Exercise 5.1.38

Theorem: Let be category having all products and equalizers. Then has all limits.

Proof: Let be a diagram. We can define the following objects in :

Now we define maps . is a product, and as such it comes with its projection maps . Likewise comes equipped with projections for each map . For any given , define a map by . Define by . Finally, let be the unique morphism such that for all morphisms , and similarly for .

Now, let be the equalizer of and . Write . We claim that is precisely a limit cone over the whole diagram .

First, let’s verify that it is in fact a cone over . Given we require , or . Indeed, observe that , and then since is an equalizer. Finally, , and , giving the desired equality.

To show it unique, let be an arbitrary cone over . There is a unique map such that . We claim that . Indeed, consider that for some , we have . Meanwhile, . Since for all , we have . Then since equalizes and , we have a unique map with . This is the desired unique map, making the limit of the diagram .