Proposition 4

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

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

Let

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

p'p \tag{4}

${B'}^f$ B

be a cartesian square, and let $B_0,B'_0,X_0$ be subspaces of B, $B'$ and X respectively. Suppose that we have $f(B'_0)\subset B_0,p(X_0)\subset B_0$ and put $X'_0=(\overset{-1}{p}')(B'_0)\cap (\overset{-1}{f}')(X_0)$ . Then the square

{X'_0}^{f'_0}X_0

(4')p'_{_0}p_{_0}

{B'_0}^{f_0}B_0

(where the mappings $f_0,f'_0,p_0,p'_0$ are deduced from $f,f',p,p'$ respectively by passing to the subsets) is cartesian.

Consider the canonical mapping $h: X'\rightarrow B'\times_BX$ deduced from commutative diagram (4). Since square (4) is Cartesian, $h$ is a

homeomorphism. By construction, one has $X'_0=\overset{-1}{h}(B'_0\times_{B_0}X_0)$ and the mapping $h_0: X'_0\rightarrow B'_0\times_{B_0}X_0$ deduced from commutative diagram ( $4'$ ) is deduced from $h$ by passing to subsets. It is therefore a homeomorphism and square ( $4'$ ) is Cartesian (Prop. 2).

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.

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