Corollary 1

Tag 01LS, permanent. Cite it as [Bourbaki, Tag 01LS].

Topologie algébrique, chapter I, § 1. Produits fibrés et carrés cartésiens

Let X be a B-space, let $p$ be its projection, and let F be a topological space. The square

\begin{array}{ccc} X\times F & \overset{\mathrm{pr}_1}{\longrightarrow} & X \\ {\scriptstyle p\times \mathrm{Id}_F}\big\downarrow & & \big\downarrow{\scriptstyle p} \\ B\times F & \overset{\mathrm{pr}_1}{\longrightarrow} & B \end{array} \tag{8}

is cartesian.

Let P be a topological space consisting of a single point. Corollary 1 follows from Prop. 5 applied to the cartesian squares

\begin{array}{ccccccc} X & \overset{\mathrm{Id}_X}{\longrightarrow} & X & & F & \longrightarrow & P \\ {\scriptstyle p}\big\downarrow & & \big\downarrow{\scriptstyle p} & \text{and} & {\scriptstyle \mathrm{Id}_F}\big\downarrow & & \big\downarrow{\scriptstyle \mathrm{Id}_P} \\ B & \overset{\mathrm{Id}_B}{\longrightarrow} & B & & F & \longrightarrow & P \end{array}.

Let B and $B'$ be topological spaces and let $f: B'\rightarrow$ B be a continuous mapping. Let I be a set and, for each $i\in I$ , let $X_i$ be a B-space, $X'_i$ a $B'$ -space, and $f'_i: X'_i\rightarrow X_i$ a continuous mapping such that the square

{X'_i}^{f'_i}X_i

(9)

${B'}^f$ B

be commutative. There exists a unique continuous mapping

f':\prod_{i\in IB'}X'_i\rightarrow \prod_{i\in IB}X_i

such that pr $_i\circ f'=f'_i\circ$ pr $_i$ for every $i\in I$ and such that the square

$i\prod\in IB'X'if'i\prod\in I$ B $Xi$

(9')

${B'}^f$ B be commutative (this last condition resulting from the others if I $=\not\emptyset$ ) $:$ it is the mapping deduced from the mapping

f\times \prod_{i\in I}f'_i: B'\times \prod_{i\in I}X'_i\rightarrow B\times \prod_{i\in I}X_i

by passing to the subsets. With these notations:

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

Cited by

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Cites

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