Exercise 1.3.34


Theorem: Equivalence of categories is an equivalence relation

Proof: Let , and be categories.

Clearly, the identity functor is an equivalence of categories, so equivalence is reflexive.

Suppose is an equivalence. Then we have and natural isomorphisms and . along with the natural isomorphisms and is an equivalence from to , so equivalence is symmetric.

Finally, suppose and are equivalences. We have and along with all of the following natural isomorphisms:

Consider , and . We need to show that is naturally isomorphic to . We need to define . In fact, we will write it as the vertical composition of and a new natural transformation . To find an expression for , consider the following diagram:

We can then horizontally compose the natural transformations here (the identity natural transformation on the left and right), to yield a natural transformation – abusing notation a bit. To be clear, we define .

Then, is a natural isomorphism from to . We proceed similarly to find . Consider:

so define the natural transformation . Then, let .

Finally, then, we have a pair of functors and along with and that make the functors into an equivalence of categories, and such an equivalence is transitive.