Lemma 3

Let X and Y be topological spaces, let $p: X\rightarrow Y$ be a continuous proper and open mapping, and let $f: X\rightarrow \mathbf{R}$ be a continuous mapping. The mapping $g: Y\rightarrow \overline{\mathbf{R}}$ given by $y\mapsto$ sup$_{x\in\overset{-1}{p}(y)}f(x)$ is continuous.

Let $b\in Y$. Since $p$ is proper, its fibre $\overset{-1}{p}(b)$ is a quasi-compact space; hence we have $g(b)\in \mathbf{R}\cup \{-\infty \}$.

Let $m\in \mathbf{R}$ such that $g(b)< m$. For every $a\in \overset{-1}{p}(b)$, one has $f(a)\leqslant$ $g(b)< m$; let $V_a$ be a neighbourhood of $a$ in X such that $f(x)< m$ for every $x\in V_a$. The union V of the sets $V_a$ is a neighbourhood of $\overset{-1}{p}(b)$ in X. By Lemma 5 (I, p. 75), there exists a neighbourhood W of $b$ in Y such that $\overset{-1}{p}(W)\subset V$. For every $y\in W$, one has $g(y)\leqslant m$. This proves that $g$ is upper semi-continuous at $b$.

Let us now prove that $g$ is lower semi-continuous at $b$. We may suppose that $g(b)\in \mathbf{R}$. Let $m\in \mathbf{R}$ such that $m < g(b)$. Let $a\in \overset{-1}{p}(b)$ such that $m < f(b)$; let then V be a neighbourhood of $a$ in X such that $f(x)> m$ for every $x\in V$. It follows that $g(y)> m$ for every $y\in p(V)$. Since $p$ is open, $p(V)$ is a neighbourhood of $b$, so that $g$ is lower semi-continuous.

The lemma is thus proved.