Talk:Proper map

This article is rated Start-class on Wikipedia's content assessment scale. It is of interest to the following WikiProjects:

Mathematics Low‑priority This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.MathematicsWikipedia:WikiProject MathematicsTemplate:WikiProject Mathematicsmathematics Low This article has been rated as Low-priority on the project's priority scale.

Article doesn't say a proper map needs to be continuous. Reference (Lee's Smooth Manifolds) also does not assume continuity. If our proper maps are not continuous, we can't claim the definition is equivalent to a continuous closed map with compact fibers even for Y Hausdorff and locally compact. Maybe we can say these definitions are equivalent for continuous f. Hence the question. Are people even talking about noncontinuous proper maps? (except for Lee) — Preceding unsigned comment added by 147.251.81.229 (talk) 17:03, 10 August 2024 (UTC)Reply

The statement

• If f : X → Y is a proper continuous map and Y is a compactly generated Hausdorff space then f is closed.

is one I proved myself. I'm not aware of a reference, but surely one exists. At any rate, it is an improvement over the prior version (which had first-countable in place of CGHaus) both in generality and correctness (the Hausdorff condition is necessary here). -- Fropuff 07:15, 6 December 2006 (UTC)Reply

"An equivalent, possibly more intuitive definition is as follows: we say an infinite sequence of points {pi} in a topological space X escapes to infinity if, for every compact set S ⊂ X only finitely many points pi are in S. Then a map f : X → Y is proper if and only if for every sequence of points {pi} that escapes to infinity in X, {f(pi)} escapes to infinity in Y."

I'm pretty sure this is only true in a metric space. —Preceding unsigned comment added by 202.161.2.95 (talk) 03:17, 2 June 2008 (UTC)Reply

I agree that it is NOT TRUE the way it is stated, though I'm unsure under what restrictions this is true. For example, the map f:[0,1] -> R that is the identity on [0, 1/3] and [2/3,1] but is identically equal to [1/2] on (1/3, 2/3) is not proper yet it trivially satisfies this sequence condition. Of course this map is not continuous, but it appears that may or may not be part of the definition. Either way the claimed equivalence is false in the stated generality. ````

Another point concerning the equivalent conditions. Does anyone have a reference for the claim that for X locally compact a continuous map f: X → Y is proper iff it is closed and all its fibres are compact? In Bourbaki this equivalence is only proved for X Hausdorff and Y locally compact Hausdorff. TheLaeg (talk) 10:24, 9 February 2011 (UTC)Reply

Don't you people think that the equivalence between a set being compact and the map to a point being proper should be taken out? Is just a redundancy — Preceding unsigned comment added by 85.241.87.82 (talk) 18:42, 13 July 2013 (UTC)Reply

The proof does not seem to use any of the extra assumptions about X or Y (I'm not sure which ones it is supposed to use, as the rest is chaotic...), and it also looks dead wrong to me: in the last passage, it uses ${\displaystyle f^{-1}(\cup _{i=1}^{s}V_{k_{i}})\subset \cup _{\lambda \in \Gamma }U_{\lambda }}$, when in fact the opposite inclusion is true, I think (at any rate, the inclusion is not justified). — Preceding unsigned comment added by 87.99.27.160 (talk) 03:22, 20 February 2016 (UTC)Reply