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For a T0 space $X$, evaluation against all continuous maps $g:X\to S$ is a homeomorphism onto its image under the mapinto a power of the Sierpiński space. The $T_0$ axiom makes it injective, and the coordinate inverse images of $\{1\}$ recover every open subset of $X$.

A $T_0$ space $E$ is exponentiable in ${\mathbf{Top}}_0$ if and only if the functor ${\mathbf{Top}}_0(-\times E,S)$ to sets is representable, where $S$ is the Sierpiński space. Every $T_0$ space is an equalizer of maps between powers of $S$, so a representing object for maps into $S$ constructs exponentials for every target by products and equalizers.

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, thenhas 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 $0\times E\cong 0$. 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 mapis 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. Consequentlyis 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 giveExpress 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.

The closure $\overline{Z}$ is the intersection of all closed subsets containing $Z$; equivalently, every neighbourhood of a point of $\overline{Z}$ meets $Z$. The subspace is dense when $\overline{Z}=X$. A space is Hausdorff when distinct points have disjoint neighbourhoods.

For Hausdorff $Y$, the diagonal ${\Delta }_Y$ is closed in $Y\times Y$. Thusis closed. Since it contains the dense set $Z$, it equals $X$.

The conclusion fails without Hausdorffness. Let $X=Y=\{0,1\}$ have the Sierpiński space topology $\{\varnothing ,\{1\},X\}$ and let $Z=\{1\}$, which is dense. The identity map and the constant map with value $1$ are continuous and agree on $Z$, but differ at $0$.