In Exercise 12 of Section 26 we established that if $p$ is closed and $U$ containing $p^{-1}(y)$ is open, then there is some neighborhood $V$ of $y$ such that $p^{-1}(V)\subset U$ . Neither continuity nor surjectivity of $p$ is required for this fact. More generally, we can state the same fact for any subset $B\subset Y$ instead of just one point $y$ (take the union of all such neighborhoods for each point in $B$ ).

(a) Take two different points $a,b\in Y$ , their preimages are disjoint compact nonempty (surjectivity) subspaces of a Hausdorff space, which can be separated by open neighborhoods $U,V$ . Now, using the fact above, find neighborhoods of $a$ and $b$ such that their preimages are in $U$ and $V$ . Since $U$ and $V$ are disjoint, so are the neighborhoods of $a$ and $b$ (we again use surjectivity here). (How did we use the continuity of $p$ ?)

(b) Let $y\in Y$ , and $B$ be a closed subset of $Y$ such that $y\notin B$ . Then, $C=p^{-1}(y)$ is compact, and $A=p^{-1}(B)$ is closed (continuity), where both $A$ and $C$ are nonempty (surjectivity). Since $X$ is regular, we can separate each point in $C$ and the set $A$ by some open neighborhoods. Since $C$ is compact, we can take a finite number of such open neighborhoods covering $C$ (let $U$ be their union), and the corresponding finite intersection of open neighborhoods of $A$ is an open neighborhood $V$ of $A$ disjoint from $U$ . Now, we use the same fact above to find neighborhoods of $y$ and $B$ such that their preimages are in $U$ and $V$ , correspondingly. Since $p$ is surjective and $U$ and $V$ are disjoint, the neighborhoods of $y$ and $B$ are disjoint as well.

(c) Let $y\in Y$ . $A=p^{-1}(y)$ is a non-empty compact set (surjectivity). Each point of $A$ has a neighborhood that lies within a compact subspace of $X$ . All such neighborhoods cover $A$ , and there is a finite subcovering $U$ , a neighborhood of $A$ . The corresponding finite union $C$ of compact subspaces of $X$ is compact and contains $U$ . Once again, we use the fact above to find a neighborhood $V$ of $y$ such that $p^{-1}(V)$ is in $U$ . Therefore, the neighborhood lies within $p(C)$ ($p$ is surjective), which is compact as the continuous image of a compact set.

(d) Using the hint, let $\{B_{n}\}$ be a countable basis for $X$ . For each $N$ , a finite subset of $\mathbb{Z}_{+}$ , define $B_{N}=\cup_{n\in N}B_{n}$ , and let $V_{N}$ be the largest open set in $Y$ such that its preimage is in $B_{N}$ , i.e. $V_{N}$ is the union of all open sets of $Y$ having preimages in $B_{N}$ . Let $U_{N}=p^{-1}(V_{N})$ , open by continuity of $p$ . There are countably many sets in $\{V_{N}\}$ . We show that they form a basis for the topology of $Y$ . Take $y\in V$ , where $V$ is open in $Y$ . Let $A=p^{-1}(y)$ (compact, nonempty by surjectivity), $U=p^{-1}(V)$ (open by continuity). For each point of $A$ choose a basis neighborhood contained in $U$ , and then, a finite subcovering $B_{N}$ of $A$ contained in $U$ . Using the fact above, there is a neighborhood of $y$ such that its preimage is contained in $B_{N}$ , therefore, $y\in V_{N}$ , $A\subset U_{N}\subset B_{N}\subset U$ , and $V_{N}$ is a basis neighborhood of $y$ contained in $V$ (again, by surjectivity of $p$ ). Done.