Exercise 5.1.42


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.