Re: Local Homeomorphisms

On 17.04.2008 12:25, Jannick Asmus wrote:
On 17.04.2008 11:45, William Elliot wrote:

To provide a far simpler one: Z -> Z, x -> 0, Z equipped with the
discrete topology.

Oh, I get it, local homeomorphisms don't have to be injective.

Correct.

I am so sorry for my bad English, I just literally translated the
expression from my mother tongue. I meant a topological covering _map_.

Your English is good. Problem is translating technical expressions.
By your email address, you are German? I'm American.

Yes, I am German.

If U,V are open and U homeomorphic f(U), V homeomorphic f(V),
is not U \/ V homeomorphic to f(U) \/ f(V) = f(U \/ V) ?

The simple example given above shows this is incorrect.

Sorry about the confusion. Yes - if you mean to take *one* of the
counter examples above and patch two open subsets U and V such that the
map on U u V is *not* injective.

Not even if U and V are overlapping?
Do not know what you are saying here. I suspect that you were not having
the counter examples above in mind when you said this.

What if U /\ V is not empty? The simple counter example Z, fails.

I've a simple three point T0 space that's a counter example.
If I warp the line around S^1, ie
f:R -> S^1, x -> (sin x, cos x)

This is one of the prominent examples of covering maps. Note that R is a connected simply connected manifold.

Cf. http://en.wikipedia.org/wiki/Covering_space

BTW: Every covering map X -> Y of connected topological manifolds with Y simply connected is injective.

then I can find a counter example,
U = (-e, pi + e), V = (-pi - e, e)

for most small e. That's because the union isn't injective.

The first example, exp(z) isn't a bijection.
The second example, f:C\0 -> C\0, z -> z^k is a bijection?

No. There are roots of unity different from 1 if k>1, hence f cannot be bijection. If k is even, 1 and -1 are mapped to 1.

I don't know enough about complex analysis to fathom this example.

Consider C as R^2 and have a look at the total differential in every point. This is just calculus. Note that the differential is always invertible ...

However, I doubt that a connected space for which every point is a cut
point is other than simply connected. Hopefully I won't have to include
that hypothesis to get the conclusion that seems likely. That continuous
bijection and perhaps connected will suffice.

The result I'm wishing for is
a bijective local homeomorphism is a homeomorphism.

This is clear: A local homeomorphism is an open map or - in other words - the inversion map f^-1 is continuous.

Locally compact is another property of the domain space.

Apparently not needed.

HTH.

Best wishes,
J.


--
Best wishes,
J.
.

Relevant Pages

  • Re: Local Homeomorphisms
    ... then is f a homeomorphism? ... discrete topology. ... map on U u V is *not* injective. ... bijection from a connected space. ...
    (sci.math)
  • Re: Local Homeomorphisms
    ... The key difference between local homeomorphisms and covering ... + open map is equivalent to Wikipedia's definition. ... Open map seems like a strong extra requirement. ...
    (sci.math)
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