How can I show $I\simeq L$ where $L$ is a direct summand of $X$ such that $L\subseteq \textrm{ker}(g)$?

Question

Consider the exact sequence of $R$-modules ($R$ is ring with identity $1_R$): $$ 0\longrightarrow K\oplus I\stackrel{f}{\longrightarrow} X\stackrel{g}{\longrightarrow} Y\longrightarrow 0,$$ where $I$ is an injective module. How can I show $I\simeq L$ where $L$ is a direct summand of $X$ such that $L\subseteq \textrm{ker}(g)$?

The only place I could use the hypothesis $I$ is injective is to get a map $v:X\longrightarrow I$ such that $vf=p_2$ where $p_2:K\oplus I\longrightarrow I$ is the projection. I don't know what else I can do.

Thanks.

Answer 1

You are almost there. There is a section $s:I\to K\oplus I$ such that $p_2s=id_I$. All you need to do is multiply the equality $vf=p_2$ on the right by $s$, to obtain $$v(fs)=id_I.$$

Thus $fs$ is a section whose image is in $\ker g$, and the claim is proved.