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So I'm trying to understand the proof on page 63

In the proof he says that if $\tilde{f}_{1}(y) \not = \tilde{f}_2 (y)$, then $\tilde{U}_{1} \not = \tilde{U}_2$. But,how is this true? You can't deduce this from anything. Certainly, there must be something magical happening for him to deduce this. Surely, they the maps could be taken the points to different places in $\tilde{U}_1, \tilde{U}_2$, but they are still equal.

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It tends to be nicer to phrase this type of questions more along the line of «I do not understand such line in Hatcher» than «Hatcher seems to be wrong»... :) – Mariano Suárez-Alvarez May 21 '12 at 3:09
I would not normally upvote this question, but while the choice of words is somewhat unfortunate, I can’t see downvoting the question on that account; accordingly, I’ve upvoted. – Brian M. Scott May 21 '12 at 4:10
Prof Hatcher still maintains the book, which seems to be used by everyone and their dog; so it sounds unlikely that there's hocus pocus in the first chapter! – Dylan Moreland May 21 '12 at 4:43

You must use the fact that $\tilde{f}_1$ and $\tilde{f}_2$ are lifts of $f$. If $\tilde{U}_1$ and $\tilde{U}_2$ are the same, then $\tilde{f}_1(y)=\tilde{f}_2(y)$ (i.e. they can't go to different places in the same neighborhood). This is simply because they both map to $f(y)$ under $p:\tilde{X}\to X$, and the neighborhoods were chosen to be small enough so that each $\tilde{U}_\alpha$ contains exactly one preimage of $f(y)$.

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The idea is that different lifts of the function $f$ are determined by the choice of one point.

For example, consider the standard covering of $S^1$ by $\mathbb{R}$ (see page 29 in the text). Let $\gamma:[0, 1]\to S^1$ be the path traversing the circle once in the counter-clockwise direction. Then $\gamma$ can be lifted to any interval $[n, n+1]$ on the real line. Suppose $\gamma_2$ starts at $1$ and spirals up to $2$, and $\gamma_1$ starts at $0$ and spirals down to $1$.

The statement says that if the two lifts do not agree at a point, then the neighborhoods that are mapped homeomorphically to $S^1$ are disjoint.

Think about the pre-images of $-1$ under the covering map, corresponding to $\gamma(\frac{1}{2})$. Then $\gamma_2(\frac{1}{2})=\frac{3}{2}$, while $\gamma_1(\frac{1}{2})=\frac{1}{2}$. In particular, the neighborhoods of these points that map homeomorphically onto the corresponding neighborhood of $\gamma(\frac{1}{2})=-1$ are disjoint.

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