On the image of a module homomorphism being direct summand

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 $N$ ?

• if you have an element $b\otimes m$ in $N$, there are very few thinga you can do to obtain an element of $M$... – Mariano Suárez-Álvarez Dec 3 '17 at 19:28

Define a retraction $N\to g(M)$ sending $b\otimes m$ to $1\otimes bm$.
• why is $bm \in g(M)$ ? – user Dec 3 '17 at 19:17