Please help me to prove the following result:
Let $R$ be a ring with $1$ and $M$ a uniserial left $R-$module and $\alpha,\beta\in End_R(M)$.
$\alpha\beta$ is surjective if and only if both $\alpha$ and $\beta$ are surjective
This is what i wrote:
The implication if both $\alpha$ and $\beta$ are surjective then $\alpha\beta$ is surjective is evident since the composition of two surjective functions is surjective:
Let $\alpha:M \to M$ and $\beta:M \to M$ be surjective functions.
We wish to show that $\alpha\circ \beta:M \to M$ is also surjective.
Now, we know that for every $c \in M$, there is a $b \in M$ such that $c = \alpha(b)$, by surjectivity of $\alpha$. And we know that for that same $b$, there exists an $a \in M$ such that $b=\beta(a)$, by surjectivity of $\beta$.
So what we have shown is that for every $c \in M$, there is an $a \in M$ such that $c=\alpha(\beta(a))=\alpha \circ \beta(a)$, i.e., $\alpha \circ \beta$ is surjective.
For the other inmplication let us suppose that $\alpha\beta$ is surjective. Let $a\in M$ by sujectivity of $\alpha\beta$ there exist $b\in M$ such that $\alpha\beta(b)=a$ hence there exist $c=\beta(b)\in M$ such that $\alpha(c)=a$ then $\alpha$ is surjective
But i am not able to prove that $\beta$ is also surjective. Please help me to do so. Thanks in advance.
The key is to use the uniserial condition this way: since $M$ is uniserial, either $\ker(\alpha)\subseteq Im(\beta)$ or $Im(\beta)\subseteq \ker(\alpha)$. The latter case is obviously impossible when $M$ is nonzero and $\alpha\beta$ is surjective.
Now suppose $x\notin Im(\beta)$.
As you've already seen, there exists $b\in M$ such that $\alpha(\beta(b))=\alpha(x)$.
That means $\beta(b)-x\in \ker(\alpha)$.
So $\beta(b)-x\in\ker(\alpha)\subseteq Im(\beta)$. But this implies $x\in Im(\beta)$, a contradiction.
Therefore $\beta$ must be surjective.