Is the following claim in "Elements of Representation Theory I" true?

Question

I looked at the proof of Theorem 1.13 of the book in the title some days ago and bumped into many gaps, which I thought I would be able to fill in. This happened in some other parts of the text, and eventually I managed to work it out.e Unfortunately, I was not able to do so this time, and I've run out of ideas. The authors take a short exact sequence

$$0 \longrightarrow L \stackrel f\longrightarrow M \stackrel g\longrightarrow N \longrightarrow 0$$

and assume that $f$ and $g$ are left (resp.) right almost split (this means $L\to M$ is not a section, and any non-section $L\to V$ factors through $f$, and dually for $g$). The authors now want to show that if $f$ and $g$ are both almost split and if $L$ and $N$ are both indecomposable, then $f$ and $g$ are left (resp.) right minimal: any endomorphism of $M$ that fixes $f$ on the left (resp. $g$ on the right) is an automorphism.

What is surprising is that the authors do not explain in any way how these four (!!!) conditions ($f$ left almost split, $g$ right almost split, $L$ and $M$ indecomposable) are used at all. They just implicitly seem to assume that if $h:M\to M$ fixes $f$ on the left, i.e. $hf= f$ then $h$ fixes $g$ on the right. Then, by the Three Lemma, $h$ is an automorphism.

Since we already have $1_L$ and $h$, there is a unique map $\alpha: N\longrightarrow N$ that gives us a morphism of short exact sequences from this sequence to itself (namely, pick a preimage $m$ of $n$, then send it to $gh(m)$). The authors then claim that this map is the identity of $N$, but I fail to see how this is true. Eventually it must be true that this map is at least an automorphism of $N$. Does anyone know how the authors arrive to this conclusion? If not, perhaps there is a way to salvage the proof and/or argument?

Edit: I've added a proof below, which in particular gives no indication that $\alpha$ must necessarily be the identity of $N$. So perhaps a more precise question is: is it really true that the diagram can be completed to an automorphism of the exact sequence itself using $h$, i.e. that if $h$ fixes $f$ on the left then it fixes $g$ on the right?

Answer 1

Proof: one can uniquely complete the diagram with a map $\alpha : N \longrightarrow N$. By the Snake Lemma, $\alpha$ is an automorphism if and only if $h$ is an automorphism, so assume $\alpha$ is not an automorphism.

Since $\alpha$ is not a retraction, because $N$ is indecomposable, and since $g$ is right almost split, we see that there exists $\beta : N \longrightarrow M$ such that $g\beta = \alpha$. Since $gh = \alpha g$, we see that $h - \beta g$ has image in the kernel of $g$, so for each $m\in M$ one has $h(m) - \beta g(m) = f s(m)$ for some unique $s(m) \in N$. But then $hf - \beta gf = f = fsf$, and since $f$ is injective we get $sf = 1_N$, which means $f$ is a section, which is a contradiction. Thus, $g$ right almost split and $N$ indecomposable implies that $f$ is left minimal.