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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 , and I was stuck on Theorem 4.44 as follows

A complete abelian category with a projective generator is fully abelian.


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.


\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

  1. 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.
  2. 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.

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The category should be assumed to be cocomplete. The proof uses infinite direct sums. – Martin Brandenburg Aug 17 '13 at 7:34
Welcome to Math.SE :-) I suggest you choose a nickname (your real name or a fantasy name), so that other users may more easily recognize you later. – magma Aug 17 '13 at 8:42
up vote 2 down vote accepted
  1. You are right, it is just an endomorphism. And it is clear how the endomorphisms of $R$ look like.
  2. Use the universal property of cokernels.
share|cite|improve this answer
Thanks , I had forgotten about that part. – epsilon_0 Aug 17 '13 at 10:53
Also about the infinite direct sum existence , the book defines left-complete:infinite products exist , right-complete:infinite sums exist, complete as both left complete and right complete. So I did not see any discrepancy. This book is 47 years old though. I should get a newer edition – epsilon_0 Aug 17 '13 at 10:55
Alright, I didn't know this old terminology. In fact this isn't used anymore. But Freyd's book is a classic. – Martin Brandenburg Aug 17 '13 at 12:43

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