**Exercise 5.1.42**

Let

be a pullback square.

**Theorem**:
If is monic, so is .

*Proof*:
Let be maps with .
Since the square above commutes, we have , and thus .
Thus, is a cone over .
Thus, we have a unique with (1) , and (2) .
Taking trivially makes both statements true.
makes (1) true by supposition.
To see that it satisfies (2) as well, consider the following:
Since , it follows that .
But , and we have .
Finally, since is monic we have , so indeed makes (2) true.
Since is unique, we must have , so is monic.