# What is the inverse image sheaf necessary for in algebraic geometry?

Given a continuous map $f \colon X \to Y$ of topological spaces, and a sheaf $\mathcal{F}$ on $Y$, the inverse image sheaf $f^{-1}\mathcal{F}$ on $X$ is the sheafification of the presheaf $$U \mapsto \varinjlim_{V \supseteq f(U)} \Gamma(V, \mathcal{F}).$$ If $X$ and $Y$ happen to be ringed spaces, $f$ a morphism of ringed spaces, and $\mathcal{F}$ an $\mathcal{O}_X$-module, one then defines the pullback sheaf $f^* \mathcal{F}$ on $X$ as $$f^{-1}\mathcal{F} \otimes_{f^{-1} \mathcal{O}_Y} \mathcal{O}_X.$$ However, I cannot think of any other usage of the inverse image sheaf in algebraic geometry. Moreover, if $X$ and $Y$ are schemes and $\mathcal{F}$ is quasicoherent, there is an alternate way of defining $f^* \mathcal{F}$. Given $f \colon \mathrm{Spec} B \to \mathrm{Spec} A$, and $\mathcal{F} = \widetilde{M}$, where $M$ is an $A$-module, one defines $f^* \mathcal{F}$ to be the sheaf associated to the $B$-module $M \otimes_A B$. To extend this to arbitrary schemes, it is necessary to prove that it is well-defined; but I still think it is easier to work with than the other definition, which involves direct limits and two sheafifications of presheaves (the inverse image, and the tensor product). I have not checked, but I imagine that something similar can be done for formal schemes.

Hence, my question:

What uses, if any, does the inverse image sheaf have in algebraic geometry, other than to define the pullback sheaf?

A closely related question is

In a course on schemes, is there a good reason to define the inverse image sheaf and the pullback sheaf for ringed spaces in general, rather than simply defining the pullback of a quasicoherent sheaf by a morphism of schemes?

To go from the first question to the second question, I suppose one must also address whether there are $\mathcal{O}_X$-modules significant to algebraic geometers that are not quasicoherent.

Edit: I think the question deserves a certain amount of clarification. Several people have given interesting descriptions or explications of the inverse image sheaf. While I appreciate these, they are not the point of my question; I am, specifically, interested to know whether there are constructions or arguments in algebraic geometry that cannot reasonably be done without using the inverse image sheaf. So far, the answer seems to be that such things exist, but are not really within the scope of, say, a one-year first course on schemes. There are other constructions (such as the inverse image ideal sheaf) that do not, strictly speaking, require the inverse image sheaf, but for which it may be more appropriate to use the inverse image sheaf as a matter of taste.

• This exact question came up in the Wisconsin reading group for Ravi's notes! – JSE Sep 30 '10 at 15:46
• The other place I'm aware of where it appear in Hartshorne's algebraic geometry book is in the discussion of blow-ups. See page 163. This definition can be describe by $f^*$ though too as Hartshorne points out in 7.12.2. – Karl Schwede Sep 30 '10 at 15:50
• I really like the way it's covered in the stacks project. Hartshorne's definition crams the sheafification and the inverse image presheaf into one functor. It becomes much clearer when viewed as a composition. – Harry Gindi Sep 30 '10 at 16:43
• Charles, how do you propose to define pullback maps in sheaf cohomology for a map between general (perhaps non-noetherian and non-separated) schemes, since one can't expect to get by using quasi-coherent injective sheaves of modules? There is something nice about having one concept of ringed space cohomological pullback which unifies all others (e.g., in deRham theory, topology, etc.). – BCnrd Sep 30 '10 at 17:20
• What about the definition of Y-linear differential operators on a Y-scheme X/Y (as in EGA 4 Ch. 16, for example)? Or, for that matter, what about the cotangent complex of X/Y? Alternatively, if one is thinking about sheaves of sets on Y that are not O_Y-modules, then, of course, there is nothing but the inverse image. One might consider sheaves of sets that are not O_X-modules when formulating moduli problems or, to change the context a bit, when working with etale cohomology. – user2490 Sep 30 '10 at 18:33

By some coincidence, I have a student going through this stuff now, and we got to this point this just yesterday.

The definition of $f^{-1}$ is certainly disconcerting at first, but it's not that bad. You'd like to say $$f^{-1}\mathcal{F}(U) = \mathcal{F}(f(U))$$ except it doesn't make sense as it stands, unless $f(U)$ is open. So we approximate by open sets from above. A section on the left is a germ of a section of $\mathcal{F}$ defined in some open neighbourhood of $f(U)$, where by germ I mean the equivalence class where you identify two sections if they agree on a smaller neigbourhood. Even if you're still unhappy with this, the adjointness property tells you that it is the right thing to look at.

Also, some of us work with non-quasicoherent sheaves (e.g. locally constant sheaves or constructible sheaves), so it's nice to have a general construction.

Addendum: In my answer yesterday, I had somehow forgotten to mention the etale space or sheaf as a bunch of stalks $$\coprod_y \mathcal{F}_y\to Y$$ viewpoint discussed by Emerton and Martin Brandenburg. Had you started with this "bundle picture", we would be having this discussion in reverse, because pullback is the natural operation here and pushforward is the thing that seems strange.

• Bingo -- there are so many sheaves of interest apart from quasi-coherent sheaves (e.g., etale topology, or just injective sheaves on modules on a general scheme!), and even when proving things about ringed-space pullback it cleans things up to separate the issues in topological pullback from the issues in the tensoring step. Plus, later in life one meets complex-analytic spaces, rigid-analytic spaces, formal schemes, etc., and so developing good habits early on makes later adjustments straightforward (as opposed to having to "unlearn" bad definitions and redo all the basic proofs). – BCnrd Sep 30 '10 at 17:17
• Charles, why the objection to sheafifying? It's an absolutely basic fact of life with sheaves (e.g., quotients). There are plenty of non-scheme geometric theories (e.g., manifolds) where one forms tensor products of sheaves (e.g., with vector bundles) and uses quotients, etc. for which there is no crutch of affines and one has to sheafify to get the big picture. Moreover, the sheaf-theoretic unification of topology and function theory in complex variables is marvelous, and there's no crutch of affine opens or quasi-coherence there either. Over-reliance on quasi-coherence misses too much. – BCnrd Sep 30 '10 at 18:22
• Sheafification is probably best thought of as the left adjoint to the inclusion $Sh\hookrightarrow Psh$. You can derive pretty much all of its properties from that fact. The hands-on construction is really just a proof of existence. – Harry Gindi Sep 30 '10 at 18:41
• Harry, it is important to be able to "compute" things in both proofs and examples. For example, how do you prove using abstract adjoint nonsense stuff that sheafification preserves monicity (which involves maps in the "wrong" direction relative to the adjunction property)? For this and other reasons, I think it is a mistake to disregard the hands-on construction too much. – BCnrd Sep 30 '10 at 18:59
• Dear Charles: you're right that computing sheafification is often impossible over "general" open sets, and so when we can nail it down on specific opens that is good for proofs, calculations, etc. But theoretically it is important to not develop a theory which suppresses it (e.g., when teaching/learning about sheaves, getting a grip on sheafification is an essential part of the process, both where it's easy to compute on the nose and, when not, how to deal with it). For example, good exercise that $\Omega^4_M$ is sheafification of presheaf 4th wedge power of $\Omega^1_M$ for manifolds $M$. – BCnrd Oct 1 '10 at 4:36

Here is a fairly polemical answer, in a similar spirit to Brian's:

Sheafification is not a painful process: you take a presheaf, and you think about how you need to change it so that the stalks are the same, but sections can be glued. It is very natural.

The inverse image is also naturally understood in the same kind of terms: you have a sheaf $\mathcal F$ on $X$, and you would like to make a sheaf on $Y$ whose stalk at $y$ is equal to the stalk of $\mathcal F$ at $f(y)$ (i.e. $(f^{-1}\mathcal F)_x = \mathcal F\_{f(x)}$). If you ponder how you can make a rigorous construction with these properties, you will be led to the inverse image. (It's essentially taking the fibre product of $\mathcal F$ over $X$ with the map $f:Y \to X$, and indeed thinking about the inverse image is good practice for developing intuitions about fibre products in lots of other contexts.)

Using the crutch of affines and quasi-coherent sheaves discourages thinking about the (fairly simple and natural) local picture of a sheaf as a bunch of stalks glued together. A lot of the power of the geometric ideas in algebraic geometry comes from thinking geometrically, so one doesnt' want to be discouraging thinking about sheaves in this way; rather, you want to encourage it.

As for applications, Donu notes some in his answer.

Let me note another here: if $\mathcal I$ is an ideal sheaf on $X$, then $f^{-1}\mathcal I$ is naturally a subsheaf of $f^{-1} \mathcal O_X$ (because $f^{-1}$ is exact, as one sees immediately by looking on stalks and using the fact that $f^{-1}$ doesn't change stalks!), and one often wants to look at the ideal sheaf in $\mathcal O_Y$ generated by this. This is not the same (typically) as $f^*\mathcal O_Y$. (Just as, if $I$ is an ideal in $A$ and $B$ is an $A$-algebra, $B\otimes_A I$ is typically is not isomorphic to the ideal in $B$ generated by $I$.)

Now there are other ways to describe this ideal sheaf in $\mathcal O_Y$ (e.g. it is the image of the natural map $f^*\mathcal I \to \mathcal O_Y$), but the description of it in terms of $f^{-1}\mathcal I$ is convenient and very natural.

• Dear Emerton, certainly you have made a good case for thinking of sheaves on spaces as a bunch of stalks glued together, but unless I'm quite mistaken, doesn't this picture fail to generalize to sheaves on more general sites? – Harry Gindi Oct 1 '10 at 3:00
• If the site doesn't have enough points, then perhaps so. But it's a bit like the Grothendieck "functor of points" view-point: eventually we can think of any morphism as being a "point" of its source, but in the beginning, its good to understand geometric objects as being collections of points (in the more conventional sense) sweeping out some shape. This lets when one develop some geometric intuition (which is, after all, the point of the "functor of points" metaphor: one wants to use it as a tool through which geometric intuition can be channelled) before moving onto more abstract things. – Emerton Oct 1 '10 at 3:12
• Returning more directly to your comment: in the etale site one has enough points, and so the picture I'm emphasizing continues to make sense. (And this is one of the things that makes the etale site so pleasant.) On the other hand, if one wants to work in the crystalline site/topos (just to give an example where a more topos-theoretic viewpoint is needed), one has to deal with constructions that are much more involved than forming $f^{-1}$, and so the advice that one should learn this first (and not skip over it) still seems to be pretty reasonable. – Emerton Oct 1 '10 at 3:16
• Dear Emerton, you don't mean that $f^{-1}$ is not useful in the crystalline site/topos though, right? The functor $f^{-1}$ is the "more important half" of the adjunction defining a geometric morphism (induced by a morphism of underlying sites). – Harry Gindi Oct 1 '10 at 5:54
• Dear Harry, No, I don't mean that. What I have in mind, rather, is that the crystalline site/topos is one of the first contexts in which topos-theoretic issues really matter. E.g. the morphism of crystalline topoi induced by a morphism $f:X \to Y$ does not arise from a corresponding morphism of sites (unlike in the etale case), and, when computing crystalline cohomology, there is no final object in the crystalline site (again, unlike in the Zariski or etale case), and so to compute the cohomology, one really has to use the final object in the topos. – Emerton Oct 1 '10 at 6:15

One quick answer is that the stalk of a sheaf $F$ at a point (say, given by an inclusion $f\colon pt \to X$) is just $f^{-1}F$.

Donu Arapura (and BCnrd) already made this point, but I want to emphasize it: algebraic geometry employs a whole universe of sheaves that do not have $\mathscr{O}_X$ actions, and in those cases, the inverse image is the pullback of choice. Standard examples include:

1. Sheaves of solutions to a system of linear algebraic differential equations, in other words, flat sections of a quasicoherent sheaf with respect to a connection. Sometimes these are reasonably familiar, e.g., when they are locally constant.
2. $\ell$-adic sheaves on (the étale site of) a variety over a finite field of characteristic $p$ - this was the first toolset for proving the Weil conjectures.
3. Sheaves of sets, for studying representability and so on.
4. Sheaves of commutative monoids, in log geometry.
5. Sheaves of closed differential forms (which appear when studying e.g., characteristic classes related to twisted differential operators)

I've definitely seen the inverse image employed in the first 4 cases, and I wouldn't be surprised if it appeared in the fifth.

• I would also add : 6. Sheaves of groups (such as the multiplicative group whose first cohomology group is the Picard group). – ACL Mar 14 '13 at 21:55
• ACL, that is a very nice example. – S. Carnahan Mar 15 '13 at 9:21

I prefer the definition of $f^\*$ as a left-adjoint to $f_* : Mod(X) \to Mod(Y)$. The formula involving inverse image is then basically abstract nonsense using a transitivity argument with constant sheaves, at least philosophical. The proof of existence is another issue, but it follows from rather general facts of category theory (Kan extensions).

Anyway, your question was about the use of $f^{-1}$ in algebraic geometry. An example is the reduced structure sheaf on a closed subset of a locally ringed space. You take the vanishing ideal $I$ and then pull back $\mathcal{O}_X / I$ along the inclusion map, which is a priori just a continuous map. You can also view this as a module pull back, but only if you have already defined the structure sheaf.

Also, I think it is very important to learn the somewhat old-fashioned view on sheaves, namely as sections of the etale space. Then you quickly arrive at the question to which sheaf corresponds to restriction of the etale space to a subset, which is not necessarily open. Well, it is just the pullback with respect to the inclusion map.

Finally, it is good to know that the morphism $f^{\#} : \mathcal{O}_Y \to f_{*} \mathcal{O}_X$, appearing in the definiton of a morphism of a ringed spaces, corresponds to a morphism $f^{-1} \mathcal{O}_Y \to \mathcal{O}_X$, from which you get the stalk maps directly.

Roughly speaking, an element of $\varinjlim_{V\supseteq f(U)} \Gamma(V, \mathcal{F})$ is just a section in an open neighborhood of $f(U)$ (with some proper identifications).

On the other hand, a section $s\in \Gamma(U, f^{-1}\mathcal{F})$ is given by an open cover $\{V_i\}_{i\in I}$ of $f(U)$ and a section $s_i\in \Gamma(V_i, \mathcal{F})$ for each $i\in I$ for which we require that $$s_i\mid_{V_i\cap V_j \cap f(U)} = s_j\mid_{V_i\cap V_j \cap f(U)}.$$ Here equality means stalk-wise equality. In other words, $s_i=s_j \in \mathcal{F}_y$ for every $y\in V_i\cap V_j \cap f(U)$. Given two such collections $\{(s_i, V_i)\}$ and $\{(t_j, W_j)\}$, we say they are equal if they match stalk-wise for every point $y\in f(U)$.

The difference is just that one substitute a "global" statement where a section in $\Gamma(U,f^{-1}\mathcal{F})$ is required to extend to a nbhd of the whole of $f(U)$ for a "local" version of essentially the same statement and require the sections to glue together on $f(U)$.

For example, if we have $V_1\cup V_2 \supseteq f(U)$ and sections $s_i$ over the respective open sets. We might have $s_1$ and $s_2$ matches on $f(U)\cap V_1\cap V_2$ but differs at some point $y\not\in f(U)$, so they fail to glue to a section on $V_1\cup V_2$. The good thing here is that in this case the support of $s_1-s_2$ is closed in $V_1\cap V_2$, after shrinking our open sets $V_1$ and $V_2$, the sections do match on the intersection. Of course you expect this method to fail when infinite open sets are involved. This also leads to the idea that once compactness is involved, then one may have hope for this to work. For example, see Lemma 1 of Akhil Mathew's Note.