:36
• Am I correct that in the case of smooth projective varieties, this argument reduces the problem (via Stein factorization) to studying surjective, non-flat, maps of smooth projective varieties with $f_* \mathscr{O}_X \simeq \mathscr{O}_Y$ (i.e. geometrically connected fibers)? Commented Nov 30, 2016 at 3:31
• @AlexYoucis: You're right, the direct summand conjecture is probably a bit of overkill; it just occurred to me as a useful general result underlying this type of question. I'm not sure about your more refined questions/suggestions right now, but will try to think about them. Commented Nov 30, 2016 at 3:53
• @Dorebell How do you know that the finite map you factor through is regular, even if the base and target of the original map are? Also, have you tried just asking Brian? Commented Nov 30, 2016 at 3:56