If $C = A \oplus B$ then $A \cong C/B$.

Question

I'm not sure if the following proof works.

Prop. Let $A,B,C$ be left $R$-modules such that $C = A \oplus B$. Then $A \cong C/B$.

Proof. Recall that since $C = A \oplus B$, $C/B \cong (A + B)/A \oplus B/B$. Then by the second isomorphism theorem,$$(A+B)/A \cong A/(A\cap B) = A/\{0\} = A.$$ Therefore $A \cong C/B$.

Answer 1

You have a s.e.s $$0\rightarrow B\overset{i}{\rightarrow} A\oplus B\overset{\pi}{\rightarrow} A\rightarrow 0$$ Where $i$ is the canonical inclusion, and $\pi$ is the canonical projection. Since $\text{im}(i)=\ker(\pi)$, and $\pi$ is surjective we have by the 1st isomorphism theorem $$A\cong\frac{A\oplus B}{\ker(\pi)}=\frac{A\oplus B}{\text{im}(i)}\cong\frac{A\oplus B}{B}=\frac{C}{B}$$

Answer 2

I'd go for the first isomorphism theorem:

Let $\pi:C\rightarrow A$ be the projection $\pi(a+b):=a$ with $a\in A$, $b\in B$. Then $\ker\pi=B$ with $\pi$ surjective, so by the first isomorphism theorem, $C/B=C/\ker\pi\cong$im$\pi=A$.

Answer 3

I think to adapt and clarify what you wrote:

$$\frac{C}{0\oplus B}=\frac{A\oplus B}{0\oplus B}\cong \frac{A\oplus 0}{(A\oplus 0)\cap (0\oplus B)}=\frac{A\oplus 0}{0\oplus 0}\cong A.$$

By the second isomorphism theorem. The thing on the far left is the same thing as $C/B$ when identifying $0\oplus B$ as a subgroup of $C$.