Corollary 2

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

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

If the square (9) is cartesian for every $i\in I$ , the square ( $9'$ ) is cartesian.

From Prop. 5 one deduces a cartesian square

B'\times \prod_i{X'_i}^{Id_B\times}\prod_{_i}^{f'_i}B\times \prod_iX_i

B'\times (B')^{If\times}\prod_{_i}^fB\times B^I

Let $\Delta_{B'}$ and $\Delta_B$ denote the diagonals of $B'\times (B')^I$ and $B\times B^I$ . The diagram

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

\Delta_{B'}\Delta_B

deduced from the preceding one by passing to the subspaces is cartesian (I, p. 9, Prop. 4). It is identified with diagram ( $9'$ ).

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

Nothing.

Cites

Nothing.