Naturality in the Adjoint Property of Tensor Products In the Lecture Notes in Algebraic Topology by Davis and Kirk, one has the following result

Proposition 1.14 (Adjoint Property of Tensor Products) There is an isomorphism of $R$-modules $$\operatorname{Hom}_R(A \otimes_R B, C) \simeq \operatorname{Hom}_R(A, \operatorname{Hom}_R(B,C))$$ natural in $A$, $B$, $C$ given by $\phi \leftrightarrow (a \mapsto (b \mapsto \phi(a \otimes b) ) ).$

My problem is it now to understand what exactly natural in $A$, $B$, $C$ means.
When one talks about naturality, then one probably means there is a natural isomorphism. At least this is what I assume.
Let us consider the naturality in $A$ and denote the category of $R$-modules by $R\operatorname{-MOD}$. Then we can define two functors $\mathcal{F}^A, \mathcal{G}^A: R\operatorname{-MOD} \to R\operatorname{-MOD}$ by 
$$\mathcal{F}^A(X) = \operatorname{Hom}_R(X \otimes_R B, C), \quad \mathcal{G}^A(X) = \operatorname{Hom}_R(X, \operatorname{Hom}_R(B,C)).$$
Now the natural isomorphism $\eta^A: \mathcal{F}^A \Rightarrow \mathcal{G}^A$ must satisfy that for every $R$-module $X$ we have an $R$-module isomorphism $\eta^A_X: \mathcal{F}^A(X) \longrightarrow \mathcal{G}^A(X)$. This is true by the universal property (one notes that $\mathcal{G}^A(X)$ can be interpreted as an bilinear map $X \times B \to C$).
The point where I am not sure about is if $\mathcal{F}^A$ and $\mathcal{G}^A$ are actually functors. For instance, given an $R$-module homomorphism $f: X \to Y$, how does the induced morphism $\mathcal{F}^A(f): \mathcal{F}^A(X) \to \mathcal{F}^A(Y)$ look like? My first try was it to take a $g \in \mathcal{F}^A(X) = \operatorname{Hom}_R(X \otimes_R B, C)$, $x \otimes b \mapsto g(x \otimes b)$ and set $\mathcal{F}^A(f)(g) (x \otimes b) = g(f(x) \otimes b)$. But this does not make sense since $f$ does not need to be surjective and the first component of $\mathcal{F}^A(f)$ has to be defined on the whole module $Y$.
Now I am totally stuck and have no idea how to continue. Could someone please unwind my messy thoughts? That would be great.
 A: One thing you're missing is that the functor you call $\mathcal{F}^A$ is contravariant. We can compute it by unfolding notation.
Given $f : X \to Y$...
The morphism $f \otimes B : X \otimes B \to Y \otimes B$ is determined from the fact that $\otimes$ is a bifunctor. If mixing arrows and objects bothers you, note that $f \otimes B$ means the same thing as $f \otimes 1_B$.
This is, as you've presumably realized, the morphism $x \otimes b \mapsto f(x) \otimes b$.
Now, $\hom(-, C)$ is contravariant. Given any $g : U \to V$, $\hom(g, C) : \hom(V, C) \to \hom(U, C)$ is the mapping $h \mapsto h \circ g$. Note the reversal in the direction of the arrow.
So, $\hom(f \otimes B, C) : \hom(Y \otimes B, C) \to \hom(X \otimes B, C)$ is the function that sends $h : Y \otimes B \to C$ to the function $h \circ (f \otimes B) : X \otimes B \to C$. $h \circ (f \otimes B)$ is the mapping $x \otimes b \mapsto h(f(x) \otimes b)$.
And it will be true that this gives a natural isomorphism between functors
$$ \mathcal{F}^A, \mathcal{G}^A : (R\operatorname{-MOD})^{\mathrm{op}}  \to R\operatorname{-MOD} $$
It is ultimately true that both sides of the equation define naturally isomorphic functors
$$ (R\operatorname{-MOD})^{\mathrm{op}} \times (R\operatorname{-MOD})^{\mathrm{op}} \times R\operatorname{-MOD} \to R\operatorname{-MOD} $$
