Exercise 5.1.35


Take a commutative diagram in some category:

such that the right square is a pullback.

Theorem: The left square is a pullback if and only if the outer square is a pullback.

Proof: First, assume the outer square is a pullback. Let’s label things:

So we have that is a pullback over , and is a pullback over . Take an arbitrary cone over . We need to produce a unique map (dotted) so that

commutes. Notice that is a cone over . So we have a unique making:

commute. We claim that is exactly the unique map that makes (2) commute. The left triangle would commute by (3), so we need only show the top triangle of (2) – that is .

To get more information, consider that is a cone over . So we have a unique making:

commute.

Note that is another map from . We claim that if we replace by this map in (4), the diagram still commutes. To see this, note that the top triangle becomes that of (3) written slightly differently. For the left triangle, we need to verify . Note that by (1), and by (3), so .

Similarly, we can replace with in (4) and the diagram still commutes. The top triangle commutes trivially. The left triangle commutes because it is really the square:

which commutes precisely because is a cone over .

Finally then, since is unique we have . So the top of (2) commutes. So, (2) commutes if is the dotted map, and is unique, so the left square is a pullback.


Now, suppose that both the left and right square are pullbacks. With the same labeling, we are now assuming that is a pullback over , and is a pullback over . We wish to show the outer square is a pullback, so let be an arbitrary cone over . We need to produce a unique map (dotted) so that:

commutes. Since is a cone over , we have a unique making:

commute. Now, we have that is a cone over . So we have a unique making:

commute. But then the whole diagram:

commutes. Of course, (5) is just a subdiagram of this, so taking the (unique) map makes (5) commute, and the outer rectangle is a pullback.