We defined the pullback functor

We had a question of whether is a sheaf without sheafifying? No issues arose last time because

because sheafification is the left adjoint to the forgetful functor. So even though we care about the RHS when proving the adjunctionadjunction last time, we can just work with the LHS, which is easier because we have a more explicit description of the presheaf version: the isomorphism provided last time is described in terms of .

In general, the answer to whether is a sheaf without sheafifying is no by the following example.

Let and consider . What is a sheaf (of abelian groups) on ? It is just an abelian group: . Take some and let . Then for open non-empty, we have

and hence

Therefore, the pullback of by is the constant presheaf (except maybe for ). This is not a sheaf. In particular, given disjoint open sets and , taking and , there is no gluing condition as . So we obtain . This is not satisfied by the constant sheaf.

Consider

We want to compare vs. .

Let’s first see just what happens on an open set. Given and , we have

Therefore .

By the adjunction,

Therefore by Yoneda.

Let for a topological space . Then there is an inclusion map with . Hence, the pullback gives

Let . The stalk at is

There is a map for all because is defined as a direct limit. Given a morphism of sheaves , we get induced maps for all because is a functor. So taking stalks at is a functor.

Let be a topological space. Let , be sheaves on . Let be a morphism. Then is an isomorphism if and only if is an isomorphism for all .

This direction is trivial.

We need to show that is an isomorphism for all open, i.e., is injective and surjective.

Let such that . We have for all

Therefore by Fact about $ varinjlim$Fact about , for all , there exists an open such that . The collection is an open cover of . By one of the sheaf axioms, (locally globally ).

Let . Then we have for all

By Fact about $ varinjlim$Fact about , for all , there exists for some such that . We now consider . Since

then there exists such that . Consider which is an open cover of with the family . We claim that these glue to give some element . We need to check that

But this is true because both sides of the equality on the right are by . Now, because this equation holds in for all .

Zero-ness is detected at a finite point. (Atiyah-McDonald Exercise ???)

is exact () if and only if

is exact for all .

is exact.

We have a commutative diagram

Therefore

So we have an isomorphism . Now given that

is exact, then by Exercise (Criterion for exactness)Exercise (Criterion for exactness),

is exact for all . This is equivalent to

being exact. Using Exercise (Criterion for exactness)Exercise (Criterion for exactness) again in the other direction,

is exact.

sends injectives to injectives.

It has an exact left adjoint (c.f. same fact for ).

has enough injectives, i.e., every sheaf can be embedded into an injective one.

This is the same as the proof that has enough injectives.

Consider the function

where as a function of sets. If , consider . We embed an injective, and get by adjunction.

Things to check:

In the points don’t “talk to each other”. Let . For all , is open. So . The datum of for all completely and uniquely determines : is an open cover of and all intersections are empty. So for , we can put it into an injective by picking where is injective in for all . Then the sheaf built from is an injective element of .