Compete abelian categories with projectieve generators are fully abelian.

Compete abelian categories with projectieve generators are fully abelian.

So this is my first time on stackexchange so if you need more detail from
me , please ask.
I was reading the book "Abelian Categories : An Introduction to the Theory
of Functors" by Peter Freyd , in which I was stuck on Theorem 4.44 as
follows
A complete abelian category with a projective generator is fully abelian.
Proof:
Let $\mathcal{A'}$ be a small full subcategory of a complete abelian
category $\mathcal{A}$, and $\bar{P}$ be a projective generator for
$\mathcal{A}$. For each $A\in\mathcal{A}$ we consider the epimorphism $$
\sum_{(\bar{P},A)}\bar{P} \to A.$$ By taking
$I=\cup_{A\in\mathcal{A'}}(\bar{P},A) $ and defining $P=\sum_I\bar{P}$, we
obtain a projective generator $P$ such that for each $A\in \mathcal{A'}$
there is an epimorphism $P\to A$.
Define $R$ to be the ring of endomorphisms of $P$. For every$A\in
\mathcal{A}$ , the abelian group $(P,A)$ has a canonical $R$-module
structure: for $P\xrightarrow{x}A\in(P,A)$ and $P\xrightarrow{r}P $ define
$rx\in(P,A)$ to be $P\xrightarrow{r}P\xrightarrow{x}A$.
Given a map $A\xrightarrow{y}B$ , the induced map
$(P,A)\xrightarrow{\bar{y}}(P,B)$ is an $R$-homomorphism ($\bar{y}(rx) =
P\xrightarrow{r}P\xrightarrow{x}A\xrightarrow{y}B = r(\bar{y}(x))$). We
define, therefore, $F:\mathcal{A}\to\mathcal{G}^R$($\mathcal{G}^R$ is the
category of $R$-modules) by $F(A)=(P,A)$ with the canonical $R$-module
structure. $F$ is an embedding since $P$ is a projective generator.
$F|_{\mathcal{A'}}$ is known to be an exact full embedding , therefore ,
once it is known to be full. Given $A,B\in\mathcal{A'}$ and a map
$F(A)\xrightarrow{\bar{y}}F(B)\in\mathcal{G}^R$ we wish to find a map
$A\xrightarrow{y}B\in\mathcal{A'}$ such that $F(y)=\bar{y}$. Let $0\to
K\to P\to A\to 0$ and $P\to B \to 0$ be exact sequences in $\mathcal{A}$.
Notice that $F(P)=R$. We obtain the commutative diagram in
$\mathcal{G}^R$:
\begin{array}{cccccccc} 0 & \to & F(K) & \to & R & \to & F(A) & \to & 0\\
& & & & \downarrow{f} & & \downarrow{\bar{y}} & & \\ & & & & R & \to &
F(B) & \to & 0 \end{array}
where the existence of the map $f$ is insured by the projectiveness of $R$
in $\mathcal{G}^R$. Since $R$ is a ring, any automorphism on $R$ must be
equivalent to multiplication on the right by an $R$-element. We assume
then that $f(s)=sr$ for all $s\in R$, where $P\xrightarrow{r}P\in R$.
Returning to $\mathcal{A}$, the diagram
\begin{array}{cccccccc} 0 & \to & K & \to & P & \to & A & \to & 0\\ & & &
& \downarrow r & & & & \\ & & & & P & \to & B & \to & 0 \end{array}
is such that $K\to P\xrightarrow{r}P\to B=0$, since $F(K)\to
R\xrightarrow{f}R\to F(B)=0$ and $F$ is an embedding . Hence there is a
map $A\xrightarrow B$ such that
\begin{array}{ccc} P & \to & A \\ r\downarrow & & \downarrow y\\ P & \to &
B \end{array} commutes.
Hence
\begin{array}{ccc} R & \to & F(A) \\ f\downarrow & & \downarrow \bar{y}\\
R & \to & F(B) \end{array} commutes.
and since $R\to F(A)$ is epimorphic , $F(y)=\bar{y}$.
So my questions are
after we got the existence of the map $f$ the author said that it was an
automorphism. I don't see why it should be an automorphism, though the
further proof still would go through if we assume it to be just a
endomorphism as an $R$-module.
I do not see why there should exist a map $A\xrightarrow{y}B$ from
previous arguments. Please explain in some clear terms.
Other than that the proof is fairly clear.