Thread: 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

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, e₀ lies in only one of them. so when we pick the "slice" V_α to restrict p to (for the interval [0,s₁]), we pick the one containing e₀. 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 = S¹, suppose we choose e₀ = 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:

U₁ = p((-π/2, π/2))
U₂ = p((0,π))
U₃ = p((π/2,3π/2))
U₄ = p((π,2π))

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

we can pick s₁ = 3/16, so that h([0,3/16]) ⊂ U₁.
we can pick s₂ = 7/16, so that h([3/16,7/16]) ⊂ U₂.
we can pick s₃ = 11/16, so that h([7/16,11/16]) ⊂ U₃.
we can pick s₄ = 15/16, so that h([11/16,15/16]) ⊂ U₄.

finally, we see that h([15/16,1]) ⊂ U₁.

now let's look at p^-1(U₁), 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 U₂ 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 U₁, now we want the interval in p^-1(U₁) 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.

Thanks for the help and guidance, Deveno

Will now work through the post

Peter

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?

oh my goodness, you are correct! we need, of course p(h(s₁)) to be in U₁, 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 U₁.

we should be able to continue then with s₂ = 3/16, because h([1/16,3/16]) then lies in U₂.

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

[3/16,5/16] <--> U₃
[5/16,7/16] <--> U₄
[7/16,9/16] <--> U₁
[9/16,11/16] <--> U₂
[11/16,13/16] <--> U₃
[13/16,15/16] <--> U₄
[15/16,1] <--> U₁, as before.

which wraps around the circle twice, as i intended.

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