= Solution
A <Cartesian closed category> has finite <products in a category> and, for all $X,Y$, an <exponential object> $Y^X$ with a <natural bijection>
$$
\mathcal E(Z,Y^X)\cong\mathcal E(Z\times X,Y).
$$
Equivalently the <functor> $-\times X$ has a <right adjoint>. Its <counit of an adjunction> is the <evaluation map of an exponential object> $Y^X\times X\to Y$.
For the <presheaf category>, define the <exponential in a presheaf category> by
$$
\boxed{(Y^X)(C)=\operatorname{Nat}(h_C\times X,Y).}
$$
For $f:D\to C$, its restriction sends $\alpha$ to $\alpha\circ(h_f\times1_X)$. Composition of these restriction maps follows from composition of the $h_f$, so this is a <categorical presheaf>. Define <evaluation map of an exponential object> by
$$
e_C(\alpha,x)=\alpha_C(1_C,x).
$$
For $f:D\to C$, naturality of $\alpha$ gives
$$
Y(f)\alpha_C(1_C,x)=\alpha_D(f,X(f)x)
=e_D\bigl((Y^X)(f)\alpha,X(f)x\bigr).
$$
Thus $e$ is a <natural transformation>.
Given a <natural transformation> $\beta:Z\times X\to Y$, define its <currying> by
$$
\widehat\beta_C(z)_D(g,x)=\beta_D(Z(g)z,x),
\qquad g:D\to C,\quad x\in X(D).
$$
<Naturality> of $\beta$ proves that $\widehat\beta_C(z)$ is a natural transformation $h_C\times X\to Y$. For $f:C'\to C$, the same formula proves that $\widehat\beta:Z\to Y^X$ is natural. Evaluating at $g=1_C$ gives $e(\widehat\beta\times1_X)=\beta$.
Conversely, if $\gamma:Z\to Y^X$ and $\beta=e(\gamma\times1_X)$, then
$$
\widehat\beta_C(z)_D(g,x)=\gamma_D(Z(g)z)_D(1_D,x)
=\gamma_C(z)_D(g,x),
$$
by naturality of $\gamma$. Hence the two constructions are inverse and natural in $Z,Y$. Products are available pointwise, including the terminal singleton presheaf, so \b[the presheaf category is Cartesian closed]. The exponential uses <natural transformations> out of $h_C\times X$; simply exponentiating the individual values generally does not give this object.
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