For $y\in Y$, every neighborhood of $y$ in the subspace is of the form $V\cap Y$ with $V$ an ambient open neighborhood. Since $A\subseteq Y$, this meets $A$ exactly when $V$ meets $A$. Thus

A Hausdorff compactification is a topological embedding $h:X\to K$ with dense image in a compact Hausdorff space $K$. Define the domination preorder byIt is reflexive by the identity map and transitive by composition. Any comparison map is onto: its image is compact and hence closed in $K$, and contains the dense set $h(X)$. Two compactifications are equivalent if a homeomorphism $u:K\to K^{\prime }$ satisfies $u\circ h=h^{\prime }$.

Such a homeomorphism and its inverse give domination both ways. Conversely, if comparison maps exist in both directions, each composite agrees with the identity on the dense embedded copy of $X$. Two continuous maps to a Hausdorff space agreeing on a dense set agree everywhere. The comparison maps are therefore inverse homeomorphisms. Thus equivalence is exactly mutual domination.

The Stone-Čech extension property says that every continuous map from $X$ into any compact Hausdorff space extends continuously to the compactification. The extension is unique by density. For two compactifications with this property, extend each embedding through the other compactification. This gives comparison maps both ways, hence equivalence by the preceding argument.

For the perfect-map characterization by Stone-Čech remainders, the universal property first gives a unique continuous $F:\beta X\to \beta Y$ extending the composite $X\stackrel{f}{\to }Y\hookrightarrow \beta Y$. The required remainder condition is equivalent toOne inclusion is already forced by the extension property; the other forbids any point of the remainder mapping into $Y$.

Suppose first that $f$ is a perfect map. Fix $y\in Y$ and let $K=f^{-1}(y)$, a compact subset of $X$ and hence a closed subset of $\beta X$. If $p\in F^{-1}(y)\setminus K$, normality of $\beta X$ supplies an open $U$ with $K\subseteq U$ and $p\notin \overline{U}$. Because $f$ is closed,is an open neighborhood of $y$ and $f^{-1}(V)\subseteq U\cap X$. Choose an open $V^{\prime }$ of $\beta Y$ with $V^{\prime }\cap Y=V$. The open set $F^{-1}(V^{\prime })$ contains $p$, and its intersection with the dense subset $X$ lies in $U$. Every point of that open set is consequently in $\overline{U}$, contradicting the choice of $p$. Therefore $F^{-1}(y)=K\subseteq X$ for every $y\in Y$, proving the remainder condition. This argument also handles an empty fiber.

Conversely suppose the remainder condition holds. Every fiber $f^{-1}(y)=F^{-1}(y)$ is a closed subset of $\beta X$, hence compact, and lies in $X$. To show that $f$ is closed, let $A$ be closed in $X$ and set $D={\overline{A}}^{\beta X}$. The initial closure identity gives $D\cap X=A$. Compactness makes $F(D)$ closed in $\beta Y$, and the remainder condition givesIndeed any preimage in $D$ of a point of $Y$ must belong to $X$ and hence to $A$. Thus $f(A)$ is closed in $Y$. Together with compact fibers this proves

A T0 space distinguishes any two distinct points by an open set containing one and not the other. A Tychonoff space, also called a completely regular Hausdorff space, is a $T_1$ space in which a point outside a closed set can be separated from it by a continuous function: after rescaling, $f(x)=0$ and $f{|}_C=1$ with values in $[0,1]$.

The product topology has the subbasis of a topologyA map into the product is continuous exactly when each coordinate map is continuous. One direction follows by composition with the continuous product projections; the other follows by taking inverse images of the displayed subbasic sets.

Define the diagonal evaluation map $\Delta (x)=(f_{\lambda }(x){)}_{\lambda \in \Lambda }$. Its coordinates are continuous, so the map is continuous by continuity into a product space. Point separation makes it injective. For a basic open set $B=f_{\lambda }^{-1}(U)$,Every open subset of $X$ is a union of such $B$, so its image is open in $\Delta (X)$. The inverse on the image is continuous, and $\boxed{\Delta \text{ is a topological embedding}}$. If $X$ is $T_0$, any two distinct points are distinguished by an open set and then by one basic neighborhood within it. Equality of all coordinates would prevent that distinction. Thus the basis hypothesis alone implies point separation in a $T_0$ space. The unsubscripted $f$ in the printed separation hypothesis is understood as the chosen coordinate $f_{\lambda }$.

For the Sierpiński space $S=\{0,1\}$ with opens $\varnothing ,\{1\},S$, each open $U\subseteq X$ gives a continuous characteristic map ${\chi }_U:X\to S$ with ${\chi }_U^{-1}(\{1\})=U$. These inverse images form the whole topology, hence a basis. In a $T_0$ space they separate points, proving the embedding into $S^{\tau }$.

For a Tychonoff space, use all continuous maps $X\to [0,1]$. Given $x\in U$ with $U$ open, complete regularity supplies $f(x)=0$ and $f=1$ on $X\setminus U$. Then $f^{-1}([0,1/2))$ is an open neighborhood of $x$ contained in $U$. These inverse images form a basis. The diagonal theorem gives an embedding into a product of closed unit intervals.

A three-point space works even without any separation axiom. TakeFor each open $U$ of an arbitrary space, let $g_U=2$ on $U$ and $0$ elsewhere. It is continuous. For each point $a$, let $h_a=1$ at $a$ and $0$ elsewhere. It is also continuous, because its image lies in the subspace $\{0,1\}$ of $Y$, which has the indiscrete topology; equivalently $h_a^{-1}(\{2\})=\varnothing$. The $g_U$ supply every open set as an inverse image, and the $h_a$ distinguish every pair of points. Using the diagonal evaluation map for their combined family gives a universal three-point space for product embeddings:

Finally embed the given completely regular Hausdorff space into a cube $K=[0,1{]}^I$. The Tychonoff theorem makes $K$ compact; it is Hausdorff and hence normal. The image of the compact subset $C$ is compact, so closed in $K$. The Tietze extension theorem says that a continuous bounded real-valued function on a closed subset of a normal space extends continuously to the whole space. The given function on $C$ is bounded by compactness, so apply this theorem to its transported function on the image of $C$ in $K$. Restrict the extension to the image of $X$ and pull it back along the embedding. This yields $\boxed{F:X\to \mathbb{R}\text{ continuous},F{|}_C=f}$. This proves that compact subsets of Tychonoff spaces are C-embedded, using normality of the compact cube.