Remark 3

Suppose that A is closed and that the mapping $f$ is proper. We have just seen that the mapping $\beta_f$ is closed. For every $x\in X$,

$\beta^{-1}_f(\alpha_f(x)) =\overset{-1}{f}(x)$. It is therefore a quasi-compact subset of A (TG,

I, p. 75, Theorem 1), hence also of B. For every $b\in B,\beta^{-1}_f(\beta_f(b))$

equal to $\{b\}$ if $b\in B-A$, and to $\overset{-1}{f}(f(b))$ if $b\in A$; in both cases, it is a quasi-compact subset of B. The fibres of the mapping $\beta_f$ are therefore quasi-compact; consequently (loc. cit.), the mapping $\beta_f$ is proper.

The mapping $\alpha_f$ is likewise proper (TG, I, p. 72, prop. 2). It follows that the canonical mapping of Y onto $X\cup_fB$ is proper.