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Let $X$,$Y$ be topological space, surjective map $\varphi:X\rightarrow Y$ is called a covering map if there is an open cover $\{U_{\alpha}\}$ of $Y$ such that for every $\alpha$, $\varphi^{-1}(U_{\alpha})$ is a disjoint union of open sets in $X$, each of which is mapped by $\varphi$ homeomorphically onto $U_{\alpha}$.

Question. Let $X, Y$ be compact metric spaces,

Is there is $k$ such that $\varphi$ is $k$- to- $1$ at all points?

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    $\begingroup$ Well, it is definitely false if you don't assume some connectedness. $\endgroup$ – Shubhodip Mondal May 9 '16 at 14:13
  • $\begingroup$ If you assume that X is pathconnected, then it is true. The compactness gives you the finiteness of the fibre. $\endgroup$ – Shubhodip Mondal May 9 '16 at 14:16
  • $\begingroup$ @ShubhodipMondal Given that they're metric spaces, just connectedness is enough. $\endgroup$ – David C. Ullrich May 9 '16 at 14:18
  • $\begingroup$ I think in any case, as far as this result is concerned, connectedness alone is enough? $\endgroup$ – Shubhodip Mondal May 9 '16 at 14:26
  • $\begingroup$ @ShubhodipMondal If you mean connected+compact, as opposed to path-connected+compact, then yes. If you mean connected without compactness then it seems to me yes, if you allow $k$ to be an infinite cardinal: It's still true that for every $k$ the set of $x$ with a fiber of cardinality $k$ is open, and since this holds for every $k$ it's also closed. $\endgroup$ – David C. Ullrich May 9 '16 at 14:39
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Obviously not if $Y$ is not connected; you could have a covering of this part of $Y$ that simply has nothing to do with a covering of that part of $Y$.

Yes, if $Y$ is connected. First you have to show that the inverse image of each point is finite (hint: If $x_j\ne x_k$ for $j\ne k$, $\phi(x_j)=y$ for all $j$ and $x_j\to x$ then $\phi$ cannot be a covering map).

Now for every $k$ the set of points of $Y$ that have exactly $k$ inverse images is open. Since this set is open for every $k$ it follows that it is closed...

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This is not true if $X$ is not connected. But is true if $X$ is connected since the function $g$ defined on $X$ such that $g(x)$ is the cardinal of the fiber of $x$ is continuous.

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    $\begingroup$ Well, I think you should mention that the "cardinal" is finite because of compactness. $\endgroup$ – Shubhodip Mondal May 9 '16 at 14:17
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If you assume everything is connected and compact then the fiber over any point $y\in Y$ is given as the orbit of the deck-group transformation which is discrete and hence finite.

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