Definition of covering space and local homeomorphisms

Question

The definition of a covering space of some path-connected, locally path-connected, Hausdorff space $Y$ is given by Bredon (Topology and Geometry) as follows:

A path-connected, locally path-connected, Hausdorff space $X$ with some continuous map $p\colon X \to Y$ s.t. for every $y\in Y$ exists a path-connected neighbourhood $U\ni y$ (called elementary neighbourhood, not necessarily open) s.t. $\varnothing \neq p^{-1}(U)= \bigsqcup U_\alpha$, where the $U_\alpha$ (sheets) are the path-connected components of $p^{-1}(U)$. Furthermore all restrictions $f \vert_{U_\alpha}\to U$ must be homeomorphisms.

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It is stated that the covering map $p$ is a local homeomorphism, that is for all $x\in X$ exists a neighbourhood $V\ni x$ s.t. $p\vert_V\colon V\to p(V)$ is a homeomorphism and $p(V)$ is a neighbourhood of $p(x)$. Under the assumption that $p^{-1}(U)$ is locally path-connected, I could proof that: In this case the $U_\alpha$ are open (as path-connected components of $p^{-1}(U)$) and for any $x\in X$ we consider an elementary neighbourhood $U \subset Y$ of $p(x)$. Now $x$ is contained in some sheet $U_\alpha$ over $U$. As an open set this is a neighbourhood of $x$ and $U_\alpha$ is mapped homeomorphically to $U$ — a neighbourhood of $p(x)$.

1. Is it true that $p^{-1}(U)$ is locally path-connected and how could one prove that?

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I have now some questions about assumptions made in different definitions. What I want is that $p^{-1}(U)$ decomposes in these $U_\alpha$ which are path-connected components and coincide with the connected components. This would be fulfilled if $p^{-1}(U)$ was locally path-connected. I also want to adress the assumptions made about the sheets $U_\alpha$ and the elementary neighbourhoods $U$:

• Assume the $U_\alpha$ are path-connected and disjoint. I don't think this is enough for them to be the path-connected components of $p^{-1}(U)$.

• Assume the elementary neighbourhoods $U$ are open and the $U_\alpha$ are the path-connected components of $p^{-1}(U)$. Then $p^{-1}(U)\subset X$ locally path-connected as an open subset of a locally path-connected space. In particular the $U_\alpha$ are open subsets of $p^{-1}(U)$ and even of X.

• Now we drop the assumption that the $U_\alpha$ are path-connected components and just assume they are open (in $p^{-1}(U)$), disjoint and path-connected. In this case I think that they are still the path-connected components of $p^{-1}(U)$. 2. How can one proof this?

• If we even drop the assumption that the elementary neighbourhood $U$ has to be open, I don't see why $p^{-1}(U)$ is locally path-connected.

• 3. If we take the definition of Bredon, is $p^{-1}(U)$ locally path-connected? This would make the further assumptions from above unnecessary.

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In summary, it seems to me that the following properties are crucial:

• The elementary neighbourhoods $U$ must be open
• The $U_\alpha$ are path-connected components of $p^{-1}(U)$

or equivalently

• The $U_\alpha$ are path-connected, disjoint open subsets of $p^{-1}(U)$.

Are my thoughts correct? Many thanks in advance.

Answer 1

Let us fist literally quote Bredon's definition of a covering map:

A map $p : X \to Y$ is called a covering map (and $X$ is called a covering space of $Y$) if $X,Y$ are Hausdorff, arcwise connected, and locally arcwise connected, and if each point $y \in Y$ has an arcwise connected neighborhood $U$ such that $p^{-1}(U)$ is a nonempty disjoint union of sets $U_\alpha$ (which are the arc components of $p^{-1}(U)$) on which $p \mid_{U_\alpha}$ is a homeomorphism $U_\alpha \xrightarrow{\approx} U$. Such sets $U$ will be called elementary, or evenly covered.

There are a number of issues with this definition.

The first point is that usually one does not make any assumptions on the spaces $X$ and $Y$. However, the requirements "arcwise connected" and "locally arcwise connected" are standard prerequisites in many important theorems about coverings, and thus I would regard it as acceptable to incorporate them into the definition. Why Bredon restricts to Hausdorff spaces it not clear to me.

A more important point is Bredon's concept of "neighborhood" given in Definition 2.3. He explicitly emphasizes

Note that a neighborhood is not necessarily an open set.

This makes some things more complicated than working only with open neigborhoods, in particular in the definitions of "locally connected" and "locally arcwise connected". See Definition of locally pathwise connected for a discussion of various definitions of "locally pathwise connected" (which Bredon calls "locally arcwise connected").

The essence is that each point of a locally arcwise connected space in the sense of Bredon has a neigbhorhood base consisting of arcwise connected open sets.

You can use this fact to prove

Theorem. A map $p : X \to Y$ is a covering map if and only if for every $y\in Y$ there exists a arcwise connected open neighbourhood $U$ of $y$ such that $p^{-1}(U)$ is a nonempty disjoint union of sets $U_\alpha$ on which the restrictions $p \vert_{U_\alpha}: U_\alpha\to U$ are homeomorphisms.

The most serious issue is the use of the phrase "disjoint union". In his book Bredon does not really give a precise definition of what it means that a space $Z$ is the disjoint union of subspaces $Z_\alpha$.

He uses the phrase in Definition 4.1, in Problem 5.(b) on p.8 and in Problem 2. on p.9. This suggests that he simply means a partition of a space into disjoint subsets.

However, in Definition 8.11 he introduces the concept of the topological sum or disjoint union $X + Y$ of two spaces $X,Y$. He then says

In ordinary parlance, if $X$ and $Y$ are disjoint spaces, one regards $X + Y$ as $X \cup Y$ with the topology making $X$ and $Y$ open subspaces.

So, what does "disjoint union of sets $U_\alpha$" mean in his covering map definition? Actually he missed to give the following definition:

A space $Z$ is the disjoint union (or topological sum) of subspaces $Z_\alpha$ if

1. $Z = \bigcup Z_\alpha$.
2. $Z_\alpha \cap Z_\beta = \emptyset$ for $\alpha \ne \beta$.
3. The topology of $Z$ agrees with the final topology on $Z$ induced by the family of inclusions $i_\alpha : Z_\alpha \hookrightarrow Z$. In the present situation this means that a subset $U \subset Z$ is open in $Z$ iff $U \cap Z_\alpha$ is open in $Z_\alpha$ for all $\alpha$.

It is easy to see that 3. is equivalent to the requirement that all $Z_\alpha$ are open subspaces of $Z$. Requirements 1. and 2. say that the $Z_\alpha$ form a partition of $Z$.

For example, in Theorem 12.11 we have a disjoint union of open subspaces; thus it is okay to understand it in the naive sense of a partition.

In the definition of a covering map it is definitely inadequate to understand "disjoint union" in the sense of "partition". Here is an example:

Let $p : \mathbb R^2 \to \mathbb R, p(x,y) = x$, be the projection on the the first coordinate. The spaces are Hausdorff, arcwise connected and locally arcwise connected. Take $U = \mathbb R$ and $U_y = \mathbb R \times \{y\}$. These sets form a partition of $p^{-1}(U) = \mathbb R^2$ and are mapped by $p$ homeomorphically onto $U$.

But the $U_y$ do not form a disjoint sum decomposition of $p^{-1}(U)$. In fact, $p$ is certainly no covering map.

With the correct understanding of "disjoint union" Bredon's definition implies that the sheets $U_\alpha$ over a (not necessarily open ) $U$ are arcwise connected because they are homeomorphic to the arcwise connected $U$. This moreover implies that the $U_\alpha$ are the arc components of $p^{-1}(U)$. In fact, let $A_\alpha$ be the arc component containing $U_\alpha$. Assume that $U_\alpha \subsetneqq A_\alpha$. Then $A_\alpha$ cannot be connected because it is the union of the nonempty disjoint open subsets $U_\alpha$ and $A_\alpha \cap \bigcup_{\beta \ne \alpha} U_\beta$. This contradicts the fact that $A_\alpha$ is connected.

If we work with open $U$ (see the above Theorem), then clearly $U$ and all $U_\alpha$ are locally arcwise connected.

If we allow non-open $U$, then $U$ and the $U_\alpha$ need not be locally arcwise connected.

Answer 2

Local path-connectivity of $p^{-1}(U)$ is a trivial consequence of the fact $p$ restricts to a homeomorphism on each sheet. Before I discuss that, let's dispense with the definition of Bredon (which assumes too much).

"Real" definition of covering map:

A continuous function $p:X\to Y$ is a covering map if every element $y\in Y$ has an open neighbourhood (called "evenly covered") $U$ such that $p^{-1}(U)$ is a disjoint union of open sets $U_i$ such that $p|_{U_i}:U_i\to U$ is a homeomorphism.

Note no assumptions on $X$ or $Y$'s topology are needed here (path-connected, Hausdorff,...). A perhaps simpler perspective is to note that this is equivalent to saying, there exists a discrete space $F$ and a homeomorphism $h:p^{-1}(U)\cong U\times F$ such that the projection $\pi_1:U\times F\to U$ makes $p=\pi_1\circ h$. This perspective shows covering maps are just locally trivial maps with discrete fibres.

An important observation is that if $V\subseteq U$ is another neighbourhood of $y$, then $V$ is also evenly covered; this is simply because $p|_{p^{-1}(V)\cap U_i}:p^{-1}(V)\cap U_i\to V$ is a homeomorphism. Bredon doesn’t demand the evenly covered neighbourhoods be open but the two definitions are equivalent once you interpret ‘disjoint union’ as explained in Paul Frost’s answer; if $U$ is an evenly covered neighbourhood but it is not open, it contains an open subset $V$ (containing the point of interest) which must be evenly covered and you find a partition of $p^{-1}(V)$ into sheets. These sheets must all be open in $X$ since each sheet is open in the subspace $p^{-1}(V)$ (disjoint union topology) and $p^{-1}(V)$ is open in $X$.

It's true that if $Y$ is Hausdorff, then so is $X$ (the converse is false). For your question, we have the important observation that $X$ is locally path connected if and only if $Y$ is. Why? Because this is of course a local property and $Y$ locally looks similar to (several copies of) $X$.

Say $Y$ is locally path connected. Then for $x\in X$, $p(x)$ has an open evenly covered neighbourhood $U$. Let $W$ be any open neighbourhood of $x$ and denote by $U'$ the sheet containing $x$.

$W\cap U'$ is an open neighbourhood of $x$ and since $p|_{U'}:U'\to U$ is a homeomorphism $p(W\cap U')$ is open in $U$; because $U$ is open, $p(W\cap U')$ is open in $Y$ too.

There exists an open path-connected $V\subseteq p(W\cap U')\subseteq U$ which contains $p(x)$, since $Y$ is locally path connected. Denote $V':=p^{-1}(V)\cap U'$, which is an open neighbourhood of $x$. As $p$ is a homeomorphism on $U'$, the path connectivity of $V$ implies the path connectivity of $V'$. Since any (open) neighbourhood $W$ of $x$ contains a path connected neighbourhood $V'$ of $x$, $X$ is locally path connected.

Open subsets of a locally path connected space inherit the local path connectivity, so in particular $p^{-1}(U)$ is locally path connected for any open subset $U$ of $Y$. Assuming $Y$ is locally path connected, that is.

About your second question about the sheets $U_\bullet$ being the path components of $p^{-1}(U)$ when $Y$ is locally path connected. In my more general definition, this isn't always true. However we can arrange it to be true; we can always choose an evenly covered neighbourhood $U$ with this property. First observe that since the $U_\bullet$ are disjoint and open, the sheets 'disconnect' $p^{-1}(U)$ and it must be the case that each path-component of $p^{-1}(U)$ is contained in a unique sheet. To prove the claim it then suffices to show we can arrange for each sheet to be path connected.

That's easy because $U$ contains a path connected subneighbourhood $V$; $V$ is evenly covered and each sheet of $V$ is homeomorphic to $V$ so each sheet is path connected. So, $V$ is such a neighbourhood as desired.