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 ${\mathcal{C}}$ be a category. Then we recall that ${A \in \mathcal{C}}$ if for each ${X \in \mathcal{C}}$, there is a unique ${A \rightarrow X}$. Let us consider the weaker condition that for each ${ X \in \mathcal{C}}$, there exists a map ${A \rightarrow X}$.

Definition 1 Suppose ${\mathcal{C}}$ has equalizers. If ${A \in \mathcal{C}}$ is such that ${\hom_{\mathcal{C}}(A, X) \neq \emptyset}$ for each ${X \in \mathcal{C}}$, then ${X}$ is called weakly initial.

We now want to get an initial object from a weakly initial object. To do this, note first that if ${A}$ is weakly initial and ${B}$ is any object with a morphism ${B \rightarrow A}$, then ${B}$ is weakly initial too. So we are going to take our initial object to be a very small subobject of ${A}$. 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 2 If ${A}$ is a weakly initial object in ${\mathcal{C}}$, then the equalizer of all endomorphisms ${A \rightarrow A}$ is initial for ${\mathcal{C}}$.

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

Now ${A' \rightarrow A}$ is a morphism, so by the remarks above, ${A'}$ is weakly initial: to each ${X \in \mathcal{C}}$, there exists a morphism ${A' \rightarrow X}$. However, we need to show that it is unique.

So suppose given two maps ${f,g: A' \rightrightarrows X}$. We are going to show that they are equal. If not, consider their equalizer ${O}$. Then we have a morphism ${O \rightarrow A'}$ such that the postcompositions with ${f,g}$ are equal. But by weak initialness, there is a map ${A \rightarrow O}$; thus we get a composite $\displaystyle A \rightarrow O \rightarrow A'.$

We claim that this is a retraction of the embedding ${A'\rightarrow A}$. This will prove the result. Indeed, we will have constructed a retraction ${A \rightarrow A'}$, and since it factors through ${O}$, the two maps $\displaystyle A \rightarrow O \rightarrow A' \rightrightarrows X$

are equal. Thus, composing each of these with the inclusion ${A' \rightarrow A}$ shows that ${f,g}$ were equal in the first place.

Thus we are reduced to proving:

Lemma 3 Let ${A}$ be an object of a category ${\mathcal{C}}$. Let ${A'}$ be the equalizer of all endomorphisms of ${A}$. Then any morphism ${A \rightarrow A'}$ is a retraction of the inclusion ${A' \rightarrow A}$.

Proof: Consider the canonical inclusion ${i: A' \rightarrow A}$. We are given some map ${s: A \rightarrow A'}$; we must show that ${si = 1_{A'}}$. Indeed, consider the composition $\displaystyle A' \stackrel{i}{\rightarrow} A \stackrel{s}{\rightarrow} A' \stackrel{i}{\rightarrow} A .$

Now ${i}$ equalizes endomorphisms of ${A}$; in particular, this composition is the same as $\displaystyle A' \stackrel{i}{\rightarrow} A \stackrel{\mathrm{id}}{\rightarrow} A;$

that is, it equals ${i}$. So the map ${si: A' \rightarrow A}$ has the property that ${isi = i}$ as maps ${A' \rightarrow A}$. But ${i}$ being a monomorphism, it follows that ${si = 1_{A'}}$.

Theorem 4 (Freyd) Let ${\mathcal{C}}$ be a complete category. Then ${\mathcal{C}}$ has an initial object if and only if the following solution set condition holds: there is a set ${\left\{X_i, i \in I\right\}}$ of objects in ${\mathcal{C}}$ such that any ${X \in \mathcal{C}}$ can be mapped into by one of these.

The idea is that the family ${\left\{X_i\right\}}$ is somehow weakly universal together. Proof: If ${\mathcal{C}}$ has an initial object, we may just consider that as the family ${\left\{X_i\right\}}$: 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” ${\left\{X_i\right\}}$. Then the product ${\prod X_i}$ is weakly universal. Indeed, if ${X \in \mathcal{C}}$, choose some ${i'}$ and a morphism ${X_{i'} \rightarrow X}$ by the hypothesis. Then this map composed with the projection from the product gives a map ${\prod X_i \rightarrow X_{i'} \rightarrow X}$.