Bourbaki

Example 4

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

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

Let (Bi)i∈I(B_i)_{i\in I} be a family of topological spaces and let, for each i∈Ii\in I, (Xi,pi)(X_i, p_i) and (Yi,qi)(Y_i, q_i) be BiB_i-spaces. Put B=∏i∈IBiB =\prod_{i\in I}B_i, X=∏i∈IXiX =\prod_{i\in I}X_i and Y=∏i∈IYiY =\prod_{i\in I}Y_i. Endowed with the continuous mapping p=∏ipip=\prod_ip_i (resp. q=∏iqiq=\prod_iq_i), the topological space X (resp. Y) is a B-space. By the associativity isomorphism of the topological products of ∏i(Xi×Yi)\prod_i(X_i\times Y_i) onto (∏iXi)×(∏iYi)=X×Y(\prod_iX_i)\times (\prod_iY_i) = X\times Y (TG, I, p. 25, prop. 2), the subspace ∏i(Xi×BiYi)\prod_i(X_i\times_{B_i}Y_i) of ∏i(Xi×Yi)\prod_i(X_i\times Y_i) is identified with X×BYX\times_BY.

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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Machine translation of the English, by gpt-5.4, against glossary version 34. Not checked by a person.