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.)
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*}