Is $f$ a proper map if the fibers are compact?

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In summary, a map is considered proper if the inverse image of any compact set is compact, and the fibers of a proper map must also be compact sets. A map cannot be proper if its fibers are not compact, and to determine if a map is proper, one must check that the inverse image of any compact set is also a compact set. Additional conditions for a map to be proper include being continuous and satisfying the Hausdorff condition.
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Euge
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Here is this week's POTW:

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Let $f : X \to Y$ be a closed map of topological spaces such that the fibers $f^{-1}(y)$ are compact for every $y\in Y$. Prove that $f$ is a proper map, i.e., $f^{-1}(K)$ is a compact subset of $X$ for every compact subset $K$ of $Y$.
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  • #2
This week's problem was solved correctly by janssens. You can read his solution below.
The given statement is quite natural from the point of view of set-valued analysis. (There was ahttps://mathhelpboards.com/analysis-50/multi-value-function-25137.html about that topic, too.)

Let $g : Y \twoheadrightarrow X$ be the inverse image correspondence for $f$,
$$
g(y) := \{x \in X\,:\, f(x) = y\}, \qquad y \in Y.
$$

(Note that $g$ may be empty-valued if $f$ is not surjective, which we explicitly allow.)

Closedness of $f$ implies (in fact, is equivalent to) upper hemicontinuity of $g$. Moreover, the given compactness of the fibers of $f$ is equivalent to compact-valuedness of $g$. So, $g$ is a compact-valued, upper hemicontinuous correspondence and therefore maps compact subsets of $Y$ to compact subsets of $X$. By definition of $g$, this shows that $f$ itself is proper.

Remarks:

1. In order to avoid ambiguity, I use the definition of upper hemicontinuity from $\S$17.2 of Aliprantis and Border, Infinite Dimensional Analysis, 3rd edition, 2007. This is a beautiful book for mathematicians and economists alike. Accordingly, a correspondence between two topological spaces is upper hemicontinuous if the upper inverse image of any open set is open. It is one of a number of co-existing conventional definitions.

2. The compactness theorem used above is nothing more than the direct analogue of the corresponding (!) theorem for ordinary, continuous functions. In particular, ordinary functions are (as correspondences) singleton-valued, hence compact-valued.
 

FAQ: Is $f$ a proper map if the fibers are compact?

What does it mean for a map to be proper?

A map is considered proper if the inverse image of any compact set is compact. In other words, the preimage of a compact set must be a compact set.

What are fibers in the context of proper maps?

Fibers refer to the inverse image of a single point in the codomain of the map. In the case of a proper map, the fibers must be compact sets.

Can a map be proper if the fibers are not compact?

No, a map cannot be considered proper if the fibers are not compact. This is because the definition of proper maps requires the fibers to be compact sets.

How do you determine if a map is proper based on its fibers?

To determine if a map is proper, you must check that the inverse image of any compact set is also a compact set. If this condition is satisfied, then the map is considered proper.

Are there any other conditions that must be met for a map to be proper?

Yes, in addition to the fibers being compact, a proper map must also be continuous and satisfy the Hausdorff condition. The Hausdorff condition states that for any two distinct points in the domain, there must exist disjoint open sets containing each point respectively.

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