Preservation of Local Compactness

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In summary, local compactness is a topological property that guarantees each point in a space has a compact neighborhood. Preservation of this property is important in ensuring the continuity of functions between different spaces. A compact space automatically preserves local compactness, and it can be proven or disproven by examining the compactness of each point's neighborhood or by comparing it to a known locally compact space. Examples of spaces that preserve local compactness include Euclidean spaces, compact spaces, and metric spaces.
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Euge
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Let ##f : X \to Y## be a surjective continuous closed map of topological spaces such that every fiber ##f^{-1}(y)## is compact. Show that ##Y## is locally compact if ##X## is locally compact.
 
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Doesnt this have to see with Saturated subsets of quotient maps, IIRC?
 
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WWGD said:
Doesnt this have to see with Saturated subsets of quotient maps, IIRC?
I don't understand the question, could you please clarify?
 
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Euge said:
I don't understand the question, could you please clarify?
Apologies, I remember the concept of saturated subsets playing a role in proofs when dealing with fibrations and coverings. Let me try to make the argument more precise.
 
  • #5
I spent a long time on this before noticing you said f was a closed map! It doesn't seem like this has gotten a lot of traction, but it's really just chaining all the topology definitions together.

given ##y\in Y##, we need to find ##y\in U\subset K## with ##K## compact and ##U## open.

Since ##f## is subjective, there exists ##x\in X## such that ##f(x)=y##. ##X## is locally compact so there exists ##x\in V\subset L## with ##V## open and ##L## is compact.
##y\in f(V)\subset f(L)##. Since ##f## is continuous, ##f(L)## is compact. But ##f(V)## may not be open. The rest of the proof fixes this issue.

##X-V## is closed, and ##f## is a closed map, so ##f(X-V)## is closed. So ##Y-f(X-V)## is an open set, and it's a subset of ##f(L)##: if ##z\in Y-f(X-V)##, we know that ##f(a)=z## for some ##a##, and ##f(a)\notin f(X-V)## means ##a\in V## which we know is contained in ##L##. So we're close. But if ##f## is many to 1, this set may be too small, e.g. ##y## might be in ##f(X-V)##.

The final correction uses the compactness of $##f^{-1}(y)##. Let ##x_\alpha## be the pre image of ##y##, indexed by some set ##A## for notational convenience (the index set can literally be the pre image). For each ##\alpha##, ##x_\alpha \in V_\alpha \subset L_\alpha## with each ##V_\alpha## open and ##L_\alpha## compact. The ##V_\alpha##s are an open cover of ##f^{-1}(y)##, so there is a finite subcover, which is indexed by a finite set ##M\subset A##. ##X-\bigcup_M V_\alpha## is a closed set such that ##y\notin f(X-\bigcup_M V_\alpha)##. So ##y\in Y-f(X-\bigcup_M V_\alpha)## which is an open set, that is contained in ##f(\bigcup_M L_\alpha)##, by similar argument in the one index case. But ##\bigcup_M L_\alpha## is a finite union of compact sets so is compact, and then applying ##f## gives a compact subset of ##Y##.
 
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