Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 4 / 13E / b / i / Solution

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.