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I have a question about the torus:

Let $X = S^{1} \times S^{1}$, and $p$ is standard covering map ($p : \mathbb{R} \to S^{1} , p(x) = ( \cos(2\pi x) , \sin(2\pi x) )$). Restrict $p \times p : \mathbb{R} \times \mathbb{R} \to S^{1} \times S^{1}$ to $h : I^{2} \to S^{1} \times S^{1}$. Let $a_{0}(t)=(t,0)$, $b_{0}(t)=(0,t)$, $A = h(\partial I^{2})$ (boundary of $I^{2}$ where $I=[0,1]$) so why $[h(a_{0}(t))]$ and $[h(b_{0}(t))]$ are two paths form a system of free generators for $\pi_{1}(A,a)$ where $a = h(0,0)$?

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    Your question is very confused and I don't understand what you are asking. I suppose $R$ is bad notation for $\mathbb{R}$, and $(\cos 2\pi,\sin 2\pi)(x)$ is bad notation for $(\cos(2\pi x),\sin(2\pi x))$. But what is $A=h(BdI^2)$ supposed to mean? The class $[h(a_0)]$ is taken with respect to which equivalence relation? Please edit your question – 2017-02-02
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    ok , $Bd I^{2}$ means boundary of $I^{2}$ where $I=[0,1]$ and $h(a_{0})$ means $h(a_{0}(t))$ . Sorry . – 2017-02-02
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    Are you satisfied with my answer? – 2017-02-05
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    Why every loop based at point $p$ can be homotoped to either $a$ or $b$ . Even if it was happen , why $a,b$ are generators of $\pi_{1}(A,p)$ – 2017-02-06
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    Every loop can be homotoped to either $a$,$b$ or contracted to the point $p$ (i.e. homotopic to the constant map $c_{p}$, whose homotopy class is $0$). There are essentially three types of loops you can draw on $S^1\wedge S^1$ (i.e. the union of $a$ and $b$). 1) loops that go through either $a$ or $b$ without completing the circle (these can be clearly contracted to $p$). 2) loops that go around $a$ completing at least one revolution (those can be homotoped to a multiple of $a$) 3)loops that go around $b$ completing at least one revolution (which can be homotoped to a multiple of $b$) – 2017-02-06
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    $[a]$ and $[b]$ are the generators of $\pi_1(A,p)$ precisely because given any class $[\gamma]\in\pi_1(A,p)$, where $\gamma$ is a loop starting at $p$, either $[\gamma]=0$ (case 1) above), $[\gamma]=n[a]$ , for some $n\in\mathbb{Z}$ (case 2) above), or $[\gamma]=m[b]$ for some $m\in\mathbb{Z}$ (case 3) above). – 2017-02-06
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    A more conceptual way of proving this is to use additivity of $\pi_1$ with respect to the wedge product: $\pi_1(S^1\vee S^1)=\pi_1(S^1)\times\pi_1(S^1)=\mathbb{Z}\times\mathbb{Z}$. From this you conclude that $\pi_1(S^1\vee S^1)=\pi_1(A,p)$ has two generators, and the argument above shows that $a$ and $b$ are those generators. – 2017-02-06
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    Sorry in my first (long) comment I meant $S^1\vee S^1$ not $S^1\wedge S^1$. I can't edit it anymore – 2017-02-06
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    How is it now? Do you understand? – 2017-02-08
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    Yes , I think I tried by myself and understood . I also read your post . It's good . – 2017-02-08
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    Sorry to ask you again, but PLEASE can you accept the answer? It only takes a click... – 2017-02-11

1 Answers 1

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Let $a,b:I\rightarrow S^1\times S^1$ be defined by $a(t):=h(a_1(t))$ and $b(t):=h(a_0(t))$ for $t\in I$.

Let $p:=h(0,0)=(1,0)$. With these definitions you can see that the paths described above correspond to the following picture.

Generators of the fundamental group of the torus

By definition, $A=h(\partial I^2)$ is the image of $a$ and $b$ in the torus. in other words, $A$ is just the wedge product of two circles $S^1\vee S^1$.

By looking at this picture you can make the following observations:

  • The classes $[a]$ and $[b]$ are nonzero in $\pi_1(A,p)$. indeed, they cannot be continuously shrinked to a point since the "hole" of the torus prevents you from doing so.
  • We have $[a]\neq [b]$. Indeed, it is impossible to continuously deform $a$ into $b$ or vice versa due to the hole.
  • Every other single loop based at point $p$ can be homotoped to either $a$ or $b$.

This is why $a$ and $b$ are the generators of $\pi_1(A,p)=\pi_1(S^1\vee S^1,p)$.

I will leave it to you to prove everything formally.