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Suppose I have a map $f:X\rightarrow Y$ continuous, $X$ is compact, connected and also $f$ is a local homeomorphism, what condition should we include in $X$ so that $f$ becomes a covering map? Am I making any sense?

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You should probably assume that it is surjective. :) Connectedness of $X$ is not needed. The definition can be found on Wikipedia: en.wikipedia.org/wiki/Covering_map. –  tomasz Jul 1 '12 at 21:17
    
@mex you should list the definition of the covering map or give us a link in the question. –  Paul Jul 2 '12 at 8:12
    
Thank you every one –  Bunuelian Trick Jul 2 '12 at 10:30
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up vote 3 down vote accepted

If $X$ and $Y$ are Hausdorff, then compactness of $X$ and surjectivity of $f: X\to Y$ are already sufficient conditions for $f$ to be a covering map (edit: just to be clear, this is in addition to $f$ being a local homeomorphism). See this question.

Note that $f$ is necessarily surjective if $Y$ is assumed to be connected, because $f$ is closed and open (still under the hypothesis that $Y$ is Hausdorff).

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Surely $f$ must also be a local homeomorphism? –  JSchlather Jul 1 '12 at 23:14
    
@Jacob: This was assumed in the OP's question –  Juan S Jul 1 '12 at 23:50
    
@JacobSchlather: You are right, of course. I (implicitely) meant in addition to $f$ being a local homeomorphism. I have edited accordingly. –  Sam Jul 1 '12 at 23:59
    
@JuanS It was unclear to me, because the OP was already assuming that X was compact as well. –  JSchlather Jul 2 '12 at 0:07
    
Thank you every one –  Bunuelian Trick Jul 2 '12 at 10:30
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