Exercise A0.III.7.06

Proposition

Let $\varphi$ be a homomorphism of $R$-modules. Then the short exact sequence

$$0 \longrightarrow \ker \varphi \longrightarrow M \overset{\varphi}\longrightarrow N \longrightarrow 0$$

splits if and only of $\varphi$ has a right inverse.

Proof

Suppose the sequence splits. Then $M$ can be identified with the direct sum $\ker \varphi \oplus N,$ with $\varphi$ being a projection. Then the inclusion $N \to \ker \varphi \oplus N$ is a right inverse of $\varphi.$

Suppose instead that $\varphi$ has a right inverse $\varphi^{r}$. Then we can contemplate the morphism $\rho = \varphi^{r} \circ \varphi.$ Certainly

$$\rho \circ \rho = \varphi^{r} \circ \varphi \circ \varphi^{r} \circ \varphi = \varphi^{r} \circ \varphi,$$

so $\rho$ is idempotent.

By 6.3 this gives us $M \cong \im \rho \oplus \ker \rho.$ Since $\ker \rho = \ker \varphi$ (since a right inverse cannot kill further elements, not killed by $\varphi$).

If we restrict $\varphi$ to $\im \rho,$ then $\varphi$ is an isomorphism. This is because $\varphi$ is already surjective, and selecting a unique preimage via $\varphi^{r}$ for each element of $N$ forces injectivity, as these preimages live in $\im \rho$ by construction. So $\im \rho \cong N.$

$\blacksquare$