Adjointness of Hom and Tensor for non commutative ring Let $M$ be an $(A,B)$-bimodule, let $N$ be a $(B,C)$-bimodule, and let $K$ be an $(A,C)$-bimodule. Then $$\operatorname{Hom}_C(M \otimes_{B}N,K) \cong \operatorname{Hom}_B(M,\operatorname{Hom}_C(N,K))$$ as right $A$-modules, and $$\operatorname{Hom}_A(M\otimes_B N,K) \cong \operatorname{Hom}_B(N,\operatorname{Hom}_A(M,K))$$ as right $C$-modules.
Where  can I get a proof for this? I tried to do. But I am not able to complete.
 A: There isn't very much that you can do except the obvious thing:
Define a map $$f:\operatorname{Hom}_C(M \otimes_{B}N,K) \to \operatorname{Hom}_B(M,\operatorname{Hom}_C(N,K))$$
by $f(\phi) = (m \mapsto (n \mapsto \phi(m \otimes n)))$. Check that this is well-defined, i.e., that $f(\phi)$ really is a homomorphism of right $B$-modules.$^{(*)}$
Conversely, define a map $$\operatorname{Hom}_C(M \otimes_{B}N,K) \leftarrow \operatorname{Hom}_B(M,\operatorname{Hom}_C(N,K)):g$$
by $g(\psi) = (m \otimes n \mapsto \psi(m)(n))$. Check that this is well-defined using the universal property of the tensor product.
Then $$f(g(\psi))(m) = (n \mapsto g(\psi)(m \otimes n) = \psi(m)(n)) = \psi(m)$$
and $$g(f(\phi))(m \otimes n) = f(\phi)(m)(n) = \phi(m \otimes n),$$
which proves that $f,g$ are mutual inverses.

$^{(*)}$ Let's prove this. The subtlety lies in the definitions of the different module structures on Hom sets. We need to recall that the $(A,B)$-bimodule structure on $\operatorname{Hom}_C(N,K)$ is given by $$a \cdot \kappa \cdot b:n \mapsto a\kappa(bn). \tag{1}$$
And also, the right $A$-module structure on $\operatorname{Hom}_B(M,\operatorname{Hom}_C(N,K))$ is given by $$\psi \cdot a: m \mapsto \psi(a m). \tag{2}$$
For the domain, the tensor product $M \otimes_B N$ is an $(A,C)$-bimodule via $$a \cdot (m \otimes n) \cdot c = (am \otimes nc), \tag{3}$$ and $\operatorname{Hom}_C(M \otimes_{B}N,K)$ becomes a right $A$-module via $$(\phi \cdot a)(m \otimes n) = a(m \otimes n) = am \otimes n. \tag{4}$$

Okay. First, to check that $f(\phi)$ is a homomorphism of right $B$-modules:
$$\begin{align}f(\phi)(m + m'b)(n) &= \phi(m + m'b \otimes n) \\&= \phi(m \otimes n) + \phi(m'b \otimes n) \\&=\phi(m \otimes n) + \phi(m' \otimes bn) 
\\&=f(\phi)(m)(n) + (f(\phi)(m')\cdot b)(n) \tag{1 and 3}
\\&=(f(\phi)(m) + (f(\phi)(m')\cdot b))(n).
\end{align}$$
Next, to check that $f$ is a homomorphism of right $A$-modules:
$$\begin{align}
f(\phi + \phi' \cdot a)(m)(n) &= (\phi + \phi' \cdot a)(m \otimes n) \\ &=
\phi(m \otimes n) + (\phi'\cdot a)(m \otimes n) \\ &=
\phi(m \otimes n) + \phi'(am \otimes n)\tag{4} \\
&= f(\phi)(m)(n) + (f(\phi') \cdot a)(m)(n) \tag{2}.
\end{align}$$
