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Let $X,B,B'$ be topological spaces. Let $\pi\colon X\to B$ be a continuous, proper map (proper in the sense of Bourbaki-proper: closed and with quasi-compact fibres).

Let $f\colon B'\to B$ be continuous. Form the fibre product $X\times_{B} B'$. I have to prove that the projection map $\pi'\colon X\times_{B}B'\to B'$ is proper.

The fibre $\pi^{-1}(b')=\{(x,b')\in X\times B':\pi(x)=f(b') \}$ is quasi-compact, since it is homeomorphic to the quasi-compact fiber $\pi^{-1}(f(b'))=\{x\in X\mid \pi(x)=f(b') \}$ via the map $(x,b')\mapsto x$.

However, I do not know how to prove that $\pi'$ is closed. Clearly, we need to use the fact that the fibers of $\pi$ and/or those of $\pi'$ are quasi-compact, since the property of being closed is in general not preserved under base change. I am failing to do so, any help is appreciated.

rae306
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1 Answers1

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Let $A' \subseteq X \times_B B'$ be closed. We need to show that $\pi'(A') \subseteq B'$ is closed. Let $b' \in B' \setminus \pi'(A')$. If $x \in \pi^{-1}(f(b'))$ then $(b',x) \in (X \times_B B') \setminus A'$. If it were in $A'$ then $b' = \pi'(b',x) \in \pi'(A')$, contradiction.

Since $A'$ is closed, we can find open subsets $U_x \subseteq B'$ and $V_x \subseteq X$ such that $b' \in U_x$ and $x \in V_x$ and $(U_x \times V_x) \cap A' = \emptyset$. Since the fiber $\pi^{-1}(f(b'))$ is compact, there is a finite set $F$ of points $x$ in the fiber such that $V_x$ with $x$ in $F$ cover the fiber. The open set $V = \bigcup_{x \in F}V_x$ contains that fiber, and the open set $U = \bigcap_{x \in F}U_x$ contains $b'$ and $(U \times V) \cap A' = \emptyset$. Since $\pi$ is closed the subset $W = B \setminus \pi(X \setminus V)$ is open and it contains $f(b')$ ($\pi^{-1}(f(b')) \subseteq V$, so $f(b') \not\in \pi(X \setminus V)$).

By continuity of $f$ there is an open subset $U' \subseteq U$ such that $b' \in U'$ and $f(U') \subseteq W$. We are done if we can show that $U' \cap \pi'(A') = \emptyset$. Assume that this is not the case then there exists $u' \in U'$ and $x \in X$ such that $(u',x) \in A'$. Also $f(u') \in W$ and $\pi(x) = f(u')$, so that $\pi(x) \in W$. This means $x \in V$. Therefore $(u',x) \in (U' \times V) \cap A' \subseteq (U \times V) \cap A' = \emptyset$, contradiction.

Orb
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