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Use the Stone-Čech extension property: a discrete topological space $X$ embeds densely into a compact Hausdorff space $\beta X$, and every continuous map from $X$ to a compact Hausdorff space extends continuously to $\beta X$. Apply it to the characteristic function ${\chi }_A:X\to \{0,1\}$ of an arbitrary subset $A$; discreteness makes it continuous. Its extension ${\tilde{\chi }}_A$ has clopen fibers, andIndeed that fiber is closed and contains $A$, while its openness and density of $X$ show that every point in the fiber is approached by points of $A$.

If $A$ and $B$ are disjoint, the extension of ${\chi }_A$ is one on $\overline{A}$ and zero on $\overline{B}$, so these closures are disjoint. The inclusion $X\to \beta X$ therefore satisfies the criterion from part (b), and its ultrafilter-limit map is both surjective and injective. ThusThis is a bijection of underlying sets; identifying the compatible topology on the ultrafilter topological space would give the familiar stronger homeomorphism statement, but is not needed for the requested conclusion.

Take the natural numbers as the dense discrete subspace of their Stone-Čech compactification $\beta \mathbb{N}$. The ambient compact Hausdorff space is normal, and the Stone-Čech extension property makes this subspace $C^b$-embedded. It is not $C$-embedded: the continuous function $n\mapsto n$ is unbounded, whereas every continuous real-valued function on a compact space is bounded. The subspace is not closed, so this does not contradict the preceding closed-subspace theorem.

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

For a continuous map $f:X\to Y$ between Tychonoff spaces, extend it to $F:\beta X\to \beta Y$ by the Stone-Čech extension property. If $f$ is perfect, compactness of a fiber and closedness of $f$ show that no remainder point can map to a point of $Y$: separate such a point from the fiber by a neighborhood in $\beta X$, and use closedness to obtain a target neighborhood whose preimage stays inside it, contradicting density. Conversely, if $F^{-1}(Y)=X$, each fiber is compact, and for closed $A\subseteq X$ the equality $f(A)=F({\overline{A}}^{\beta X})\cap Y$ proves closedness. Equivalently $F(\beta X\setminus X)\subseteq \beta Y\setminus Y$.