I am trying to show that if we have the left splitting short exact sequence of $R$-modules $$ 0 \longrightarrow M \xrightarrow{\enspace f \enspace} N \xrightarrow{\enspace g \enspace} S \rightarrow 0 \,, $$ then there exists an isomorphism of $R$-modules $\phi \colon N \to M \oplus S$.

I know that there is $\psi \colon N \to M$ such that $\psi \circ f = \mathrm{Id}_M$, and we have the epimorphism $g \colon N \to S$, so maybe the morphism $\beta \colon N \to M \oplus S$ defined as $\beta(n) = (\psi(n), g(n))$ could work ($\beta$ would be $\phi^{-1}$).

It is easy to show that $\beta$ is a module morphism. I am having some difficulty to prove it is injective and surjective, I would appreciate some help, maybe I’ve chosen the wrong morphism.


1 Answer 1


Your morphism is correct.

For the surjectivity, if you take any $(m,s)\in M\oplus S$, take $f(m)+s'-f\circ\psi(s')$ where $s'$ is any preimage of $s$ under $g$.

For injectivity, if $(\psi(n),g(n))=(0,0)$, then $n$ is in the image of $f$ because the sequence is exact, i.e., $n=f(m)$, but then $0=\psi (n)=\psi\circ f (m)=m$ and so $n=0$.


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