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Algebraic Topology - Fundamental Group and Liftings - Munkres CH 9

In Section 54 of his book "Topology" on the Fundamental Group of the Circle, Munkres presents the following Lemma (Lemma 54.1 - see attachment)

**Lemma 54.1**

Let be a covering map, let .

Any path beginning at has a unique lifting to a path in E beginning at

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**[My question relates to the proof of the uniqueness of ]**

Munkres begins the proof as follows:

**Proof:**

Cover B by open sets U each of which is evenly covered by p.

Find a subdivision of [0.1], say , such that for each i the set lies in an open set U (Use Lebesgue number lemma).

We define the lifting step by step.

First define .

Then supposing is defined for we define ] on as follows:

The set lies in some open set U that is evenly covered by p.

Let be a partition of into slices; each is mapped homeomorphically onto U by p.

Now lies in one of these sets. ... ... etc etc ... see attachement

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**But now focussing on the uniqueness of - couldn't we define as belonging to any of the sets and make this work? **

So any of the sets would do since the covering is even.

Then the argument for uniqueness would follow (see page 342 of attachment) - so we would only have one for each of the - but this is not may idea of a unique .

Can someone please help clarify Munkres argument regarding the uniqueness of

Peter

Re: Algebraic Topology - Fundamental Group and Liftings - Munkres CH 9

the uniqueness derives from the fact that we are considering "evenly-covered" open neighborhoods U in B. this means the pre-images p^{-1}(U) are disjoint.

in particular, e_{0} lies in only one of them. so when we pick the "slice" V_{α} to restrict p to (for the interval [0,s_{1}]), we pick the one containing e_{0}. so not just any of the slices will do.

when we extend the lifted map to "the next subinterval", we pick the ONLY pre-image of that U that contains .

for example, given the path h(s) = (cos(4πs),sin(4πs)) in B = S^{1}, suppose we choose e_{0} = 0 in E = R, under the covering map:

p(x) = (cos(x),sin(x)).

for our open cover of the circle, all we need to do is choose "arc-lengths" less than 2pi radians. so let's use:

U_{1} = p((-π/2, π/2))

U_{2} = p((0,π))

U_{3} = p((π/2,3π/2))

U_{4} = p((π,2π))

now we need to pick our subintervals of [0,1].

we can pick s_{1} = 3/16, so that h([0,3/16]) ⊂ U_{1}.

we can pick s_{2} = 7/16, so that h([3/16,7/16]) ⊂ U_{2}.

we can pick s_{3} = 11/16, so that h([7/16,11/16]) ⊂ U_{3}.

we can pick s_{4} = 15/16, so that h([11/16,15/16]) ⊂ U_{4}.

finally, we see that h([15/16,1]) ⊂ U_{1}.

now let's look at p^{-1}(U_{1}), which is:

only one of these intervals contains the real number 0, namely the one with k = 0. so that's where we start, and since

we have:

on (-π/2,π/2), p is injective, so we see that on [0,3/16], that

(it's the injectivity of p on (-π/2,π/2) that allows us to uniquely "undo" the trig functions).

so that gives us the lift of h on [0,3/16]. to extend this to the next sub-interval [3/16,7/16],

we want to find the interval in the pre-image of U_{2} that contains ,

which is (0,π). again we see that , on this interval as well.

we continue in this way, obtaining a path in R that goes from 0 to 4π. the only interesting part is when we pick the slice for the last pre-image.

note that although we want a pre-image of the same open set U_{1}, now we want the interval in p^{-1}(U_{1}) that contains ,

which is no longer (-π/2,π/2) but rather (7π/2,9π/2) instead (k = 2, instead of k = 0).

yes, there are other paths in R that p would map to h, but this is the only continuous one.

Re: Algebraic Topology - Fundamental Group and Liftings - Munkres CH 9

Thanks for the help and guidance, Deveno

Will now work through the post

Peter

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Re: Algebraic Topology - Fundamental Group and Liftings - Munkres CH 9

I am working through Deveno's example - but I have a problem (which may well be an error I am making)

This is annoying since the example illustrates the theorem very well - can anyone help with this possible error?

**Problem: It does not appear to me that as is required**

My working is as follows: (See attachment Example for Munkres - Topology - Lemma 54.1)

To start we have (see Deveno's post above)

p(x) = ( cos x, sin x )

) )

Now

**Now check that **

which is not in --- Problem!!!

For diagrams see attachment Example for Munkres - Topology - Lemma 54.1

Can anyone help?

Re: Algebraic Topology - Fundamental Group and Liftings - Munkres CH 9

oh my goodness, you are correct! we need, of course p(h(s_{1})) to be in U_{1}, so my subinterval is "way too big" (i forgot about the factor of 4).

changing the first to: [0,1/16] we then have: h([0,1/16]) lies in U_{1}.

we should be able to continue then with s_{2} = 3/16, because h([1/16,3/16]) then lies in U_{2}.

it appears the subintervals (and corresponding U's) should go as follows:

[3/16,5/16] <--> U_{3}

[5/16,7/16] <--> U_{4}

[7/16,9/16] <--> U_{1}

[9/16,11/16] <--> U_{2}

[11/16,13/16] <--> U_{3}

[13/16,15/16] <--> U_{4}

[15/16,1] <--> U_{1}, as before.

which wraps around the circle twice, as i intended.

Re: Algebraic Topology - Fundamental Group and Liftings - Munkres CH 9

Thanks so much for the revised set of subintervals and Ui

will now rework the problem

So helpful to have an example like this ... I wish the various texts had more such examples or that there was supplementary texts with such examples ... Would make it so much easier for those working alone ... I wish algebraic topology had the equivalent of Project Crazy Project

thanks again

peter