Is $X\cong\textrm{Ker}(f)\oplus\textrm{Im}(f)$ for a module homomorphism $f:X\to Y$ with semisimple domain? Let $f:X\to Y$ be a module homomorphism with semisimple domain.
Does
\begin{equation*}
X\cong\textrm{Ker}(f)\oplus\textrm{Im}(f)
\end{equation*}
hold true that?
(In my previous question it was kindly pointed out to me that without the semisimple condition the answer here is "no", so basically I'm asking if $X$ being semisimple is a sufficient condition.)
 A: I believe the answer is "yes", and that this can be shown as follows.
$X$ is semisimple, so it is an internal direct sum of simple submodules:
\begin{equation*}
X
=
\bigoplus_{i\in I}X_i.
\end{equation*}
For each $i$ in $I$, $\textrm{Ker}(f)\cap X_i$ is a submodule of $X_i$, and because $X_i$ is simple it follows that $\textrm{Ker}(f)\cap X_i$ equals either $\{0\}$ or $X_i$, and so
\begin{equation*}
\textrm{Ker}(f)
=
\bigoplus_{i\in I:\textrm{Ker}(f)\cap X_i\neq\{0\}}X_i.
\end{equation*}
It follows that $X=V\oplus\textrm{Ker}(f)$, where
\begin{equation*}
V
=
\bigoplus_{i\in I:\textrm{Ker}(f)\cap X_i=\{0\}}X_i.
\end{equation*}
Since $X=V\oplus\textrm{Ker}(f)$, it satisfies
\begin{equation*}
X/\textrm{Ker}(f)
=
(V\oplus\textrm{Ker}(f))/\textrm{Ker}(f)
\cong
V.
\end{equation*}
(See e.g. Grillet's book, Abstract Algebra, p. 328.)
So
\begin{equation*}
X
=
V\oplus\textrm{Ker}(f)
\cong
(X/\textrm{Ker}(f))\oplus\textrm{Ker}(f).
\end{equation*}
The result now follows from the first isomorphism theorem which tells us that
\begin{equation*}
X/\textrm{Ker}(f)\cong\textrm{Im}(f).
\end{equation*}
