$M'\bigoplus M''\overset{\phi}\cong M\implies\phi(M'\bigoplus0)=M'$?

Question

Let $R$ be a commutative ring with unity, $M$ an $R$-module, $M',M''$ submodules of $M$, $M'\bigoplus M''=\{(m',m'':m'\in M', m''\in M''\}$, the direct sum of $M',M''$. If $M'\bigoplus M''\overset{\phi}\cong M$, then $\phi(M'\bigoplus0)\cong M'$

$\phi(M'\bigoplus0)$ is indeed $M'$, that is, $\phi(M'\bigoplus0)=M'$ as sets?

Answer 1

Yup!

$\phi \restriction M' \oplus 0$ is still an injection, so $\phi \restriction M' \oplus 0$ is a bijection from $M' \oplus 0$ to $\phi[M' \oplus 0]$ (By restricting the codomain, we make $\phi$ surjective, and thus bijective). But $M' \oplus 0 \cong M'$, and so

$$ M' \cong M' \oplus 0 \cong \phi(M' \oplus 0) $$

As desired.

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I hope this helps!