It’s been a couple of weeks since I’ve posted anything here. I’ve been trying to understand homotopy theory, especially the modern kind with model categories. The second semester of my algebraic topology course is slated to cover that, to which I am looking forward. Right now, we are learning about spectral sequences. I have also been trying to understand Tate’s thesis, unsuccessfully.

Today, I’d like to prove a fairly nontrivial result, due to Freyd, following MacLane; this is a post that, actually, I take from a recent change I made to the CRing project. This gives a sufficient condition for the existence of initial objects.

Let be a category. Then we recall that if for each , there is a *unique* . Let us consider the weaker condition that for each , there exists *a* map .

Definition 1Suppose has equalizers. If is such that for each , then is calledweakly initial.

We now want to get an initial object from a weakly initial object. To do this, note first that if is weakly initial and is any object with a morphism , then is weakly initial too. So we are going to take our initial object to be a very small subobject of . It is going to be so small as to guarantee the uniqueness condition of an initial object. To make it small, we equalize all endomorphisms.

Proposition 2If is a weakly initial object in , then the equalizer of all endomorphisms is initial for .

*Proof:* Let be this equalizer; it is endowed with a morphism . Then let us recall what this means. For any two endomorphisms , the two pull-backs are equal. Moreover, if is a morphism that has this property, then factors uniquely through .

Now is a morphism, so by the remarks above, is weakly initial: to each , there exists a morphism . However, we need to show that it is unique.

So suppose given two maps . We are going to show that they are equal. If not, consider their equalizer . Then we have a morphism such that the postcompositions with are equal. But by weak initialness, there is a map ; thus we get a composite

We claim that this is a *retraction* of the embedding . This will prove the result. Indeed, we will have constructed a retraction , and since it factors through , the two maps

are equal. Thus, composing each of these with the inclusion shows that were equal in the first place.

Thus we are reduced to proving:

Lemma 3Let be an object of a category . Let be the equalizer of all endomorphisms of . Then any morphism is a retraction of the inclusion .

*Proof:* Consider the canonical inclusion . We are given some map ; we must show that . Indeed, consider the composition

Now equalizes endomorphisms of ; in particular, this composition is the same as

that is, it equals . So the map has the property that as maps . But being a monomorphism, it follows that .

Theorem 4 (Freyd)Let be a complete category. Then has an initial object if and only if the followingsolution set condition holds:there is a set of objects in such that any can be mapped into by one of these.

The idea is that the family is somehow weakly universal *together.* *Proof:* If has an initial object, we may just consider that as the family : we can hom out (uniquely!) from a universal object into anything, or in other words a universal object is weakly universal.

Suppose we have a “weakly universal family” . Then the product is weakly universal. Indeed, if , choose some and a morphism by the hypothesis. Then this map composed with the projection from the product gives a map .

February 6, 2011 at 8:49 pm

[…] blogged about it here, and added it to the CRing project here. Basically, the point is that complete categories are prone […]

June 16, 2012 at 1:16 am

For what it’s worth, I think you mean “retract” where you say “section”.

June 16, 2012 at 6:01 am

Fixed, thanks.