Let be a topological space, let be a point of , and let be any set. In the notation of Vakil, define the skyscraper sheaf supported at by the formula Here is the singleton set. We can also define such a sheaf in other categories (abelian groups, rings, etc.), replacing the singleton set by the appropriate terminal object.
If is a containement of open sets, then the restriction is given as follows:
- if , then is the identity on ;
- otherwise, is the unique map to .
For , we have , because if both sides are the unique map to , and otherwise both sides are the identity map on . Therefore we have a presheaf of sets on (and this could also be a presheaf of other objects as well, as long as there is a terminal object).
Suppose is an open cover of some open set in , and suppose is a collection of sections such that, for any , we have . If , then the unique element of is evidently the unique gluing of the sections. If , then there is some such that ; let be the gluing of the sections. Because restriction to is the identity, the gluing is clearly unique. We want to show that for any , we have . If , then it’s obviously true. If , then , and since with these restrictions being the identity, we find , as required. This was a lot of words to say a simple thing: we have a sheaf.
The “skyscraper” in the name is explained by the following fact:
The stalk of at a point is if is in the closure of , and is the singleton set otherwise.
That’s not very hard to show. Suppose that is not in the closure of . Then there exists some open set around that does not contain . Hence any germ in the stalk at can be represented by an element in the singleton set , which means all germs are equal and the stalk may be identified with . On the other hand, suppose that is in the closure of . This means all open sets which contain also contain . The stalk is a colimit, and now we’re saying it’s a colimit over a constant diagram (every object in the diagram is ). Therefore, the colimit is .
Note that we can argue more abstractly for the first case, when is not in the closure, in a way that shows the stalk is the terminal object in other categories (abelian groups, rings, etc). The stalk is a direct limit which is computed over the directed set of all opens containing , ordered by reverse inclusion. Recall that a directed set is a poset in which every pair of elements has an upper bound. A subset of a poset is said to be cofinal in if, for every , it is possible to find some such that . For instance, when is not contained in the closure of , the set of open neighborhoods of that do not contain is cofinal in the directed set of all open neighborhoods of , ordered by reverse inclusion. Note that any cofinal set in a directed set is also directed. One can show that the direct limit computed over a directed set is equal (or more precisely, isomorphic up to a unique canonical isomorphism) to the direct limit computed over the “smaller” cofinal set. In our example, this means the stalk at is in any category with such a terminal object, because the direct limit can be computed over the cofinal set of neighborhoods not containing , and that’s a constant diagram with all objects equal to .
From the previous discussion, skyscraper sheaves look like a skyscraper towering above a point, and this mental picture is accurate when the point is closed. When a point is not closed (such a situation happens frequently in algebraic geometry), there are some points “nearby” over which the stalk is also , so it looks like a city’s downtown more than a single skyscraper.
What About the Weird Notation?
The notation is weird, but it makes sense in light of the following construction. Let be a (continuous) map of topological spaces, and let be a sheaf on . We define the pushforward of along to be the sheaf defined by the equation where denotes the inverse image (or preimage) of (it’s more often written as but I prefer the notation with a star). Because is continuous, the inverse image of an open set is an open set, so the previous equation makes sense. Given an inclusion of open sets in , the restriction from to is defined by the equation This defines a presheaf, simply because itself is a presheaf: clearly the restriction from an open set to itself is the identity, and
The fact is a sheaf is also sufficient to make its pushforward a sheaf as well. Suppose is an open set in , and is an open cover of . For each , let and suppose further that, for any , we have . We want to show the existence of a unique section such that for each . Each section is an element of , and the fact these sections all agree on overlaps together with the fact means there exists a unique with the desired property. Notice that a key part of why the pushforward is a sheaf, is the fact the inverse image preserves both arbitrary unions and intersections (union is because we need the collection to be an open cover of ).
To make sense of the notation for skyscraper sheaves, we also need to talk about the constant sheaf. Let be any set. The constant sheaf associated to , denoted , is defined by labeling each open set with the set of functions that are locally constant (i.e. around each point of there exists some open set contained in on which the function is constant – this is the same as requiring the function to be constant on connected components of ). Restriction is the usual restriction of maps, which obviously respects the presheaf condition. The sheaf axiom is not hard to check either.
Back to skyscrapers. Let be the “inclusion map” which points to . We consider as a sheaf over the topological space . Let be an open set in . If , then is the unique point of , while on the other hand if then is the empty set. Hence we see the pushforward of the constant sheaf is isomorphic in some obvious sense to the skyscraper sheaf as defined earlier.