Exercise 2.3.10


Theorem: Let be functors and , natural isomorphisms such that we have an equivalence of categories between and . Then is left adjoint to .

Proof: and are certainly natural transformations, but they may not satisfy the triangle identities. In particular, we may have that . Define a new natural transformation . We will let . Note that since and are natural isomorphisms, so is each member of the composition and thus as well. We claim that . Or, expanding out, that . Let be arbitrary and examine the component of the left-hand side at :

we want this to be equal to .

First, we write down some commutative diagrams that give some useful identities about the components of and that depend on the fact that they are isomorphisms. First, note that this square commutes:

since is is the statement of the naturality requirement for the natural isomorphism with respect to the arrow . This diagram states that . Since is an isomorphism, we can compose on the right by to get

Pull the dual trick with . That is, since this diagram commutes:

we can say that , and left-cancel to get:

Now, we write down another commutative diagram:

This one commutes because it is precisely the statement of the naturality for the natural isomorphism with respect to the arrow . Applying the identities we just derived (keeping in mind they also work for the inverses of and ), we can rewrite the right and bottom arrows.

Finally, note that the bottom arrow is an isomorphism and has an inverse . With this, simply attach an to the bottom of the diagram to get the final commuting diagram:

Following the diagram right, down, left, and down gives exactly the composite in question from earlier. But, following the left side straight down gives . Since the diagram commutes, we have the desired equality.

So, and are natural isomorphisms satisfying the triangle inequalities. So with unit and counit .