[example: Tensor Hom For Modules]
Let $R$ be a commutative ring and let $M$ be an $R$-module. We compute the adjunction
\begin{align*}
\operatorname{Hom}_R(M \otimes_R N, P)
\cong
\operatorname{Hom}_R(N,\operatorname{Hom}_R(M,P))
\end{align*}
for $R$-modules $N$ and $P$.
Given an $R$-linear map $b:M\otimes_R N\to P$, define
\begin{align*}
\Phi(b):N&\longrightarrow \operatorname{Hom}_R(M,P),\\
n&\longmapsto \bigl(m\longmapsto b(m\otimes n)\bigr).
\end{align*}
For each fixed $n\in N$, the map $m\mapsto b(m\otimes n)$ is $R$-linear because
\begin{align*}
b((m_1+m_2)\otimes n)
&=b(m_1\otimes n+m_2\otimes n)\\
&=b(m_1\otimes n)+b(m_2\otimes n),
\end{align*}
and
\begin{align*}
b((r m)\otimes n)
&=b(r(m\otimes n))\\
&=r\,b(m\otimes n).
\end{align*}
The assignment $n\mapsto \Phi(b)(n)$ is also $R$-linear, since for $m\in M$,
\begin{align*}
\Phi(b)(n_1+n_2)(m)
&=b(m\otimes(n_1+n_2))\\
&=b(m\otimes n_1+m\otimes n_2)\\
&=b(m\otimes n_1)+b(m\otimes n_2)\\
&=\bigl(\Phi(b)(n_1)+\Phi(b)(n_2)\bigr)(m),
\end{align*}
and
\begin{align*}
\Phi(b)(r n)(m)
&=b(m\otimes r n)\\
&=b(r(m\otimes n))\\
&=r\,b(m\otimes n)\\
&=\bigl(r\,\Phi(b)(n)\bigr)(m).
\end{align*}
Conversely, given an $R$-linear map $a:N\to \operatorname{Hom}_R(M,P)$, define
\begin{align*}
\beta_a:M\times N&\longrightarrow P,\\
(m,n)&\longmapsto a(n)(m).
\end{align*}
This rule is $R$-bilinear. In the first variable,
\begin{align*}
\beta_a(m_1+m_2,n)
&=a(n)(m_1+m_2)\\
&=a(n)(m_1)+a(n)(m_2),
\end{align*}
and
\begin{align*}
\beta_a(rm,n)
&=a(n)(rm)\\
&=r\,a(n)(m).
\end{align*}
In the second variable,
\begin{align*}
\beta_a(m,n_1+n_2)
&=a(n_1+n_2)(m)\\
&=(a(n_1)+a(n_2))(m)\\
&=a(n_1)(m)+a(n_2)(m),
\end{align*}
and
\begin{align*}
\beta_a(m,rn)
&=a(rn)(m)\\
&=(r\,a(n))(m)\\
&=r\,a(n)(m).
\end{align*}
Hence $\beta_a$ is balanced, so by the universal property of the tensor product there is a unique $R$-linear map
\begin{align*}
\widetilde a:M\otimes_R N\longrightarrow P
\end{align*}
such that
\begin{align*}
\widetilde a(m\otimes n)=a(n)(m)
\end{align*}
for all $m\in M$ and $n\in N$.
The two constructions are inverse. Starting with $a:N\to \operatorname{Hom}_R(M,P)$, for every $n\in N$ and $m\in M$,
\begin{align*}
\Phi(\widetilde a)(n)(m)
&=\widetilde a(m\otimes n)\\
&=a(n)(m),
\end{align*}
so $\Phi(\widetilde a)=a$. Starting with $b:M\otimes_R N\to P$, the map $\widetilde{\Phi(b)}$ satisfies
\begin{align*}
\widetilde{\Phi(b)}(m\otimes n)
&=\Phi(b)(n)(m)\\
&=b(m\otimes n)
\end{align*}
on every pure tensor $m\otimes n$; by the uniqueness part of the tensor product universal property, $\widetilde{\Phi(b)}=b$. Thus maps out of $M\otimes_R N$ are exactly $R$-linear families of maps out of $M$ indexed by $N$, which is the adjunction $M\otimes_R -\dashv \operatorname{Hom}_R(M,-)$.
[/example]