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Let $X$ and $Y$ be locally comapct Hausdorff spaces, and $f:X\to Y$ be a surjective local homeomorphism. When is $f$ a covering map?

It is well-known that when $f$ is proper, $f$ is a covering map. However I think the properness is too strong because there are many example of non-proper covering map (e.g. the universal covering of a space whose fundamental group is not finite is not proper).

Is there any other condition?

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Path-lifting property usually does the job. –  Misha Oct 11 '12 at 11:36
    
I know a covering is a Serre fibration, and, of course, has path-lifting property, but, under the above condition, is the inverse true? Could you tell me how the property works? –  junology Oct 11 '12 at 18:22
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3 Answers 3

up vote 8 down vote accepted

This answer takes a more general viewpoint than Alexandre's. The generality is in response to the small number of assumptions on the spaces involved.

First, you should assume that $Y$ is locally path connected. Only on special occasions does anything good come from looking at coverings of non-locally path connected spaces.

Covering maps are highly structured, so promoting a local homeomorphism (even with the locally compact assumption) to a covering map requires strong conditions. Even if $f$ has the following (rather strong) lifting property, $f$ is not always a covering map: For each map $g:W\to Y$ from a locally path connected space $W$ such that $g_{\ast}(\pi_1(W,w))\subseteq f_{\ast}(\pi_1(X,x))$, there is a unique lift $\tilde{g}:W\to X$, $\tilde{g}(w)=x$ such that $f\tilde{g}=g$.

Here are (necessary) conditions you must have for $f:X\to Y$ to be a covering map.

1) $f$ has unique path lifting (equivalently, if $P(X,x)$ is the set of paths in $X$ starting at $x$, then the induced function $Pf:P(X,x)\to P(Y,p(y))$ is a bijection for each $x\in X$.

2) $f$ has unique lifting of homotopies of paths (this doesn't follow from 1. and you can formulate it in the same way as a bijection on path spaces)

Properties 1) & 2) still aren't enough to promote $f$ to be a covering map. You need to strengthen unique path lifting.

3) $f$ has continuous unique path lifting if $P(X,x)$ has the compact-open topology and the induced map $Pf:P(X,x)\to P(Y,p(y))$ is a homeomorphism for each $x\in X$.

A surjective local homeomorphism with 2) and 3) and $Y$ locally path connected now has the general lifting property listed above but this is still not enough. Such a map is a semicovering (shameless plug - but it's open-access so its ok right?): Semicoverings: a generalization of covering space theory, Homology, Homotopy and Appl. 14 (2012) pp.33-63 ). There are semicoverings of the Hawaiian earring - which is locally compact - that are not coverings (example 3.8 in that paper). On the other hand, a covering map always has properties 2) and 3).

This suggests that in the end you will need to assume $Y$ is semilocally simply connected.

If you have $Y$ locally path connected, semilocally simply connected and $f$ has properties 2) and 3), then $f$ is a covering map. With these new assumptions on $Y$, I suspect you can weaken 3) to 1).

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Here is a proof that I wrote for my differential geometry class, of the fact that if $f:X\to Y$ is a surjective local homeomorphism between semilocally simply connected topological spaces (e.g., manifolds), which satisfies a 'path-lifting property', then $f$ is a covering map. You can also look at do Carmo's book "Differentiable curves and surfaces", page 383.

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In the case of surfaces, the necessary and sufficient condition is the absence of asymptotic values. A curve in $\gamma:[0,1)\to X$ is called an asymptotic curve if the limit $\gamma(t)$ as $t\to\infty$ is $\infty$ (the infinite point of one point compactification of $X$), but $f(\gamma(t))$ has a limit in $Y$ as $t\to 1$. This limit is called an asymptotic value. A covering is a local homeomorphism that has no asymptotic values. The only references for this that I know are for surfaces, but this should not be hard to prove in general under appropriate conditions. Probably $X$ and $Y$ have to be locally connected, in addition to what you said about them.

EDIT. Here is a proof. We assume that $Y$ is connected, and that every point $y\in Y$ has a path connected simply connected base of neighborhoods. Let $f:X\to Y$ be a local homeomorphism.

A path in a space is a continous map from $[a,b]$ or from $[a,b)$ to this space.

We say that the path begins at $\gamma(a)$. Let $\gamma:[a,b]\to Y$ be a path. A lifting of $\gamma$ is a path $\Gamma:[a,b]\to X$ such that $\gamma=f\circ\Gamma$.

If $\Gamma_1$ and $\Gamma_2$ are two liftings of the same path, and $\Gamma_1(a)=\Gamma_2(a)$ then $\Gamma_1=\Gamma_2$. (This is because $f$ is a local homeomorphism).

Let $x_0\in X$ be an arbitrary point and $y_0=f(x_0)$. Assuming that $f$ is a local homeomorphism without asymptotic values, we prove that every path beginning at $y_0$ has a lifting beginning at $x_0$. Let $\gamma$ be a path parametrized by $[a,b]$, beginning at $y_0$. Let $$c=\sup\{ p>0:\gamma\vert_{[a,p]}\;\mbox{has a lifting beginning at}\; x_0\}.$$ As $f$ is a local homeomorphism, we conclude that $c>a$. If $c=b$, the statement is proved. So assume that $c\in(a,b)$. Then, because of the uniqueness of lifting, a semi-open path $\gamma\vert_{[a,c)}$ has a lifting $\Gamma$.

I claim that $\Gamma(t)\to\infty$ as $t\to c$. Indeed, let $x_1\in X$ be a limit point of this path. Let $U$ be a neighborhood of $x_1$ where $f$ is a homeomorphism. Then $V=f(U)$ is a neighborhood of $\gamma(c)$, and we can extend our lifting by putting $\Gamma(t)=\phi(\gamma(t))$, for $t$ in a neighborhood of $c$, where $\phi$ is the inverse to the homeomorphism $f\vert_U$. This extends our lifting slightly beyond $c$, so we obtain a contradiction. Thus $\Gamma(t)\to\infty$ as $t\to c$. Then $\Gamma\vert_{[a,c)}$ is an asymptotic curve and $\gamma(c)$ is an asymptotic value. Again a contradiction.

Thus every path in $Y$ has a unique lifting, and this implies that $f$ is a covering.

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could you please point out where is the reference for the above-mentioned result for surfaces? –  Changyu Guo Oct 11 '12 at 12:45
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