Is a "subfunctor generated by x" really a subfunctor?

Question

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?

Answer 1

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.