I found this example (Example 2.3 in Facchini's paper Krull-Schmidt Fails for Serial Modules) and couldn't quite understand the proof. The statement goes like this:
Let $U$ be non-zero uniserial right $R$-module. Then $U$ is a module such that End($U$) is local if at least one of the following conditions holds: (a) $U$ is projective; (b) $U$ is injective; (c) $U$ is artinian; (d) $U$ is noetherian; (e) $R$ is commutative; (f) $R$ is right noetherian.
I am particularly interested in the proof of (e) and (f). The proof is indeed short and it goes like this:
$Proof.$ (e) and (f). If a uniserial module $U$ is such that End($U$) is not local, then it has a surjective non-injective endomorphism $f$ and an injective non-surjective endomorphism $g$. These endomorphisms induce two isomorphisms $F: U/ ker(f)\to U$ and $G : U \to g(U)$. It follows that $g(U)/g(ker(f))\cong U$, so that $U$ is a shrinkable module in the sense that it is isomorphic to a proper submodule of a proper quotient of itself. But since $U$ being a uniserial right module over a right Noetherian ring or a commutative ring is supposed to be nonshrinkable (this fact can be found in the paper Uniserial modules: sums and isomorphisms of subquotients, Corollary 3). This yields a contradiction.
Question #1: For the existence of $f$ and $g$: Because End($U$) is not local, for any $\varphi \in$End($U$), we know that $\varphi$ and $Id - \varphi$ are not units, i.e. not invertible. Am I right? But how do we know that $f$ is surjective only and $g$ is injective only?
Question #2: Why is $g(U)/g(ker(f))\cong U$ true? I feel like there is some universal property going on here, or is it simply because $g$ is injective so $g(U/ker(f)) = g(U)/g(ker(f))$?
Any help would be awesome! Thanks.