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- Jun 22, 2012

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**Lemma 54.1**

Let [tex] p: E \rightarrow B [/tex] be a covering map, let [tex] p( e_0 ) = b_0 [/tex].

Any path [tex] f : [0.1] \rightarrow B[/tex] beginning at [tex] b_0 [/tex] has a unique lifting to a path [tex] \tilde{f} [/tex] in E beginning at [tex] e_0[/tex]

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**[My question relates to the proof of the uniqueness of [tex] \tilde{f} [/tex] ]**

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 [tex] s_0, s_1, .... s_n [/tex] , such that for each i the set [tex] f([s_i, s_{i+1} ])[/tex] lies in an open set U (Use Lebesgue number lemma).

We define the lifting [tex] \tilde{f} [/tex] step by step.

First define [tex] \tilde{f} (0) = e_0 [/tex].

Then supposing [tex] \tilde{f} (s) [/tex] is defined for [tex] 0 \leq s \leq s_i [/tex] we define [tex] \tilde{f} [/tex]] on [tex] [s_i , s_{i+1}[/tex] as follows:

The set [tex] f([s_i, s_{i+1} ])[/tex] lies in some open set U that is evenly covered by p.

Let [tex] \{ V_{\alpha} \} [/tex] be a partition of [tex] p^{-1} (U) [/tex] into slices; each [tex] \{ V_{\alpha} \} [/tex] is mapped homeomorphically onto U by p.

Now [tex] \tilde{f} (s_i) [/tex] lies in one of these sets. ... ... etc etc ... see attachement

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**But now focussing on the uniqueness of [tex] \tilde{f} [/tex] - couldn't we define [tex] \tilde{f} (s) [/tex] as belonging to any of the sets [tex] \{ V_{\alpha} \} [/tex] 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 [tex] \tilde{f} (s) [/tex] for each of the [tex] \{ V_{\alpha} \} [/tex] - but this is not may idea of a unique [tex] \tilde{f} [/tex].

Can someone please help clarify Munkres argument regarding the uniqueness of [tex] \tilde{f} [/tex]

Peter