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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.

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.