On the Auslander--Reiten triangles and irreducible morphisms I'm reading "Triangulated categories in the Representation Theory of Finite Dimension Algebras" by Dieter Happel, but I don't understand the proof of Proposition 4.3 in Chapter 1. So, please ask you my question.
The claim is that if $X \to Y\to Z\to X[1]$ is an AR triangle, then the morphisms $X\to Y$ and $Y\to Z$ are irreducible.
But I don't understand why the morphism $X \to Y$ (resp. $Y \to Z$) is not retraction (resp. section). 
Why is the middle term $Y$ not zero?
 A: To answer the questions, let me recall the definition of an Auslander-Reiten triangle. The triangle
$$
X\xrightarrow{u} Y\xrightarrow{v} Z\xrightarrow{w} X[1] 
$$
is an Auslander-Reiten triangle iff 


*

*$X$ and $Z$ are indecomposable,

*$w$ is non-zero, and

*if $f:W\to Z$ is not a retraction, then there exists $f':W\to Y$ such that $f=v\circ f'$.


The answer to your first question lies in the second condition.  Indeed, $w$ is zero if and only if $u$ is a section.  Also, $w$ is zero if and only if $v$ is a retraction.  This is Lemma I.1.4 of the book by Happel mentioned in your question.  Therefore, in an Auslander-Reiten triangle as above, $u$ is never a section and $v$ is never a retraction.
The answer to your second question is that $Y$ can be zero in an Auslander-Reiten triangle.  Let $k$ be a field, and consider $\mathcal{D}:= \mathcal{D}^b(\operatorname{mod}k)$, the bounded derived category of the category of finite-dimensional vector spaces over $k$.  In this category, I denote by $k$ the complex with the vector space $k$ in degree $0$ and the zero space in other degrees. Then
$$
k[-1]\xrightarrow{} 0\xrightarrow{} k\xrightarrow{id_k} k 
$$
is an Auslander-Reiten triangle.  Indeed, conditions 1 and 2 are trivially satisfied, and since any non-zero morphism $W\to k$ is a retraction in $\mathcal{D}$, condition 3 is empty.
