Consider the category . A functor is the following data:

A cone of is an object of that fits into the following commutative diagram:

The limit is a universal cone such that

So it is equivalent to the pullback or fibre product .

“Oppositely”, if we have a functor , a colimit of is a universal cocone, called a pushout or coproduct

Let be a small category (i.e., its collection of objects is a set). If -products in exist, then -shaped limits exist in whenever equalisers do. Dually, if -coproducts and coequalisers exist in , then -shaped colimits exist in .

Let be a functor. Since , then -products exist. So we can consider the diagram

The maps and are given by (uniquely determined by the universal property of products)

Now we have

Claim:

is a universal cone for .
Proof. We have a diagram

The diagram commutes because .

has all small limits and colimits.

Let be a category and be a small category. Let be a functor.

Let be a topological space. Then has all small limits and colimits, and they are computed “pointwise”, i.e.,

Let be a finite category (finite objects and morphisms), and let be a filtered category. Let be a (bi)functor. If limits and colimits exist in each argument, then the following map is bijective:

We begin with the maps

for . There are induced maps

and

In other words, taking a limit over , we get maps

So by universality, we have an induced map

Let be a topological space, and . Let be a basis at , i.e.,

We take (which is filtered) and consider the (bi)functor

for a given functor . That is, is the data of

But now, if we unravel what the statement is saying, we want:

Now recalling our definition of a stalk in terms of colimits, the above colimits are just talks. Hence, the statement is just saying that we should have a bijection

Very concretely, we want the stalk of the equaliser to be the equaliser of the stalks.

Note now that we are working in .We have

Certainly is injective. To show that the top map is injective, it suffices to show that the left map is injective.

We know that is a subpresheaf (the equaliser is a monomorphism). So it suffices to show the the following.
Claim: If as presheaves, then .
Proof. Consider with the same image in . Then there exists and such that . So . So in . □

Conversely, let . This means that , and there exists , such that has the same image in under and (equality in the stalk is detected on some open). So .