A question on endomorphisms of an injective module This is a homework question I am to solve from TY Lam's book Lectures on Modules and Rings, Section 3, exercise 23.

Let $I$ be an injective right $R$-module where $R$ is some ring.
Let $H= \operatorname{End}(I),$ the endomorphisms on $I$.
I need to show that given $f,h \in H$, if $\ker h \subseteq \ker f$, then $f \in H\cdot h$. That is, there is some endomorphism $g \in H$ such that $g\circ h = f$.

I can see that if we have that $f$ is one to one, then this will force $\ker h = \{0\}$ so we will have that $h$ must be an injection from $I$ to $I$ with filler $g:I \rightarrow I$ such that $h \circ g = f$.
How would I go about handling the case where I do have $h$ guaranteed to be an injection?

Hint. $0\to\operatorname{Im}(h)\to I$ and define $\bar f:I/\ker h\to I$ by $\bar f(\bar x)=f(x)$.
