Exactness of the tensor product considering the tensor products of abelian groups: could you tell me if (and why?) the following is true?
For any free abelian group $A$ the functor $A\otimes (-)$ is exact.
Thanks!
[I extracted this Proposition from a step in a proof of some lectures notes that I dont understand.]
 A: Yes, this is true.  Let $S$ be a basis for $A$; then for any group $G$, there is a natural isomorphism $A\otimes G\cong G^{\oplus S}$, where $G^{\oplus S}$ denotes the direct sum of copies of $G$ indexed by $S$, i.e. the group of all functions $S\to A$ which are $0$ at all but finitely many points (this is because $A\cong\mathbb{Z}^{\oplus S}$, $\mathbb{Z}\otimes G\cong G$, and tensor products distribute over direct sums). Suppose $G\stackrel{\alpha}\to H\stackrel{\beta}\to K$ is exact.  Then the induced sequence $$G^{\oplus S}\stackrel{\alpha_*}\to H^{\oplus S}\stackrel{\beta_*}\to K^{\oplus S}$$ is also exact.  To prove this, suppose $\beta_*(h)=0$ for some $h=(h_i)_{i\in S}\in H^{\oplus S}$.  Then $\beta(h_i)=0$ for each $i$, so for each $i$ there exists some $g_i\in G$ such that $\alpha(g_i)=h_i$.  Furthermore, we can choose $g_i$ to be $0$ whenever $h_i=0$.  Since $h_i=0$ for all but finitely many $i$, this means $g_i=0$ for all but finitely many $i$, so the $g_i$ define an element $g=(g_i)\in G^{\oplus S}$.  This $g$ then satisfies $\alpha_*(g)=h$.
