A question on endomorphisms of an injective module

Question

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?

Answer 1

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

Answer 2

Define $$h':I/\ker(h)\rightarrow I, \ \ x+\ker(h)\mapsto h'(x+\ker(h))=h(x)$$ and $$f':I/\ker(h)\rightarrow I, \ \ x+\ker(h)\mapsto f'(x+\ker(h))=f(x)$$

$f'$ is well defined because of $\ker(h)\subseteq \ker(f)$, so $h'$ is. It is clear that $f'$ and $h'$ are right $R$-module homomorphism. Further $h'$ is monic. Since $I$ is injective, there exists some $g\in H$ suct that $f'=g\circ h'$. Actually $g$ is the element which you look for, that is, $f=g\circ h$ holds. It can be proved elementwise.