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Can someone post a proof of the statement that if $X$ is compact then the covering map $q:E\rightarrow X$ is finitely sheeted given that $E$ is compact as well.

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    Take a local trivialisation over a finite cover of $X$, hence an open cover of $E$ which has a finite subcover. Then since I presume you are using classical logic, a subset of a finite set is finite...2012-11-06
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    Possible [duplicate](http://math.stackexchange.com/questions/157208/if-pe-to-b-is-a-covering-space-and-p-1x-is-finite-for-all-x-in-b-s/157224#157224).2012-11-06
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    @JacobSchlather - not quite, this is the implication E+X compact => finite sheets. The other was finite sheets implies both or neither of E and X are compact Hausdorff.2012-11-06
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    @DavidRoberts I'm not sure what a 'local trivialisation' is?2012-11-06
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    http://en.wikipedia.org/wiki/Fiber_bundle2012-11-06
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    There should be some other way without mention of local trivialisation...I guess I'm looking for an answer that relies only on the elementary definitions and properties of covering maps.2012-11-06
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    The definition of a covering map includes a local trivialisation, just minus the assumption that all fibres are isomorphic.2012-11-06
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    Alternatively (and assuming Hausdorff): $ E_x = q^{-1}(x) $ is closed and so compact in $E$, and is also discrete, since $q$ is a covering. Hence $E_x$ is finite.2012-11-06
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    @ Ronnie Brown: I don't think we can assume Hausdorff2012-11-06

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