= Solution
An object $E$ in a finite-product category is <exponentiable> when $-\times E$ has a right adjoint $(-)^E$. The terminal object is exponentiable. If $E$ and $F$ are exponentiable, then
$$
-\times(E\times F)\cong(-\times E)\times F
$$
has the composite of their right adjoints as a right adjoint. Hence the <product of exponentiable objects is exponentiable>, including the empty product.
Suppose $0$ is both initial and terminal. Since $-\times E$ is a left adjoint for exponentiable $E$, it preserves the initial object, so $0\times E\cong0$. Since $0$ is terminal, $0\times E\cong E$. Therefore the <zero object is the only exponentiable object in a pointed category>.
Let $G=\mathbf{Top}(X,S)$ for a <T0 space> $X$ and the <Sierpiński space> $S$. The evaluation map
$$
e:X\longrightarrow S^G,\qquad e(x)=(g(x))_{g\in G}
$$
is injective because characteristic maps of open sets distinguish distinct points. Every open $U\subseteq X$ equals $g_U^{-1}(1)$ for its characteristic map $g_U:X\to S$, so the subspace topology induced by $e$ is the original topology. This is the <Embedding of a T0 space into a power of the Sierpiński space>.
A subspace inclusion between $T_0$ spaces is a <regular monomorphism>, hence an <equalizer>. Embedding its codomain into another power of $S$ and composing the parallel pair preserves the equalizer because the embedding is monic. Consequently
$$
X\longrightarrow S^G\rightrightarrows S^H
$$
is an equalizer for suitable sets $G,H$.
If $E$ is exponentiable, $\mathbf{Top}_0(-\times E,S)$ is represented by $S^E$. Conversely, suppose it is represented by $R$. Products give
$$
\mathbf{Top}_0(Y\times E,S^G)
\cong\mathbf{Top}_0(Y,R^G).
$$
Express any $T_0$ space $X$ as the displayed equalizer $S^G\rightrightarrows S^H$ and take the corresponding equalizer $R^G\rightrightarrows R^H$. Since hom-functors preserve limits, this equalizer represents $\mathbf{Top}_0(-\times E,X)$. Thus $-\times E$ has a right adjoint on every target, proving the <Exponentiability criterion in the category of T0 spaces>.
Back to article page
