We know that if $M$ is an $R$-module and $N\subset M$ is a submodule then $M/N\simeq M'$ may not imply $M\simeq N\oplus M'$.In our module theory class,it has been proved that if $M/N\simeq R^k$ then $M\simeq N\oplus R^k$.This is in fact equivalent to saying that if $\exists$ a surjective module homomorphism $f:M\to R^k$ then $M\simeq Ker(f)\oplus R^k$.The proof uses the fact that $\exists $ a module homomorphism $g:R^k\to M$ such that $f\circ g=\text{Id}_{M'}$.Here $g$ is given by $g(a_1,a_2,...,a_k)=a_1x_1+...+a_kx_k$ where $x_i\in f^{-1}(e_i)$.So the conclusion of the theorem remains valid if $R^k$ is replaced by $M'$,such that $\exists g:M'\to M$ module homomorphism satisfying $f\circ g=\text{Id}_{M'}$.In this case,we say $f$ splits.
So,we have the following theorem:
Theorem:
If $f:M\to M'$ be an onto module homomorphism such that $f$ splits,then $M\simeq M'\oplus Ker(f)$.
Now,I am thinking about the converse of this statement:
Converse: If $f:M\to M'$ be an onto homomorphism such that $M\simeq M'\oplus Ker(f)$ then $f$ splits.
Can someone please help me to find out if the converse is true and how to prove it?It is easy to give an injective map from $M'\to M'\oplus Ker(f)$ which is the inclusion map and due to the isomorphism,we get an injective map $M'\to M$.But,I think that any injection may not work,I have to take some particular injection that will make $f\circ g=\text{Id}_{M'}$.Can someone please help me find the map?
As written, the converse is false:
For each $\newcommand{\Z}{\mathbb{Z}} a \in \Z$, we write $\overline{a}$ to denote the element $a+2\Z$ of $\Z/2\Z$. Consider the group homomorphism \begin{align} \newcommand{\N}{\mathbb{N}} f \colon \Z \oplus (\Z/2\Z)^\N & \to (\Z/2\Z)^\N, \\ (a,(x_1,x_2,\dots)) & \mapsto (\overline{a},x_1,x_2,\dots). \end{align} Obviously, $f$ is onto and its kernel is $2\Z \oplus \{(\overline0,\overline0,\dots)\} \cong 2\Z$, so $$ \Z \oplus (\Z/2\Z)^\N \cong 2\Z \oplus (\Z/2\Z)^\N \cong (\ker f) \oplus (\Z/2\Z)^\N. $$ Note that if $g \colon (\Z/2\Z)^\N \to \Z \oplus (\Z/2\Z)^\N$ is a group homomorphism, then $g(\overline1,\overline0,\overline0,\dots)$ must be an element of $\Z \oplus (\Z/2\Z)^\N$ of order $2$, hence $$ g(\overline1,\overline0,\overline0,\dots) = (0,(x_1,x_2,\dots)) $$ for some $(x_1,x_2,\dots) \in (\Z/2\Z)^\N$. So, if $f \circ g = \text{id}$, applying $f$ to the above equality yields that $(\overline1,\overline0,\overline0,\dots) = (\overline{0},x_1,x_2,\dots)$, which is absurd.
The above example is taken, essentially, from this answer.
But don’t worry, this can be fix easily. The thing is that if $f \colon M \to M’$ is an split $R$-module epimorphism (onto homomorphism), then there is not just an isomorphism $M \cong (\ker f) \oplus M’$, but an isomorphism $\varphi \colon (\ker f) \oplus M’ \to M$ making the diagram $$ \require{AMScd} \begin{CD} \ker f @>{\iota_1}>> (\ker f) \oplus M’ @>{\pi_2}>> M’ \\ @VV{\text{id}}V @VV{\varphi}V @VV{\text{id}}V \\ \ker f @>>i> M @>>f> M’ \end{CD} \tag{$*$} $$ commutative ($i$ is the inclusion map, $\iota_1$ is the canonical inclusion into the first factor, and $\pi_2$ is the projection onto the second factor). Namely, $$ \forall x \in \ker f, \forall y \in M’ \quad \varphi(x,y) = i(x)+g(y) $$ where $g \colon M’ \to M$ is such that $f \circ g = \text{id}$.
In this case, the converse holds. That is, if there is an isomorphism $\varphi \colon (\ker f) \oplus M’ \to M$ such that $(*)$ commutes, then $f$ splits: Just define $g \colon M’ \to M$ to be the composition $$ M’ \stackrel {\iota_2} \longrightarrow (\ker f) \oplus M’ \stackrel {\varphi} \longrightarrow M. $$