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A function $f:X\to Y$ between topological spaces is continuous exactly when the inverse image of every closed set in $Y$ is closed in $X$. The defining open set criterion is equivalent, because $f^{-1}(Y\setminus A)=X\setminus f^{-1}(A)$. This criterion concerns inverse images; a continuous map need not be a closed map.

For a nonempty subset $A$ of a topological space $X$, identify all points of $A$ and leave each point of $X\setminus A$ in its own equivalence class. The resulting quotient space is denoted $X/A$. For every subset $K\subseteq X$, its saturation of a subset isThe quotient topology therefore gives a useful test: $q_A(K)$ is a closed set if and only if the displayed saturation of a subset is a closed set of $X$. In particular, every closed set disjoint from $A$ has closed image, even if $A$ is not closed. When $A$ is closed, $q_A$ is a closed map, because both possibilities in the formula are then closed sets for closed $K$.

Let $X$ be a compact Hausdorff space and $R\subseteq X\times X$ a closed equivalence relation. The quotient space $X/R$ is compact and Hausdorff. Indeed, the saturation $R[C]$ of any closed $C\subseteq X$ is the projection of the compact set $R\cap (C\times X)$, so is closed. Thus the quotient map is a closed map.

Two different equivalence classes are disjoint compact subsets of $X$ and have disjoint open neighbourhoods $U,V$. To see the latter assertion, separate each pair of points by disjoint neighbourhoods and use compactness twice to extract finite covers. Now $U^{\prime }=X\setminus R[X\setminus U]$ and $V^{\prime }=X\setminus R[X\setminus V]$ are open saturated subsets containing the respective classes, and lie inside $U,V$. Their quotient images are disjoint open neighbourhoods, proving the Hausdorff condition. Compactness follows from the continuous quotient map. This applies to finite gluing along closed boundary pieces when the generated equivalence relation is closed; that last condition must be checked.

Let $f:X\to Y$ be a continuous map that is a closed map and whose fibres are compact. Then $f^{-1}(K)$ is compact for every compact set $K\subseteq Y$, without a Hausdorff hypothesis. Given an open cover, cover each fibre by a finite union $V_y$ of its members. The set $W_y=Y\setminus f(X\setminus V_y)$ is open, contains $y$, and has $f^{-1}(W_y)\subseteq V_y$. A finite subcover of $K$ by the $W_y$ uses only finitely many original cover members and covers $f^{-1}(K)$. Empty fibres cause no exception.

Let $K\subseteq X$, where the ambient topological space $X$ is Hausdorff and $K$ is compact in its subspace topology. Then $K$ is a closed set in $X$: the complement $X\setminus K$ is an open set in $X$. The ambient space need not be compact. For example, $[0,1]$ is a compact closed interval in the non-compact Hausdorff real line.^[ref]

The ambient Hausdorff hypothesis matters even when $K$ itself is Hausdorff. In the Sierpiński space $S=\{0,1\}$, the open sets are $\varnothing ,\{1\},S$, so the closed sets are $\varnothing ,\{0\},S$. Thus $K=\{1\}$ is compact and Hausdorff in its one-point subspace topology, but is not closed in $S$: its complement $\{0\}$ is not an open set. Every open neighborhood of $0$ contains $1$, so separation from $K$ fails. In contrast, the discrete topology on the same underlying set is Hausdorff and makes both singletons closed.

The diagram lists every open set and closed set, so the distinction can be checked directly. All subsets in both panels are compact: for any open cover of a finite set, choose one covering open set per point, obtaining a finite subcover. Compactness alone therefore cannot distinguish the two panels.

Here is the finite-selection mechanism in the general proof. For $x\in X\setminus K$ and each $k\in K$, the Hausdorff property gives disjoint open neighborhoods $U_k$ of $k$ and $V_k$ of $x$. The $U_k$ give an ambient open cover of $K$. Select a finite subcover $U_{k_1},\dots ,U_{k_m}$ and setThen $K\subseteq U$, $x\in V$, and $U\cap V=\varnothing$. The finite intersection makes $V$ an open set; an infinite intersection would not guarantee this. Thus every point of $X\setminus K$ has an open neighborhood contained in that complement, proving that $K$ is closed. The empty set is already closed.

This theorem combines with the continuous image of a compact space theorem to make every continuous map from a compact space to a Hausdorff space a closed map: a closed subset of its domain is compact, and its image is compact and hence closed in the target. This is the reason the inverse in the compact-to-Hausdorff continuous bijection theorem is continuous. The converse implication, from closed to compact, needs additional hypotheses: the real line is closed in itself but not compact; in a compact space, every closed subset is compact.

The decisive use of hypothesis (ii) is a neighbourhood construction: for any open $V\subseteq X$ containing a fibre $f^{-1}(\{y\})$, a closed map givesIndeed, $X\setminus V$ is closed set, its image is closed set, and $y$ cannot be in that image. A point of $f^{-1}(W)$ cannot lie outside $V$. Applying this to the finite union covering each compact fibre supplies a whole neighbourhood of that fibre's value using the same finitely many cover elements. Compactness of $K$ selects finitely many such neighbourhoods and hence a finite subcover of $f^{-1}(K)$, proving the desired compact-preimage theorem for closed maps.

The closed map assumption cannot simply be omitted. Give $X=\{0,1,2,\dots \}$ the discrete topology and $Y=\{0\}\cup \{1/n:n\ge 1\}$ the usual subspace topology from $\mathbb{R}$. The bijection $f(0)=0$, $f(n)=1/n$ is continuous, and all its fibres are compact singletons. But $Y$ is compact whereas $f^{-1}(Y)=X$ is an infinite discrete space, which is not compact. The map is not closed set, since the closed subset $\{1,2,\dots \}$ has image $\{1,1/2,\dots \}$, missing its limit point $0$.

The labels (i) and (ii) in the original PDF are hypotheses of one theorem, not separate requests. Here is the finite-cover step furnished by the compact fibres. Start with any open cover $\{U_i{\}}_{i\in I}$ of $f^{-1}(K)$, with the $U_i$ open in $X$ (a relative cover can be lifted to one of this form). For each $y\in K$, compactness gives a finite set $I_y\subseteq I$ withAn empty fibre needs no cover elements: take $I_y=\varnothing$ and $V_y=\varnothing$.

The complementary set $A_y=X\setminus V_y$ is closed set. By the closed map hypothesis, its image is closed set, sois open, contains $y$, and satisfies $f^{-1}(W_y)\subseteq V_y$. These open sets cover the compact set $K$. Select finitely many, $W_{y_1},\dots ,W_{y_m}$. The finitely many original cover elements with indices in $I_{y_1}\cup \cdots \cup I_{y_m}$ then cover $f^{-1}(K)$. ThereforeThis proves the unheaded conclusion in the PDF as well as explaining the role of hypothesis (i). No Hausdorff assumption is needed for this compact-preimage theorem for closed maps.

A closed set $C\subseteq X$ is compact because it is closed in the compact space $X$. Its image $f(C)$ is compact by the fact that a continuous image of a compact space is compact, and every compact subset of the Hausdorff space $Y$ is closed. Thus the continuous bijection $f$ is a closed map. Its inverse $f^{-1}:Y\to X$ is consequently continuous, so $f$ is a homeomorphism. This is the compact-to-Hausdorff continuous bijection theorem.

Let $X$ be compact and $q:X\to X/R$ the quotient map. If $(U_i{)}_{i\in I}$ is an open cover of $X/R$, then $(q^{-1}(U_i){)}_{i\in I}$ is an open cover of $X$. A finite subfamily covers $X$; because $q$ is surjective, the corresponding $U_i$ cover $X/R$. Thus every quotient of a compact space is compact.

The Hausdorff property need not survive. On the real line, defineThe quotient $\mathbb{R}/\mathbb{Q}$ has more than one point. If two nonempty open subsets of the quotient were disjoint, their inverse images would be disjoint nonempty open subsets of $\mathbb{R}$ invariant under rational translation. But any two nonempty open intervals acquire an intersection after one is translated by a suitably chosen rational number, so two such saturated open sets cannot be disjoint. Distinct quotient points cannot be separated, and the quotient is not Hausdorff. This is the non-Hausdorff quotient of the real line by rational translation.

Finally let $f:X\to Y$ be a continuous bijection, with $X$ compact and $Y$ Hausdorff. Every closed subset $C\subseteq X$ is compact. Its continuous image $f(C)$ is compact, and every compact subset of a Hausdorff space is closed. Hence $f$ is a closed map. For every closed $C\subseteq X$,is closed in $Y$, so $f^{-1}$ is continuous. Therefore $f$ is a homeomorphism. This is the compact-to-Hausdorff continuous bijection theorem.

A homeomorphism is a bijection for which both the function and its inverse are continuous maps.

Let $f:X\to Y$ satisfy the given assumptions. For a closed subset $F$ of $X$, the closed subset of a compact space theorem makes $F$ compact. The continuous image of a compact space theorem makes $f(F)$ compact. Because $Y$ is Hausdorff, its compact subset $f(F)$ is closed. Thus $f$ is a closed map.

For $g=f^{-1}:Y\to X$, the inverse image of a closed set $F\subseteq X$ iswhich is closed in $Y$. The closed-set criterion for continuity therefore makes $g$ continuous. Hence $f$ is a homeomorphism, proving the compact-to-Hausdorff continuous bijection theorem.

Use the equivalence relation induced by a function:Equality in $Y$ supplies reflexivity, symmetry, and transitivity, so this is an equivalence relation. Define $\bar{f}:X/\sim \to Y$ by $\bar{f}([x])=f(x)$. It is well-defined; it is injective by the definition of $\sim$ and surjective because $f$ is surjective.

For an open set $W\subseteq Y$,is open by continuity of $f$. The quotient topology therefore makes ${\bar{f}}^{-1}(W)$ open, proving that $\bar{f}$ is continuous.

The continuous image of a compact space theorem applied to $p$ makes $X/\sim$ compact. If $C$ is closed in $X/\sim$, it is compact, and $\bar{f}(C)$ is compact in the Hausdorff space $Y$, hence closed. Thus $\bar{f}$ is a closed map. Its inverse function is continuous, since its preimage of $C$ is precisely $\bar{f}(C)$. We have proved that $\bar{f}$ is a homeomorphism, including the compact-to-Hausdorff continuous bijection theorem used in the argument.

False. Give $X=\mathbb{R}$ its usual topology, identify all points of $(0,1)$ with one another, and leave every other point in a singleton equivalence class. Take the closed set $K=\{1/2\}$.

Its image $p(K)$ is a singleton in the quotient space, but its full preimage, also called the saturation of a subset, isThis is not closed in $\mathbb{R}$, since $0$ belongs to its closure. In the quotient topology, a subset is closed exactly when its preimage is closed, by applying the defining open-set rule to complements. Thus $p(K)$ is not closed, and this quotient map is not a closed map.

For the canonical projection of a quotient set $p:X\to X/\sim$, the saturation of $A\subseteq X$ is the union of all equivalence classes meeting $A$, namely $p^{-1}(p(A))$. A subset is called saturated when it equals its saturation. The quotient topology says that $p(A)$ is closed exactly when the saturation of $A$ is closed, and similarly for openness. A closed set need not have closed saturation: collapsing $(0,1)\subset \mathbb{R}$ to one point makes the saturation of $\{1/2\}$ equal to $(0,1)$. Hence a quotient map need not be a closed map.