Morphisms of Sheaves

We’ve got sheaves now, so naturally we move on to morphisms between them. We begin by fixing X a topological space and take \mathscr{F},\mathscr{G} to be two presheaves on X. A morphism of presheaves is an abelian group homomorphism \phi(U):\mathscr{F}(U)\to\mathscr{G}(U) for each U\subset X open, that satisfies an extra condition.

This extra condition is that, whenever V\subseteq U are open sets, \begin{array}{ccc}\mathscr{F}(U)&\stackrel{\phi(U)}{\to}&\mathscr{G}\\\downarrow \rho_{UV}&&\downarrow\rho'_{UV}\\\mathscr{F}(V)&\stackrel{\phi(V)}{\to}&\mathscr{G}(V)\end{array} commutes. What I mean when I say that it commutes is that \rho'_{UV}\circ\phi(U)=\phi(V)\circ\rho_{UV}, and this can be visualized by saying that whatever path you take from \mathscr{F}(U) to \mathscr{G}(V), you get the same map. A morphism of sheaves is then just a morphism of presheaves.

Now, recall that stalks were made up of equivalence classes of pairs (U,s) with s\in\mathscr{F}(U). A morphism \phi:\mathscr{F}\to\mathscr{G} of presheaves induces a map on the stalks by \phi(U,s)=(U,\phi(U)(s)).

A nice fact is that a morphism of sheaves \phi:\mathscr{F}\to\mathscr{G} is an isomorphism (ie, has a two-sided inverse) if and only if the induced maps on the stalks are all isomorphisms. This is NOT true for presheaves, though, and it really points out that a sheaf collects local data, and we can understand it by looking near individual points.

Now, anything we can do with abelian groups, we can do with presheaves. We can define \ker\phi to be the presheaf given by \ker\phi(U)=\ker(\phi(U)), and the same with images and cokernels (that would be \mathscr{G}(U)/\mathrm{im}(\phi(U))). Now, if \mathscr{F} and \mathscr{G} are sheaves, then \ker\phi is also a sheaf, but this isn’t necessarily true for the image and cokernel. We need the following:

Theorem: Let \mathscr{F} be a presheaf. There exists a sheaf \mathscr{F}^+ and a morphism \theta:\mathscr{F}\to\mathscr{F}^+ such that for any sheaf \mathscr{G} and any morphism \phi:\mathscr{F}\to\mathscr{G}, there is a unique morphism \psi:\mathscr{F}^+\to \mathscr{G} such that \phi=\psi\circ\theta. This pair is unique up to unique isomorphism and is called the sheaf associated to \mathscr{F}.

The point is that for any presheaf, there’s a “smallest” sheaf that contains it. The idea is to basically build up new sections by gluing together the sections on smaller open sets in every allowable way so that the sheaf axiom holds. All I’ll do is define \mathscr{F}^+, and point out that the proof isn’t that terrible once you have the definition.

We define \mathscr{F}^+(U) to be the collection of functions s:U\to \coprod_{p\in U} \mathscr{F}_p to the union of the stalks satisfying two conditions:

  1. For each p\in U, we have s(p)\in \mathscr{F}_p.
  2. For each p\in U, there exists V\ni p open and contained in U and an element t\in \mathscr{F}(V) such that for all q\in V, the germ t_q of t at q is equal to s(q).

We’ll also call this the sheafification of \mathscr{F}, and whenever we get something that isn’t a sheaf from an operation on sheaves, we’ll often just refer to it as a sheaf when we REALLY mean the sheafification. This is something that is commonly done, and shouldn’t cause much confusion (I hope).

Another nice thing that we should hope is true: the sheafifcation of something that is already a sheaf is itself.

So now we define can take the cokernel and image of a sheaf map to be their sheafifications, and so we can regard them as sheaves.

Now we can define things that I probably could have gotten away with no one (who didn’t already know this stuff) calling me on, but that it’s important to be explicit about. A subsheaf of a sheaf is just a second sheaf such that for each open set, it assigns a subgroup and the restriction maps are the induced ones. We say that a morphism of sheaves is injective if it has kernel zero, that is, the kernel is the sheaf which assigns to each open set the trivial group. We call a morphism surjective if the image (sheaf) is equal to the target sheaf.

More generally, we say that a sequence of sheaves \ldots\to\mathscr{F}^{i-1}\stackrel{\phi^{i-1}}{\to}\mathscr{F}^i\stackrel{\phi^i}{\to}\mathscr{F}^{i+1}\to\ldots is exact if for each i, we have \ker\phi^i=\mathrm{im}\phi^{i-1}. So a sequence 0\to\mathscr{F}\stackrel{\phi}{\to}\mathscr{G} is exact if and only if \phi is injective
and \mathscr{F}\stackrel{\phi}{\to}\mathscr{G}\to 0 is exact if and only if \phi is surjective.

So now that we’ve discussed morphisms of sheaves, there are two more operations on all sheaves that are worth mentioning before we move on. Instead of looking at two sheaves, let’s look at f:X\to Y a continuous map of toplogical spaces. If \mathscr{F} is a sheaf on X, we define f_*\mathscr{F}(U)=\mathscr{F}(f^{-1}(U)). We call this the direct image sheaf. We can also define the inverse image sheaf, which is the sheaf associated to the presheaf which, if given \mathscr{G} on Y, assigns U\mapsto \lim_{V\supseteq f(U)} \mathscr{G}(V) taken over sets containing f(U). If you don’t know limits, check out John Armstrong’s definition, but generally the inverse image sheaf isn’t very helpful, later we’ll discuss a more useful variant of it.

About Charles Siegel

Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
This entry was posted in AG From the Beginning, Algebraic Geometry. Bookmark the permalink.

9 Responses to Morphisms of Sheaves

  1. Alberto says:

    In the statement of the theorem I guess you mean $latex\psi:\mathscr{F}^+\to \mathscr{G}$…

  2. Charles says:

    Yeah, that was a typo. Corrected.

  3. Anonymous says:

    Inverse image sheaf definition should be corrected to the sheafification of the presheaf you define; The presheaf itself isn’t necceserily a sheaf.

  4. Ah, yes, sorry. Corrected.

  5. Pingback: Direct Image Sheaves Under Proper Maps « Rigorous Trivialities

  6. Anonymous says:

    A morphism of sheaves \phi is injective if and only if the homomorphism \phi(U) is injective for all U open. Now, if the homomorphism \phi(U) is surjective for all U open then the morphism \phi is surjective, why is false the inverse?

  7. The image of a morphism of presheaves where the domain and codomain are sheaves is not necessarily a sheaf. So you can hit a subpresheaf that isn’t a sheaf, and so we have to sheafify, and if the sheafification is everything, then it will be surjective. To get this, the proper condition is surjective at every stalk.

  8. mlbaker says:

    A morphism of sheaves is an abelian group homomorphism \phi(U) : \mathscr{F}(U) \to \mathscr{G}(U) for each open set U, not \phi(U) : \mathscr{F} \to \mathscr{G}

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