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?

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

This is fundamentally correct for me. However, you didn't prove that α and β are homomorphisms. Also, I would make the proof of (3) a bit clearer, along the following lines:

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 $g(b')=c$.

  • $\begingroup$ Thank you. I realised that I didn't include the proof that they are homomorphisms, but this is rather easy and the proof became a bit lengthy so I didn't include it. $\endgroup$ – user661541 Oct 4 at 13:37
  • $\begingroup$ OK, I'll remove it (it was rather an observation than a critic). $\endgroup$ – Bernard Oct 4 at 13:38
  • $\begingroup$ No worries, leave it. Other readers can find this useful. $\endgroup$ – user661541 Oct 4 at 13:39
  • 1
    $\begingroup$ Done,$\phantom i$ mylord! $\endgroup$ – Bernard Oct 4 at 13:42
  • 1
    $\begingroup$ You're welcome. Always glad to help! $\endgroup$ – Bernard Oct 4 at 13:54

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