# Algebraic Topology - Liftings - Munkres - Chapter 9

#### Peter

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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 $$p: E \rightarrow B$$ be a covering map, let $$p( e_0 ) = b_0$$.

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

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[My question relates to the proof of the uniqueness of $$\tilde{f}$$ ]

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

We define the lifting $$\tilde{f}$$ step by step.

First define $$\tilde{f} (0) = e_0$$.

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

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

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

Now $$\tilde{f} (s_i)$$ lies in one of these sets. ... ... etc etc ... see attachement

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

Can someone please help clarify Munkres argument regarding the uniqueness of $$\tilde{f}$$

Peter