Exercise 2.3.11


Theorem: Let have a left-adjoint . Suppose that for at least one , has more than one element. Then for each , the unit is an injective map of sets.

Proof: We will first establish that, under the given hypothesis, must be faithful. To see this, suppose we have such that . The goal is to show that . Indeed, if , there exists some such that . Now, let be such that , and let be elements. Define with the property and . Then, clearly, . But, each have a transpose . In particular, that transpose can be given explicitly as:

and similarly . Since , their transposes are not equal either, and , which means . So is faithful.

We now show that faithfulness of implies the desired property of the unit. Fix an . Saying that is injective amounts to saying that, for any , implies . So, lets assume we have such a and with . Note that is an arrow in , and as such it has a transpose . Furthermore, this transpose can be calculated explicitly as:

where the last equality follows precisely due to the triangle identity, which says that . Similarly we obtain . Then, since , their transposes are equal, and we have . Since is faithful, we have . So, is injective.