I’m currently working through Mac Lane’s and Moerdijk’s “Sheaves in Geometry and Logic”, and came upon the following two sentences in a proof (p. 43):

Note that when is representable the corresponding category of elements has a terminal object—the element of . Therefore the colimit of the composite will just be the value of on the terminal object.

It was not immediately obvious to me why the colimit of a functor should be the image of the terminal object, though I got the intuition that we automatically had a cocone. Sheldon Axler gives the following advice to students in the preface of his “Linear Algebra Done Right”:

You cannot expect to read mathematics the way you read a novel. If you zip through a page in less than an hour, you are probably going too fast.

Taking this advice to heart, I set out to prove:

Proposition. Let be a functor and a diagram with terminal object . Then the colimit exists and .

In other words, if a diagram has a terminal object, then the colimit is trivial: the colimit is the image of the terminal object.

Proof. Let be the natural transformation given by where is the unique morphism in given by the terminality of . Then this clearly defines a cocone from to .

Let be any other cocone on . Then there exists a morphism from to , namely, . Moreover, we claim this morphism makes the following diagram commute for all in :

Indeed, since the are , this is immediate from the naturality of . We claim is the unique such map. Let be any other morphism making the diagram commute. Then in particular, the diagram

commutes. But . So .

We thus conclude that the colimit exists and is exactly . Q.E.D.

Corollary. Let be a functor and a diagram with initial object . Then the limit exists and .