I was reading something in Munkres' book on algebraic topology and at some point I got confused.
He sais that if $B$ is a subgroup of $C$ and if $A$ is free, then the map :
\begin{equation} A \otimes B \to A \otimes C \end{equation} Induced by the inclusion is injective. But isn't this map always injective? Even if $A$ is not free?
In general the tensor product does not preserve injective maps. For example, consider $A=\mathbb Z/2\mathbb Z$, $C=\mathbb Z$ and $B=2\mathbb Z$. Then the inclusion induces the zero map $$\mathbb Z/2\mathbb Z\otimes 2\mathbb Z\to\mathbb Z/2\mathbb Z\otimes \mathbb Z$$ since $$\overline a\otimes 2b\mapsto\overline a\otimes 2b=\overline{2a}\otimes b=0\otimes b=0.$$ Here the bar means the class in the quotient $\mathbb Z/2\mathbb Z$. Notice that $A$ is not free. We use the term flat for those groups (in general $R$-modules) for which the tensor product is an exact functor. Thus free implies flat.