In the Lecture Notes in Algebraic Topology by Davis & Kirk, we consider the Kronecker pairing which is defined by
$$H^n(C;R) \times H_n(C;R) \to R \\ ([\varphi],[\alpha]) \mapsto \varphi(\alpha)$$
where $H^n(C;R)$ is the cohomology and $H_n(C;R)$ is the homology with coefficients over a ring $R$.
Then it is said
Deduce by taking the adjoints that the Kronecker pairing defines a map $$H^n(C;R) \to \operatorname{Hom}(H_n(C;R),R).$$
I assume that this map is defined by $[\varphi] \mapsto ([\alpha] \mapsto \varphi(\alpha))$.
My question: What exactly does "taking the adjoints" mean? Does it has something to do with adjoint functors? Or maybe with the Adjoint Property of Tensor Products which is 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 $\phi \leftrightarrow (a \mapsto (b \mapsto \phi(a \otimes b) ) ).$
It might be obvious but I just don't see it. I am thankful for any explanation!
In this situation, "taking the adjoint" means the following two steps. (1) By the universal property of the tensor product, view the Kronecker pairing as a homomorphism $H^n\otimes H_n\to R$. (2) Apply the adjoint property that you quoted to convert this homomorphism to a homomorphism $H^n\to\text{Hom}(H_n,R)$. Strictly speaking, "taking the adjoint" seems to refer to (2), but apparently (1) is to be understood. I suppose the author viewed (1) as so evident that it didn't need to be mentioned separately.