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In algebraic geometry, flat morphisms are usually associated with the intuition that they have "continuously varying fibers". Is there a notion in topology formalizing the same intuition? Consider for example a map $p \colon P \to X$ of good (locally compact Hausdorff, say) topological spaces and assume for the moment that it is proper (i.e. has compact fibers). If the fibers vary continuously, then the following property should be true for example:

(*) If $f \colon P \to \mathbf{R}$ is a continuous function, then the induced function $$\widetilde{f} \colon X \to \mathbf{R}, \qquad \widetilde{f}(x) := \sup_{q \in p^{-1}(\{x\})}f(q)$$ is also continuous.

This is true for example if the fibers are constant, i.e. if $p$ is a trivial bundle $p \colon X \times F \to X$. In that case, the map $f \colon X \times F \to \mathbf{R}$ induces a continuous map $X \to C(F, \mathbf{R})$, the space of continuous functions equipped with the compact-open topology (which is the topology of uniform convergence) and the map $\widetilde{f}$ factors as $X \to C(F, \mathbf{R}) \xrightarrow{\sup} \mathbf{R}$.

But there are also examples where the fibers are not (locally) constant: For example the map $\mathbf{C} \to \mathbf{C}$, $z \mapsto z^2$ has continuously varying fibers in the intuitive sense (as $z$ approaches the origin, the two points in the fiber get closer to each other). This is also a flat map in algebraic geometry.

For an example which is not flat in the intuitive sense and where also the above property is clearly wrong, consider the first projection $$p \colon \{(x,y) \in \mathbf{R} \times [0,1] \mid x \cdot y = 0 \} \to \mathbf{R}.$$ (This example is of course also inspered by algebraic geometry.)

I should perhaps add that I am aware of the related questions The topological analog of flatness? and Flatness in Algebraic Geometry vs. Fibration in Topology. But I am interested in the purely point-set-topological setting and in fact I care specifically about the property (*) (I have a specific example of a continuous map $p$ in mind for which I hope (*) to be true and I'm trying to gain some intuition.)

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  • $\begingroup$ I think the only reasonable topological analog is that of an open map. Note that open maps are closed under base-change in the topological category. It's a fun question if open proper maps satisfy your condition * and I wish I had energy to think about it right now... $\endgroup$ Commented Sep 8, 2021 at 18:23
  • $\begingroup$ @DustinClausen Thank you so much! I could verify my property (*) for proper surjective open maps of topological spaces (as long as the definition of „proper” includes „closed”, I don't need any condition on the spaces). I also found a not-so-easy theorem which guarantees that my map $p$ is open, so my initial problem is solved. $\endgroup$ Commented Sep 11, 2021 at 15:59
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    $\begingroup$ @JakobWerner Great, I'm glad my comment was helpful and you could solve your problem. About why I think openness is the only reasonable topological analog of flatness, I'm not really sure how to make precise this feeling of mine. But in several examples I've seen the notion of openness play the same role as the notion of flatness. And the only general topological consequence of flatness that I'm aware of is openness. $\endgroup$ Commented Sep 12, 2021 at 11:23
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    $\begingroup$ @JakobWerner you could write an answer to your own question outlining the solution, crediting Dustin for the comment that helped you. Then accept it, and the software will record this problem is solved. $\endgroup$ Commented Nov 25 at 6:52
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    $\begingroup$ @DavidRoberts Thanks, I added an answer. $\endgroup$ Commented Nov 25 at 10:16

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As suggested by David Roberts, here is the proposition I ended up proving, thanks to Dustin's remarks in the comments:

Recall that a map $p \colon P \to X$ of topological spaces is proper if and only if it is closed and has quasi-compact fibers.

Proposition. Let $p \colon P \to X$ be a surjective, proper and open map of topological spaces and let $\phi \colon P \to \mathbf{R}$ be a continuous function. Then the induced function $$\widetilde{\phi} \colon X \to \mathbf{R}, \qquad x \mapsto \sup_{y \in p^{-1}(x)} \phi(y)$$ is well-defined and continuous.

Proof. Since the fibers $p^{-1}(x)$ are quasi-compact, the function $\phi$ attains a supremum (in fact, a maximum) on $p^{-1}(x)$, so the map $\widetilde{\phi}$ is well-defined.

Let $x_0 \in X$ with $s_0 := \widetilde{\phi}(x_0) = \sup_{y \in p^{-1}(x_0)} \phi(y) \in \mathbf{R}$. Let $\epsilon > 0$ and consider the standard neighborhood $U_\epsilon := \{ t \in \mathbf{R} \mid |s_0 - t| < \epsilon \}$ of $s_0 \in \mathbf{R}$. We will construct a neighborhood $U_0$ of $x_0 \in X_0$ such that $\widetilde{\phi}(x) \in U_\epsilon$ for $x \in U_0$.

Consider an arbitrary $a \in p^{-1}(x_0) \subset P$ and the open neighborhood $$ V(a) := \{ y \in P \mid |\phi(y) - \phi(a)| < \epsilon / 2 \} $$ of $a$ in $P$. We have $p^{-1}(x_0) \subset \bigcup_{a \in p^{-1}(x_0)} V(a)$, so by compactness there exists a finite subset $A \subset p^{-1}(x_0)$ such that $p^{-1}(x_0) \subset \bigcup_{a \in A} V(a)$. Now let $U_0 \subset X$ be an open neighborhood of $x_0$ such that

  1. $p^{-1}(U_0) \subset \bigcup_{a \in A} V(a)$,
  2. for $x \in U_0$ and $a \in A$ we have $p^{-1}(x) \cap V(a) \neq \emptyset$.

For example, we can take $U_0 := X \setminus p(P \setminus \bigcup_{a \in A} V(a)) \cap \bigcap_{a \in A} p(V(a))$. This set is open since $p$ is both open and closed.

We claim that this $U_0$ works. Indeed, given $x_1 \in U_0$, let $s_1 := \widetilde{\phi}(x_1) \in \mathbf{R}$. We write $s_0 = \phi(y_0)$, $s_1 = \phi(y_1)$ for some $y_0 \in p^{-1}(x_0)$, $y_1 \in p^{-1}(x_1)$. We have to show that $| s_0 - s_1 | < \epsilon$. By 1., we have $y_0 \in V(a_0)$, $y_1 \in V(a_1)$ for some $a_0, a_1 \in A$. We know that $p^{-1}(x_0) \cap V(a_1) \neq \emptyset$ (because $a_1$ is contained in this set) and $p^{-1}(x_1) \cap V(a_0) \neq \emptyset$ by 2.. Hence we can pick some $b_0 \in p^{-1}(x_1) \cap V(a_0)$ and $b_1 \in p^{-1}(x_0) \cap V(a_1)$. Then we have $$ \begin{split} s_0 - s_1 &= \phi(y_0) - s_1 \\ &= \phi(y_0) - \phi(b_0) + \underbrace{\phi(b_0) - s_1}_{\le 0 \text{ by def. of } s_1} \\ &\le |\phi(y_0) - \phi(b_0)| \\ &\le \underbrace{|\phi(y_0) - \phi(a_0)|}_{\le \epsilon / 2 \text{ since } y_0 \in V(a_0)} + \underbrace{|\phi(a_0) - \phi(b_0)|}_{\le \epsilon / 2 \text{ since } b_0 \in V(a_0)} \\ &\le \epsilon. \end{split} $$ By a symmetric argument, we also get $s_1 - s_0 \le \epsilon$. ∎

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    $\begingroup$ Here is an alternative argument that might point to some useful related literature: Viewed as a set-valued map, the upper hemicontinuity of $p^{-1}$ is equivalent to $p$ being closed, and the lower hemicontinuity to $p$ being open. The result follows then from Berge's maximum theorem. $\endgroup$ Commented Nov 27 at 8:12

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