# Is a “subfunctor generated by x” really a subfunctor?

I am reading Freyd's Abelian Categories, and Essential Lemma 7.12 says:

Let $$\mathcal{A}$$ be an abelian category, and $$Ab$$ be the category of abelian groups. Let $$M \rightarrow E$$ be an essential extension in $$[\mathcal{A}, Ab]$$. If $$M$$ is a mono functor, then so is $$E$$.

Here is the first half of the proof from the book (this is the part I am concerned with):

Suppose $$E$$ is not mono, so there is a monic $$A' \rightarrow A$$ in $$\mathcal{A}$$ such that $$EA' \rightarrow EA$$ is not monic in $$Ab$$. There is $$0 \neq x \in EA'$$ with $$(EA' \rightarrow EA)(x) =0$$; we construct the subfunctor $$F \subset E$$ generated by $$x$$ as follows. (This is the construction I have a problem with.)

Define F on objects as $$F(B) = \{ y \in EB: \text{ there exists } A' \rightarrow B \text{ in } \mathcal{A} \text{ such that } (EA' \rightarrow EB)(x) = y\}$$, from which it follows that for $$B' \rightarrow B$$, $$(EB' \rightarrow EB)(FB') \subset FB.$$ Indeed, if $$y \in FB'$$ then there is $$A' \rightarrow B'$$ in $$\mathcal{A}$$ with $$(EA' \rightarrow EB')(x)=y$$. Then $$A' \rightarrow B' \rightarrow B$$ witnesses that $$(EB' \rightarrow EB)(y) \in FB$$.

So we may define $$F(B' \rightarrow B)$$ by restriction: $$F(B' \rightarrow B) = FB' \rightarrow FB, y \mapsto (EB' \rightarrow EB)(y).$$

$$F$$ is clearly a set-valued functor, but is seen to be a group-valued functor once it is established that $$FB$$ is a subgroup of $$EB$$, and this is indeed the case. ($$F$$ is the image of the transformation $$H^{A} \rightarrow E$$ such that $$\eta(1_A) = x$$.)

My question: by definition groups have at least one element, so how can $$F$$ be a group-valued functor, unless $$F(B)$$ is always a nonempty set? And I don't see why $$F(B)$$ should be nonempty. If $$\mathcal{A}$$ is $$Ab$$ and $$M$$ is any representable $$Hom(X,-)$$, then $$E$$ is exact, and in particular preserves initial objects. Then $$E(0) = \emptyset$$, so $$F(0) = \emptyset$$ can't be a group. What's going on?

The proof is fine, as I think is almost always the case with Freyd. We have $$0\in F(B)$$ because $$E(0:A'\to B)(x)=0$$. Your remark in the paragraph My question is confusing abelian groups with sets and also makes a false claim that every representable is exact. Representables are, rather, left exact; failure of right exactness leads to the theory of the functor $$\mathrm{Ext}$$. A very well-known counterexample to general exactness is $$\mathrm{Hom}(\mathbb{Z}/2,-)$$, which does not preserve the exact sequence $$0\to\mathbb{Z}\to \mathbb{Z}\to \mathbb{Z}/2\to 0$$.
In any case, a representable functor does preserve the initial object, since in abelian categories the initial and terminal objects coincide and are both the zero object! In particular, $$\emptyset$$ is not an abelian group, so is not the value of any representable functor in abelian groups.
• Thanks, the zero element of course, silly of me to forget it. Just to remark though, I wasn't claiming Hom was exact, I was assuming the truth of the lemma to say that since Hom is mono, so is $E$. But $E$ is injective mono and hence exact. But you're right my 'counterexample' confuses Set and Ab (it was 4am). Thanks! – SSF Dec 16 '18 at 1:03