Verification on Solution for a Split Exact Sequence Question

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.

Answer 1

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 $\;\operatorname{Im f}\cap\ker g=\{0\}$.

Answer 2

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 $\ker g$.