About irreducible morphisms I have asked the following question in Mathematics stack: https://math.stackexchange.com/questions/2202032/about-irreducible-morphisms. But there is no response, so I repost it here.
A morphism $f: X\to Y$ in mod A is called irreducible if 


*

*f is not a section,

*f is not a retraction,

*and whenever $f = gh $ for some morphisms $h: X \to Z$ and $g: Z \to Y$, then either $h$ is a section or $g$ is a retraction.
Now I am reading Schiffler's book "Quiver Representation", 

Proposition 7.4 (2)  states the fact that if $f: X\to Y$ is an irreducible morphsim, then $f$ admits no nontrivial factorization.
I have a questions:


*

*In the proof of Proposition 7.4 (2), $f$ is injective, then $f$ is not surjective, (Proposition 7.4 (1)) and thus $h$ cannot be a retraction. I don't know why $h$ is not a retraction.

 A: I have already posted this answer on stack exchange but I will repost it here. I am also reading Schiffler's Book on an independent study basis. I had a similar question and I think I may be able to answer one of the proposer's. I am new not only to Overflow but also to the subject, so if anyone sees an error or knows of a simpler proof please do not hesitate to let me know.
$\textbf{Theorem:}$
Let $f:X \rightarrow Y$ be an irreducible morphism. Let $f = hg$ where $h:Z \rightarrow Y$ and $g:X \rightarrow Z$ are two morphishms. If $f$ is injective, thus not surjective, then $h$ can not be a retraction.
$\textbf{Another Theorem Used in Proof:}$
Suppose $f:N \rightarrow M$ is an injective $R$-module morphism and $f(N)$ is a direct summand of M. Then $f$ has a left inverse.
$\textbf{Proposed Proof:}$ 
Suppose $f=hg$ is as above and for a contradiction, that $h$ is a retraction. Then there exists a morphism $h':Y \rightarrow Z$ such that $hh' = 1_Y$, so $h$ is surjective. Since $hg$ is injective, we have that $g$ is injective but since $hg$ is not surjective, we have that $g$ is not surjective. Thus there exists a $z \in Z$ such that $z\in coker g$, so $coker g \neq \emptyset$. Then we can decompose $Z$ as $Z = g(X) \oplus coker g$. Further, since $h$ is surjective, we can decompose $Y$ as $Y = h(g(X)) \oplus h(coker g) = f(X) \oplus h(coker g)$. Since $f:X \rightarrow Y$ is an injective $A$-module morphism and $f(X)$ is a direct summand of $Y$, we conclude there exists a morphism $f':Y \rightarrow X$ that is a left inverse of $f$. We have, $f'f = 1_X$, thus $f$ is a section. This contradicts the assumption of the irreducibility of $f$. Therefore, if $f$ is injective, thus not surjective, then $h$ can not be a retraction.
