Let $M$ be a semisimple $A$-module for some algebra $A$, and $N$ another $A$-module. Let $f:M\twoheadrightarrow N$ be an epimorphism. I can construct the following short exact sequence: $$0\rightarrow \ker(f) \hookrightarrow M \twoheadrightarrow N \rightarrow 0$$ Since $M$ is semisimple and $\ker(f)$ is a submodule of $M$, we have that $M=\ker(f)\oplus L$ for some other submodule $L$ of $M$. But apparently, I can take this $L$ to be $N$. Why is this?
EDIT 1 Similar questions have been asked before on the site, such as this one. However, the answer doesn't convince me (you can check my comment there to see why). Another approach to the problem would also be helpful. I feel that I don't see the full potential of the splitting lemma and to me it's not such a powerful tool, although I am sure that in reality it is. Any comments on this would also be highly appreciated.
EDIT 2 I also would like to know, if $f$ were not an epimorphism, would we be able to write $f=\ker(f)\oplus \text{im}(f)$?
I claim that the restriction of $f$ to $L$ is an isomorphism $L\to N$, so that $N$ is isomorphic to $L$. For injectivity, note that $\ker(f)\cap L=0$, so if $a\in L$ and $f(a)=0$ then $a=0$. For surjectivity, suppose $a\in N$. Since $f:M\to N$ is surjective, there exists $b\in M$ such that $f(b)=a$. Since $M=\ker(f)+L$, we can write $b=c+d$ for $c\in\ker(f)$ and $d\in L$. But then $f(b)=f(c)+f(d)=f(d)$, so $d$ is an element of $L$ such that $f(d)=a$. Thus the restriction of $f$ to $L$ is surjective.