Proving that Induction Maps Projective modules to Projective modules. I'm trying to prove that that induction maps projective modules to projective modules but am getting a little stuck. I've read Module induced from projective is projective but I don't quite understand the details as he only talks about free modules can someone elaborate slightly on how this directly relates to projective modules?
 A: The general proof technique is the following: say that an object $P$ in a category, not necessarily abelian, is projective if $\text{Hom}(P, -)$ preserves epimorphisms. (There are several definitions of projectivity which are equivalent in abelian categories but not in general; this one turns out to make the statement I want true.)

Lemma: Let $F : C \to D$ and $G : D \to C$ be an adjoint pair, with $F$ the left adjoint and $G$ the right adjoint. If $G$ preserves epimorphisms, then $F$ preserves projective objects.

Proof. Suppose $P \in C$ is a projective object. Then
$$\text{Hom}(F(P), -) \cong \text{Hom}(P, G(-))$$
and by hypothesis both $G(-)$ and $\text{Hom}(P, -)$ preserve epimorphisms, hence so does $\text{Hom}(P, G(-))$. $\Box$

Corollary: Free modules are projective.

Proof. The free module functor $F : \text{Set} \to \text{Mod}(R)$ is left adjoint to the forgetful functor $G : \text{Mod}(R) \to \text{Set}$, which preserves epimorphisms because epimorphisms in $\text{Mod}(R)$ are quotient maps and quotient maps are surjective on underlying sets. $\Box$

Corollary: If $f : R \to S$ is a ring homomorphism then the extension of scalars / induction functor $(-) \otimes_R S : \text{Mod}(R) \to \text{Mod}(S)$ preserves projective objects.

Proof. Extension of scalars is left adjoint to restriction of scalars, which preserves epimorphisms because epimorphisms in both $\text{Mod}(S)$ and $\text{Mod}(R)$ are precisely the maps which are surjective on underlying sets, and restriction of scalars preserves underlying sets. Alternatively, restriction of scalars has a right adjoint (exercise), and so preserves colimits, hence pushouts, hence epimorphisms (exercise). $\Box$
The proof suggested by Tyrone in the comments also works but this technique is more general; it applies even in categories with no obvious notion of free object. 
