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?
Necessary and sufficient condition for being a covering map
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general-topology
algebraic-topology
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3You should probably assume that it is surjective. :) Connectedness of $X$ is not needed. The definition can be found on Wikipedia: http://en.wikipedia.org/wiki/Covering_map. – 2012-07-01
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0@mex you should list the definition of the covering map or give us a link in the question. – 2012-07-02
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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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0Surely $f$ must also be a local homeomorphism? – 2012-07-01
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0@Jacob: This was assumed in the OP's question – 2012-07-01
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0@JacobSchlather: You are right, of course. I (implicitely) meant in addition to $f$ being a local homeomorphism. I have edited accordingly. – 2012-07-01
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0@JuanS It was unclear to me, because the OP was already assuming that X was compact as well. – 2012-07-02