Lemma

Let $T$ and $X$ be two topological spaces, $\pi$ a proper continuous mapping (GT, I, §10, No. 1, Def. 1) of $T$ into $X$. Let $g$ be a lower semi-continuous numerical function defined on $T$. For every $x \in X$, let $f(x)$ be the infimum of the function $g(t)$ in the set $\overline{\pi}^{-1}(x)$ (infimum equal to $+\infty$ if $\overline{\pi}^{-1}(x) = \varnothing$; cf. S, III, §1, No. 9). Then $f$ is lower semi-continuous on $X$.

For every (finite) real number $a$, denote by $B_a$ the set of $x \in X$ such that $f(x) \leq a$, and by $A_a$ the set of $t \in T$ such that $g(t) \leq a$; it all comes down to showing that $B_a$ is closed (GT, IV, §6, No. 2, Prop. 1). Now, $A_a$ is closed (same ref.) and the proper mapping $\pi$ is closed (GT, I, §10, No. 1, Prop. 1); we are thus reduced to proving that $\pi(A_a) = B_a$. The obvious relation $f(\pi(t)) \leq g(t)$ for all $t \in T$ implies that $\pi(A_a) \subset B_a$. On the other hand, let $x \in B_a$; the set $\overline{\pi}^{-1}(x)$ is quasi-compact (GT, I, §10, No. 2, Th. 1) and nonempty, therefore there exists a $t \in \overline{\pi}^{-1}(x)$ such that