Proposition 8

Tag 01M1, permanent. Cite it as [Bourbaki, Tag 01M1].

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

Let

\begin{array}{ccc} X' & \overset{f'}{\longrightarrow} & X \\ {\scriptstyle p'}\big\downarrow & & \big\downarrow{\scriptstyle p} \\ B' & \overset{f}{\longrightarrow} & B \end{array}

be a cartesian square. If the mapping $p$ is open ( resp. is proper, resp. has in a neighbourhood of each point a continuous section), the same is true of $p'$ .

By remark 2, I, p. 16, it is enough to prove the proposition for cartesian squares of the following type:

\begin{array}{ccccccc} \overset{-1}{p}(A) & \overset{j}{\longrightarrow} & X & & X\times F & \overset{\mathrm{pr}_1}{\longrightarrow} & X \\ {\scriptstyle p_A}\big\downarrow & & \big\downarrow{\scriptstyle p} & \text{and} & {\scriptstyle p\times \mathrm{Id}_F}\big\downarrow & & \big\downarrow{\scriptstyle p} \\ A & \overset{i}{\longrightarrow} & B & & B\times F & \overset{\mathrm{pr}_1}{\longrightarrow} & B \end{array},

where F is a topological space, A a subspace of B, and $i,j$ the canonical injections. If the mapping $p$ is open, the mappings $p_A$ and $p\times$ Id $_F$ are open (TG, I, p. 30, prop. 2 and TG, I, p. 34, prop. 8). If the mapping $p$ is proper, the mappings $p_A$ and $p\times$ Id $_F$ are proper (TG, I, p. 72, prop. 3 and TG, I, p. 72, def. 1). If U is an open subset of B and $s: U\rightarrow X$ a continuous section of $p$ over U, the mapping $s|(A\cap U)$ is a continuous section of $p_A$ over $A\cap U$ and the mapping $s\times$ Id $_F$ is a continuous section of $p\times$ Id $_F$ over $U\times F$ .

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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Cites

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