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Let $X\subset \mathbb P^n$ be a projective variety, $B$ a projective variety, and $V\subset B\times \mathbb P^n$ a family over $B$. Denote the fiber over a point by $V_b$. Exercise 4.4 in Harris's Algebraic Geometry: A First Course asks the reader to show the subset

$$\{b \in B : X\subset V_b\}$$

is closed in $B$. This seems to be equivalent to showing that the projection map from the second factor is open. I don't know how to prove that fact without invoking concepts not yet developed in the book (see here, for example).

Is there an elementary way to do this problem?

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Potato
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1 Answers1

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It is clear that the variety $X \subset \mathbb{P}^n$ has no role in the claim we make. So let us try to prove the statement first for a point $x \in \mathbb{P}^n$. Consider the closed set $F_x = \{(b,x) \mid b \in B \} \subset X \times \mathbb{P}^n$. We have $F_x \cap V = \{(b,x) \mid (b,x) \in V \}$, which is closed in $F_x$. On the other hand the map $F(x) \rightarrow B$ induced by projection is a homemorphism. Thus the image of $F(x) \cap B$ in $B$, which is the set we are interested in, is closed in $B$... call it $B(x)$. Now we have $ \{b \in B \mid X \subset V_b \} = \bigcap _{x \in X} B(x) \subset B$ is closed as it is an intersection of closed sets.

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