manipulation of internal hom

Question

Let $\text{hom}(-,-):\mathcal{C}^{op}\times\mathcal{C}\rightarrow\mathcal{C}$ denote the internal Hom functor associated to a closed category $\mathcal{C}$, and let $X$, $Y$, and $Z$ be objects in $\mathcal{C}$. True or false:\begin{equation*}\text{hom}(\text{hom}(X,Y),Z)\stackrel{?}{=}\text{hom}(Y,\text{hom}(X,Z))\end{equation*}

Answer 1

False.

A Cartesian Closed Category is an example of a Closed Category, and an example of this is $\mathbf{Set}$, where the internal hom functor corresponds to the exponential functor, so $\mathrm{hom}(X,Y)=Y^X$. Then the LHS of the equation becomes $Z^{(Y^X)}$ while the RHS is $(Z^X)^Y$.

Whether you're asking for equality or just isomorphism doesn't matter, as it is not in general true that $$Z^{(Y^X)}\cong (Z^X)^Y$$ To see this, we only need to look at cardinality. Suppose we have $X=Z=\mathbf{2}=\{0,1\}$ and $Y=\mathbf{3}=\{0,1,2\}$. Then the cardinality of $\mathbf{2}^{(\mathbf{3}^\mathbf{2})}$ is $2^{(3^2)}=2^9$ while the cardinality of $(\mathbf{2}^\mathbf{3})^\mathbf{2}$ is $(2^3)^2=2^6$. It is clearly not the case that $2^6=2^9$, and thus there is no bijection between $\mathbf{2}^{(\mathbf{3}^\mathbf{2})}$ and $(\mathbf{2}^\mathbf{3})^\mathbf{2}$.

Without resorting to a specific example immediately, we could also note the identity $(Z^X)^Y\cong Z^{X\times Y}$ in a Cartesian Closed Category. If we did have an isomorphism, then we'd have $$\mathrm{hom}(\mathrm{hom}(X,Y),Z)\cong \mathrm{hom}(Y,\mathrm{hom}(X,Z))\cong \mathrm{hom}(X\times Y,Z)$$ By the Yoneda Lemma if these are natural isomorphisms (as I'm sure is intended to be communicated by the proposed identity), then we'd find that $$Y^X\cong \mathrm{hom}(X,Y)\cong X\times Y$$ which is much more clearly false in general. Moreover, this is precisely why the above counter-example is a counter-example.