Talk:Covering space

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User:Charles Matthews There is now some overlap between the content here and at local homeomorphism.


The way covering map has been defined allows it not to be surjective (the condition holds vacuously for points with empty pre-image); the usual definition has a covering map being surjective. I think surjective should be added to the definition since that's what is needed for most purposes.

Just noticed that one property that a covering map is supposed to have, according to whoever made the page, is being surjective. So I'll add 'surjective' to the definition.


I removed this paragraph of mine:

The composition of two covering maps need not be a covering map: consider the unit circle S1 as a subset of the complex plane, and for any natural number n define pn : S1 → S1 by pn(z) = z−n. Consider the map p : S1 × N → S1 × N by p(z,n) = (pn(z),n). If N is equipped with the discrete topology and S1 × N carries the product topology, then p is a covering map. The natural projection q : S1 × N → S1 defined by q(z,n) = z is obviously a covering map. The composition qp : S1 × N → S1 is not: no matter how small an open set U you pick in S1, there will always be an n large enough so that pn−1(U) = S1 which cannot be isomorphic to U.

The last statement, pn−1(U) = S1, is false, and that kills the whole argument. I don't know if the composition of two covering maps is always again a covering map. AxelBoldt 14:52, 23 Nov 2003 (UTC)

Huh? I'm often muddle-headed and confused, but ... the last statement is perfectly true. What's false is the statement that p is a covering map. The problem being that p restricted to to the inverse image S1 does not produce a homeomorphism to U. That is, one can always find an n large enough so that pn(S1) is not equal to U; thus p was never a covering to begin with. Changing -n to +n in the definition would make p into a covering. Interesting example, though. Non-trival fundamental group. linas 15:47, 3 Apr 2005 (UTC)
The composition of two covering maps need not be a covering map. To devise a counterexample is an exercise in the Munkres book (which I don't have ready to hand, so I can't give a page citation). It is also an exercise (with a hint) in Allen Hatcher's book Algebraic Topology, which is available online at http://www.math.cornell.edu/~hatcher/AT/ATpage.html. (See exercise 6 on page 79.) The question comes up frequently in discussions at the "Topology Atlas" web site (http://at.yorku.ca/topology/); see e.g. http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraic_topologist;task=show_msg;msg=0002.0001. If p and q are (composable) covering maps and every fiber q-1(x) is finite, then the composition qp is a covering map; see http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2002&task=show_msg&msg=0348.0001 for a proof. --Logician1989 19:41, 10 March 2006 (UTC)

Would someone who knows what is meant by the "opposite" of a group like to make a stub/redirection? I can't find anything on this.

Special case of dual (category theory); anyway like defining g*h = hg.

Charles Matthews 09:06, 26 Feb 2004 (UTC)

Ah yes, of course. I've added links - that ok?

[edit] Deck transformations and fixed points

In the first paragraph of "Deck transformation group, regular covers", it is claimes that a non-trivial deck transformation has no fixed points. I think that's false. Take C to be the disjoint union of X, X and X and p to be the "disjoint union" of id, id and id (in the obvious way). Then you could define f to swap the first to copies of X. This is a deck transformation with all the points of the third copy of X as fixed points.

Maybe the statement is true if C is simply connected. -- Sven

If a topological space is path connected and locally path connected and its covering space Y is path connected, then the statement is true. The statement in Allen Hatcher's algebraic topology text is the following: each point in Y has a neighborhood U such that all the images g(U) are disjoint for varying deck transformations g. That is, the only element of the group of deck transformations that has a fixed point is the identity element. Orthografer 17:24, 26 January 2006 (UTC)

[edit] The Opposite of a Group

Don't you think it's overly pedantic to say the fundamental group of a space is isomorphic to the opposite of the group of deck transformations of its universal cover? After all, every group is canonically isomorphic to its opposite, via

g \mapsto g^{-1}.

In a detailed textbook, it might make sense to introduce the concept of the opposite of a group... but in an encyclopedia article, more people will be confused than helped - since most people will not know that every group is isomorphic to its opposite!

So, I urge that we delete the bit about the opposite of...

John Baez 02:52, 23 March 2006 (UTC)

I agree. In fact, it doesn't even show up in one textbook which I consider pretty detailed, Hatcher's algebraic topology book (searchable pdf). I'm deleting it. Why would something so trivial deserve an entire english word? Those things are getting rare scarce in math... Orthografer 14:19, 23 March 2006 (UTC)

[edit] Universal cover

The author or authors of this note have been cavalier with the hypotheses. The definition of universal cover given here is such that there is no universal cover. For instance let X=\mathbb{R}\times D and let B=\mathbb{R}, where D is any set with the discrete topology. The covering projection is p:X \rightarrow \mathbb{R} is p(x,n) = x. Now choose D of greater cardinality than D and let X'=\mathbb{R}\times D' with projection p':X'\rightarrow \mathbb{R} given by p'(x,n) = x. Notice that p lifts to p' but it is not a covering map because no lift is onto. —Preceding unsigned comment added by 64.6.88.31 (talk • contribs)

Hi, I added the math tags to make it easier to read - revert if this annoys. It seems that the only definition of universal cover is at the beginning of the article, and it is the standard one: a covering space X→B is a universal cover if X is simply connected. In your example, p is not in general a covering map because it does not satisfy the path lifting property (for example when D is an interval, the preimage of an interval could be straight or curvy). In order to address your comment, I would need to know what hypotheses you are referring to. Further, in your example, which of X or X' is a universal cover? Orthografer 04:17, 19 August 2006 (UTC)

I have a technical question. What if a space is path connected and simply connected, but not locally path connected? (You can usually make up an example from something that's connected but not path connected by adding an appropriate path going around). Then technically this space would be the universal cover of itself, yet it is not locally path connected. It contradicts the conditions of having a universal cover, listed here... 146.186.132.190 (talk) 19:23, 29 February 2008 (UTC)