Proposition 11

Let

${X'}^{f'}$ X

$p'p$

${B'}^f$ B

a cartesian square.

a) Suppose the mapping $f$ strict and surjective. Then, if the mapping $p'$ is open ( resp. closed, resp. proper), the same is true of $p$.

b) Suppose that the mapping $f$ is closed and surjective. Then, if the mapping $p'$ is strict, the same is true of $p$.

c) Suppose that the mapping $f$ is universally strict and surjective. Then, if the mapping $p'$ is strict, the same is true of $p$.

Let us first remark that for every subset A of X, one has

In fact, if $x'\in (\overset{-1}{f}')(A)$, then $f(p'(x')) =p\circ f'(x')\in p(A)$. Conversely, if $b'\in \overset{-1}{f}(p(A))$, let $x\in A$ be such that $f(b') =p(x)$. By definition of a cartesian square, there exists a unique $x'\in X'$ such that $f'(x') =x$ and $p'(x') =b'$, so that $b'\in p'((\overset{-1}{f}')(A))$.

Let us prove a). Suppose first that $p'$ is an open (resp. closed) mapping and let A be an open (resp. closed) set in X. The set $(\overset{-1}{f}')(A)$ is open (resp. closed) in $X'$. Hence the set $p'((\overset{-1}{f}')(A))$ is open (resp. closed) in $B'$. By relation (20), it is also saturated with respect to the equivalence relation defined by $f$. Since the mapping $f$ is supposed surjective, one then has $p(A) =$

$f(p'((\overset{-1}{f}')(A)))$, and since it is strict, the set $p(A)$ is open (resp. closed) in B. The mapping $p$ is therefore open (resp. closed).

In order that a continuous mapping be proper, it is necessary and sufficient that it be closed and that its fibres be quasi-compact (TG, I, p. 75, th. 1). If the mapping $p'$ is proper, the mapping $p$ is closed by what precedes. Let us study its fibres: if $b\in B$, let $b'\in B'$ be such that $f(b') =b$. By the corollary of I, p. 10, the mapping $f'$ induces a homeomorphism $X'_{b'}\rightarrow X_b$. Since $p'$ is proper, $X'_{b'}$ is quasi-compact, hence $X_b$ is so also. The mapping $p$ is therefore proper.

Let us prove b) and c). Let Y = $p(X)$ and $Y'=p'(X')$ ; relation (20) applied to X implies that $Y'=\overset{-1}{f}(Y)$. Let $g$ denote the mapping of $Y'$ into Y induced by $f$. In case b), the mapping $g$ is strict by remark 1, I, p. 18. In case c), it is strict by virtue of definition 6 since the square

$Y'$ Y

$B'$ B

is cartesian.

Let $q$ and $q'$ denote the mappings of X into Y and of $X'$ into $Y'$ induced by $p$ and $p'$, so that the square

${X'}^{f'}$ X

$q'q$

${Y'}^g$ Y

is cartesian. Since the mappings $g$ and $q'$ are both strict and surjective, their composite $g\circ q'$ is strict. Hence $q\circ f'$ is strict and $p\circ f'$ is itself strict. Since $f$ is surjective, $f'$ is so also and $p$ is strict by prop. 9, b) of I, p. 18.