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I have some confusion about the statement in Allen hatcher Book

Page No:$33$

Theorem $1.10:$ For every contnious map $f:S^2 \to \mathbb{R}^2$ there exist a pair of antipodal points $x$ and $-x $ in $S^2$ with $f(x)=f(-x)$

In the theorem of the proof it is written that

In particular, we have $$\widetilde{h}(1)= \widetilde{h}(1/2) + q/2=\widetilde{h}(0) +q $$.This means that $h$ represent q times a generator of $\pi_1(S^1)$

Here im not getting the meaning of q times a generator of $\pi_1(S^1)$

I'm thinking two meanings

My thinking $1$: we know that $\pi_1(S^1) \cong \mathbb{Z}$ and $\mathbb{Z}$ has only two generator $-1$ and $ +1$

q times a generator of $\pi_1(S^1)$ mean generator of $\mathbb{Z}$ are $-q$ and $+q$

Thinking $2$ :$h$ represent q times a generator of $\pi_1(S^1)$ mean

$\widetilde{h}: [0,1] \to \mathbb{R}$ defined by $\widetilde{h}(s)=qs$ which is simply a straight line from $0$ to $q$

I don't know which one is correct

jasmine
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    Thinking 2 is close: $(h : [0, 1] \to S^1) = (p : \Bbb{R} \to S^1) \circ (\widetilde{h} : [0, 1] \to \Bbb{R})$, where the map $p$ is what is called the universal cover of $S^1$ by $\Bbb{R}$ and is described by Hatcher a few pages earlier. – Rob Arthan Jun 07 '21 at 23:50
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    $h$ is a loop in $S^1$ and $\tilde{h}$ is its lift to the universal cover? Context helps. – Thorgott Jun 07 '21 at 23:52
  • ok@RobArthan that mean generator is a degree of $h$ Am i right ? – jasmine Jun 08 '21 at 00:38
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    From a purely algebraic point of view, every abelian group $A$ (such as $\pi_1 S^1$) has a natural $\mathbb Z$-module structure. For example if $n$ is a positive integer then $n \cdot a$ is just $a$ added to itself $n$ times (with a definition by induction if you want to be formal). – Lee Mosher Jun 08 '21 at 00:50
  • thanks for that @LeeMosher sir that mean q times a generator of $\pi_1(S^1)$ implies loops passing around the circle $q.n$ times. Am i right ? – jasmine Jun 08 '21 at 01:04
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    that's correct. If $q>0$, that is. Oh, and if $n=1$. – Lee Mosher Jun 08 '21 at 01:06

1 Answers1

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We know that there is an isomorphism $\phi : \pi_1(S^1,1) \to \mathbb Z$. This isomorphism is given as follows:

Take a loop $h : I \to S^1$ sich that $h(0) = h(1) = 1 = e^{0} \in S^1$. Lift it to $\tilde h : I \to \mathbb R$ such that $\tilde h (0) = 0$; then $\phi([h]) = \tilde h (1) \in \mathbb Z$. Thus the group $\pi_1(S^1,1)$ is infinite cyclic: It has one generator which we can choose to be either $g_1 = \phi^{-1}(1)$ or $g_{-1}\phi^{-1}(-1)$. The "standard generator" is of course $g_1$ which is represented by $\omega_1 : I \to S^1$.

$q$ times a generator $g_i$ is the element $q \cdot g_i = g_i + \ldots +g_i$ ($q$ summands). Note that the definition $q\cdot g$ works in any abelian group.

Paul Frost
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