$f:M \to Q$ a monomorphism with $Q$ injective and $g:M \to N$ an essential morphism, then there exists a mono $\bar{g}: N \to Q$ such $\bar{g} g= f$.

Question

This one is based on (A) of the answer of this previous question

For a projective cover $(\sigma, P)$ of a module $M$, $P$ is indecomposable implies $M$ is indecomposable

Let $Q$ be an injective module and $f:M \to Q$ a monomorphism. Prove that if $g:M \to N$ is an essential monomorphism, there exists a monomorphism $\bar{g}: N \to Q$ such $\bar{g} g= f$.

As $g: M \to N$ is essential monomorphism I got that $g(M)$ is an essential submodule of $N$ which means that for every submodule $N' \leq N$ such $g(M) \cap N' = \lbrace 0 \rbrace$ then $N'= \lbrace 0 \rbrace$. Also, as $g:M \to N$ is injective(essential monomorphism) with $Q$ injective module by hypothesis I got that there is a morphism $\bar{g}:N \to Q$ such $\bar{g} g= f$ still cannot prove that this $\bar{g}$ is injective

First lets prove $\ker(\bar{g}) \cap g(M)= \lbrace 0 \rbrace$. Let's take $x \in \ker(\bar{g}) \cap g(M)$. As $x \in g(M)$ there is some $x' \in M$ such $g(x')=x$ but as $x \in \ker(\bar{g})$ then $g \bar{g}(x')= g (x)=0$ so by commutativity $f(x')=0$, but as $f$ is injective by hypothesis $x'=0$. Then $x=g(x')=0$. So by the essential hypothesis $\ker(\bar{g})= \lbrace 0 \rbrace$ proving $\bar{g}$ injective.