"Inverse" of an epimorphism with modules and direct sums

Question

Let $F$ be a free $R$-module, and $F = A \oplus B$. Suppose that there is a random epimorphism $\theta : M \rightarrow A$. Then $ \theta \oplus id_B : M \oplus B \rightarrow A \oplus B $ is an epimorphism to a free module, which is also projective, so it has a function $ \sigma: A \oplus B \rightarrow M \oplus B$ with $\sigma \bullet \theta \oplus id_B = id$. Then $\sigma(A) \subseteq M$.

Why is $\sigma(A) \subseteq M$ ?

Answer 1

$M \oplus 0 = \sigma ( \theta \oplus id_B(M \oplus 0)) = \sigma(A \oplus 0)$, as $\theta(M) = A$.

Answer 2

Consider $x\in A$; then $x=\theta(y)$, for some $y\in M$, so $$ \sigma(x,0)=\sigma(\theta(y),0)= \sigma\circ(\theta\oplus\mathit{id}_B)(y,0)=(y,0) $$