proving a map is a covariant functor between modules I'm trying to show that $(\cdot)\otimes_A N$ is a covariant functor from $MOD_A\to MOD_A$. Can somebody ensure I'm not oversimplifying the task (which I feel to be the case). It is clear that this functor does indeed take objects in $MOD_A$ to objects in $MOD_A$, and the crux of this is ensuring the covariant functorality when looking at how it behaves with morphisms. 
So if $f\in Mor(P,Q)$, then $f$ is characterized by $f(\sum_k a_kp_k)=\sum_ka_kf(p_k)$ for $a_k\in A$, $p_k\in P$ and $f(p_k)\in Q$. 
Suppose we define $F(\cdot):= (\cdot)\otimes_A N$, then it must be that $F(f):P\otimes_A N\to Q\otimes_A N$ and hence $F(f)(\sum_ka_kp_k\otimes n_k)=\sum_ka_kf(p_k)\otimes n_k$.
Now if in addition we have $g\in Mor(Q,R)$, then $F(g\circ f)(\sum_ka_kp_k\otimes n_k)=F(g)(\sum_ka_kf(p_k)\otimes n_k)$ (so that the $f(p_k)$ lie in $g$'s domain) $=\sum_ka_kg(f(p_k))\otimes n_k= (F(g)\circ F(f))(\sum_ka_kp_k\otimes n_k) $, hence having $F(g\circ f)=F(g)\circ F(f)$ as we wanted. 
Would this fully solve the claim?
Thanks
 A: Yeah, that's it. In principle, of course, you need to check that your formula actually defines a homomorphism, although that's hardly a problem here. Another approach just uses the universal property and the Yoneda lemma: if $f:P\to Q$, then for any $R$ and bilinear map $a:Q\times N\to R$ we get a bilinear map $af:P\times N\to R$. This is a natural transformation between the functors $\text{BiHom}(Q\times N,-)\cong \text{Hom}(Q\otimes N,-)$ and $\text{BiHom}(P\times N,-)\cong\text{Hom}(P\otimes N,-)$, which comes by the Yoneda lemma from a unique map $f\otimes N:P\otimes N\to Q\otimes N$. This is functorial since the natural transformation was; that and the naturality itself now just follow from trivial properties of composition of module morphisms. If you haven't seen any category theory yet, this proof probably doesn't look easier, but notice that I don't have to know anything about elements of the tensor product to write it down! This is a hint that it works in a much wider context.
