On the image of a module homomorphism being direct summand

Question

Let $f: A \to B$ be a ring homomorphism of commutative rings . Let $M$ be a $B$-module. Let $M_A$ be the $A$-module structure on $M$ defined by $a.m:=f(a)m,\forall a\in A, m\in M$ . Consider the $B$-module $N:=B \otimes_A M_A$ . Consider the $B$-module homomorphism $g: M \to N$ as $g(m)=1 \otimes m$ .

Then how to show that $Im g=g(M)$ is a direct summand of $N$ ?

Answer 1

Define a retraction $N\to g(M)$ sending $b\otimes m$ to $1\otimes bm$.