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?

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?