1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Your proof looks correct. $\endgroup$ – Hanul Jeon Oct 4 at 13:10
  • $\begingroup$ Thank you so much! $\endgroup$ – user661541 Oct 4 at 13:10
2
$\begingroup$

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 $g(b')=c$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.