# How to prove $\ker \beta + \text{im} \tau = B$ in the definition of split short exact sequence?

Let $0 \xrightarrow{}A \xrightarrow{\alpha} B \xrightarrow{\beta} C \xrightarrow{} 0$, be a short exact sequences of module homomorphisms.

I understand that if there exists a hom $\tau : C \to B$ such that $\beta\tau = \text{id}_C$, then $\ker \beta \cap \text{im} \tau = \{0\}$. Proof: $b \in \ker \beta \cap \text{im} \tau \implies \beta(b) = 0 = \beta(\tau(c)) = c \implies b = \tau c = 0$.

Note, I'm trying to prove that $B \approx A \oplus C$ eventually.

But what is a simple method to prove that $\ker \beta + \text{im} \tau = B$? Also this must be proven for general $R$-modules of any dimension.

Take $b \in B$. Then $\tau(\beta(b)) \in \text{im }\tau$. Then $$\beta(b - \tau(\beta(b))) = \beta(b) - \beta\tau\beta(b) = 0$$ so $c = b - \tau(\beta(b)) \in \text{ker }\beta$, and $b = \tau(\beta(b)) + c$.
• So $\beta(b - \tau\beta(b)) = 0$ so $c = b - \tau\beta(b) \in \ker \beta$, and AHHH I GET IT !!! – BananaCats Author Oct 29 '15 at 21:46