# Show splitting lemma for short exact sequences

Consider a short exact sequence of left $$R$$-modules $$\DeclareMathOperator{\id}{id}$$

$$0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0$$

I want to show the following that the following statements are equivalent:

(1) The sequence above splits, i.e. $$f(A)$$ is a direct summand of $$B$$, i.e. there is a submodule $$B' \leq B$$ such that $$B = f(A) \oplus B'.$$

(2) There is a module homomorphism $$\alpha: B \to A$$ such that $$\alpha \circ f = \id_A.$$

(3) There is a module homomorphism $$\beta: C \to B$$ such that $$g \circ \beta = \id_C$$.

Here is my attempt for $$(1) \implies (2),(3)$$. Is this correct?

Assume (1) holds with $$B = f(A) \oplus B'$$. Define

$$\alpha: B \to A: b = f(a) + b' \mapsto a$$

Because the sum is direct and $$f$$ is injective, $$\alpha$$ is well defined. For $$a \in A$$, we have

$$\alpha \circ f (a) = \alpha(f(a)) = \alpha(f(a)+0) = a$$

and thus $$\alpha \circ f = \id_A$$. This shows $$(2)$$.

Define $$\beta: C \to B$$ in the following way.

Given $$c \in C$$, we can choose $$b \in B$$ such that $$g(b) = c$$ and we can decompose $$b$$ uniquely as $$b = f(a) + b'$$. We then define $$\beta(c) := b'.$$

Perhaps more clearly, $$\beta: C \to B$$ is defined by $$\beta(g(b' +f(a)) = b'.$$

This is well defined:

Assume $$c= g(b_1) =g(b_2)$$ with $$b_1 = f(a_1) + b_1', b_2 = f(a_2) + b_2'.$$ Then

$$b_1 - b_2 \in \ker g = f(A)$$

then $$b_1 -b_2 = f(a_1) + b_1'-f(a_1) - b_2' \in f(A)$$$$\implies b_1' - b_2' \in B' \cap f(A) = 0 \implies b_1' = b_2'$$

Now, let $$c \in C$$. Choose $$b = b' + f(a)\in B$$. Then $$g(\beta(c)) = g(b') = g(b'+f(a)) = g(b) = c$$ since $$f(a) \in \ker g$$. Hence, $$g \circ \beta = \id_C$$ and $$(3)$$ follows.

Is this correct?

• Your proof looks correct. – Hanul Jeon Oct 4 at 13:10
• Thank you so much! – user661541 Oct 4 at 13:10

In the proof that any $$b$$ such that $$g(b)=c$$ has a decomposition $$b=b'+f(a)$$, where $$b'\in B'$$, add the remark that $$b'$$ is unique, in the sense that it is independent of $$b$$, and that $$g(b')=c$$.
• Done,$\phantom i$ mylord! – Bernard Oct 4 at 13:42