Verification on Solution for a Split Exact Sequence Question Question [Hungerford p.180 (III $\S$ 1) #15]: Let $R$ be any ring If $f:A\to B$ and $g:B\to A$ are $R-$module homomorphisms s.t. $gf=1_A$, then $B = \operatorname{im}(f)\bigoplus \ker(g)$.
I am assuming that by $=$ here, Hungerford means isomorphism (I know this is commonplace, but I am just asking, because I don't really know that this iso would be canonical, at least not obviously so).
Attempted Solution:
$0\to \ker(g)\xrightarrow{i}B\xrightarrow{g}\operatorname{im}(g)\to 0$
(where $i$ is inclusion) is a SES in $\mathit{RMod}$ and $\operatorname{im}(g)$ is an $R$-submodule of $A$. let $\widehat{f}:= f\mid_{\operatorname{im}(g)}$. Then $g\widehat{f} = 1_{\operatorname{im}(g)}$. The SES thus splits and we are done.
$\blacksquare$
I am having the "this is too easy/simple so I am missing something" kind of feeling right now, I don't know about exactly what though.
 A: Given a split short exact sequence
$$
0 \longrightarrow N \stackrel i\longrightarrow M \stackrel p\longrightarrow Q \longrightarrow 0,
$$
the splitting lemma tells us that the submodule $S := \ker p = \operatorname{im} i$ of $M$ has a complement, i.e., that there is a submodule $S'$ of  $M$ such that $M$ is the internal direct sum of $S$ and $S'$, so we can indeed write the equality $M = S \oplus S’$.
In fact, we can set $S' := \operatorname{im} j$ if $j : Q \to M$ is a morphism such that $pj = \operatorname{id}$, or $S' := \ker r$ if $r : M \to N$ is a morphism such that $ri = \operatorname{id}$.
The two alternatives coincide.
Thus, in your problem, from $gf = \operatorname{id}$ we see that $f$ is injective and that $g$ is surjective, so we can form two split short exact sequences:
$$0 \longrightarrow \ker g \stackrel i\longrightarrow B \stackrel g\longrightarrow A \longrightarrow 0$$
where $i$ is the inclusion, and
$$0 \longrightarrow A \stackrel f\longrightarrow B \stackrel \pi\longrightarrow B/\operatorname{im} f \longrightarrow 0$$
where $\pi$ is the canonical projection. Then just pick the one that you likes more, and the splitting lemma tells you that $B$ is the internal direct sum of $\operatorname{im} f$ and $\ker g$.
A: Hint:
To prove equality, check, that for any $b\in B$, we have $b-fg(b)\in\ker g$. This will ensure that $B=\operatorname{Im f}+\ker g$.
Next, verify that $\;\operatorname{Im f}\cap\ker g=\{0\}$.
