**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 .