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I am trying to understand the proof of the following:

Theorem: For loops $\alpha$, $\beta$ in $S^1$ with base point $1=(1,0)$, $[\alpha]=[\beta]$ if and only if $\mbox{deg}(\alpha)=\mbox{deg}(\beta)$.

I'll reproduce the forward direction of the proof (I'm fine with the other direction) in its entirety so everything makes sense, highlighting where I'm having trouble. This is from Principles of Topology by Croom.

Suppose first that $[\alpha]=[\beta]$ so that $\alpha$ and $\beta$ are equivalent loops in $S^1$. Let $F:I\times I \to S^1$ be a homotopy demonstrating the equivalence of $\alpha$ and $\beta$: $$F(\cdot, 0)=\alpha,\quad F(\cdot,1)=\beta$$ $$F(0,s)=F(1,s)=1, \,\, s \in I.$$

The Covering Homotopy Property insures the existence of a unique covering homotopy $\tilde{F}$ of $F$ such that $\tilde{F}(0,0)=0$. For $s$ in $I$, $p\tilde{F}(0,s)=F(0,s)=1$, so $\tilde{F}(0,s)$ must be an integer...

I am fine with all of the above. It is the next section that I'm just not seeing.

...Since $I$ is connected, the same integral value must occur for each value of $s$. Since $\tilde{F}(0,0)=0$, then $\tilde{F}(0,s)=0$ for all $s$ in $I$. The same argument shows that $\tilde{F}(1,s)$ must have the constant integral value $\tilde{F}(1,0)$ for all $s$ in $I$...

Could someone explain the implication of the connectedness of $I$? And whatever that value is, is it the same for $\tilde{F}(0,s)$ and $\tilde{F}(1,s)$?

...The uniqueness of covering paths insures that $\tilde{\alpha}=\tilde{F}(\cdot,0)$ and $\tilde{\beta}=\tilde{F}(\cdot,1)$ are the unique covering paths of $\alpha$ and $\beta$ which begin at $0$. Thus


How do we see the middle equation above, i.e. $\tilde{F}(1,0)=\tilde{F}(1,1)$?

Thus equivalent loops must have the same degree.

Perhaps more thinking will produce some clarity on the matters above, but it hasn't so far, and I certainly do value the insights that the people here tend to provide that I may not be able to see myself, given my limited experience.

As always, thanks for the help.

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What is $p$ ? A covering map from what space to what space? –  user38268 Feb 12 '13 at 3:14
@BenjaLim: Sorry about that. $p:\mathbb{R} \to S^1$ is given by $p(t)=(\cos 2\pi t,\sin 2\pi t)$, $t \in \mathbb{R}$. –  Alex Petzke Feb 12 '13 at 4:03

1 Answer 1

The continuous image of a connected space is connected. Since the integers are discrete, its connected components are singletons. Thus, the image of a continuous map from a connected space to the integers is a singleton.

Thus the restricted map $s \mapsto \tilde F(0,s)$ from $[0,1] \to \mathbb{Z}$ takes just one value, which is $0$ by assumption. This says that the initial point of all the intermediate paths between $\tilde \alpha$ and $\tilde \beta$ has to be the same $0$.

The same holds of the restricted map $s \mapsto \tilde F(1,s)$, so that the final point of all the paths in the homotopy is also the same integer. In particular, this gives you your "middle equation" $\tilde F(1,0) = \tilde F(1,1)$ showing that the other endpoints of $\tilde \alpha$ and $\tilde \beta$ (the degrees) are the same.

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