I'm working through Happel's book on triangulated categories, specifically the section on Auslander-Reiten theory. The part I'm having issues with is the proposition which states that the first two maps in an AR-triangle are irreducible.
Note: I'm reading morphism composition from right to left, not left to right as Happel does.
Let $X \xrightarrow{u} Y \xrightarrow{v} Z \xrightarrow{w} X[1]$ be an AR-triangle, then $u$ and $v$ are irreducible.
The definition for a map $u$ to be irreducible is:
A morphism $u$ is $\textbf{irreducible}$ if $u$ is neither a split epimorphism nor a split monomorphism, and for any factorization $u = u_2u_1$, either $u_1$ is a split monomorphism or $u_2$ is a split epimorphism.
Well, in the AR-triangle $X \xrightarrow{u} Y \xrightarrow{v} Z \xrightarrow{w} X[1]$, the map $u$ is by assumption not a split monomorphism. Furthermore, I can follow Happel's proof that the factorization property for irreducibility is satisfied by $u$, but I can't show that it's not a split epimorphism. Here's my attempt:
Suppose $u$ is a split epimorphism, then there is some map $u':Y\rightarrow X$ such that $uu' = 1_Y$. But then $vuu' = v1_Y = v = 0$ since $vu=0$. Shifting the triangle to $$Z[-1] \xrightarrow{-w[-1]} X \xrightarrow{u} Y \xrightarrow{0} Z$$ one sees that it splits, so $X \cong Z[-1]\oplus Y$. But $X$ is indecomposable, so one of those summands must be $0$. If $Z[-1]=0$, then $-w[-1]=0$ and $w=0$, which contradicts the assumptions on AR-triangles. So $Y=0$, and we have a new triangle, isomorphic to the first: $$X \xrightarrow{0} 0 \xrightarrow{0} Z \xrightarrow{w} X[1]$$
This is where I get stuck. I don't see how to get a contradiction on the AR-triangle axioms with this last triangle. Any help would be appreciated.