I have two doubts strictly related to each other.

1) Firstly, consider the $8$-figure, namely the union of two circles in a point $x_1$. Using the Seifert-Van Kampen's theorem I proved that its fundamental group is the free group on two generators. Now say $a$ the closed path in $x_0$ around only the left circle and say $b$ the closed path in $x_0$ around only the right circle. How can I prove that the classes of $a$ and $b$ are different and that they generate the fundamental group?

2) Secondly, I want to prove that the fundamental group of the torus is $\mathbb Z\times\mathbb Z$ using the Seifert-Van Kampen's theorem. Consider the torus as the square with identified opposite edges. Say $U=T-\{y\}$ where $y$ is a point inside the square and say $x_1$ is the vertex of the identified square. Now it's ok for me that $U$ is a retract of the $8$-figure and so we know that $\pi_1(U,x_1)$ is equal to the free group on two generators. My problem is that even if I assume my point 1) I don't understand why: "classes of $a$ and $b$ generate the fundamental group of $8$-figure implies they generate $\pi_1(U,x_1)$ too". Of course now $a$ is the closed path in $x_1$ on the horizontal edge and $b$ is the closed path in $x_1$ on the vertical edge.

From this point I know how to complete the exercise.

  • $\begingroup$ 1) This is basically what the Seifert vanKampen theorem says. 2) $\pi_1(A\times B)=\pi_1(A)\times\pi_1(B)$ holds for path connected (or pointed) spaces and easy. What product is the torus? $\endgroup$ – Berci Jun 18 '16 at 23:58
  • $\begingroup$ 1) I can't see this as obvious 2) I know the formula and that $T=S^1\times S^1$ but I wanted to use the S. VK theorem as I stated. $\endgroup$ – Richard Jun 19 '16 at 0:10

The Seifert-Van Kampen theorem does not just give you the abstract fact that the figure 8 has fundamental group $\mathbb{Z} * \mathbb{Z}$. It gives you an actual formula for an isomorphism. So, look carefully at the proof you say you have, look carefully at the formula for the isomorphism given by the Seifert-Van Kampen theorem, write down the isomorphism that the theorem gives you for this particular problem, and look at the images in $\mathbb{Z} * \mathbb{Z}$ of $a$ and $b$ under that isomorphism. The formula for the isomorphism should show you that $a$ maps to a generator of the first $\mathbb{Z}$ and $b$ to a generator of the second $\mathbb{Z}$ in $\mathbb{Z} * \mathbb{Z}$.

  • $\begingroup$ Problem 1). Say $X=$figure $8$, $U=$right circle $+$ (left circle minus a point different from $x_1$), and similarly $V=$left circle$+$(right circle minus a point different from $x_1$). Now we have the inclusions: $i_1:U\to X$ and $i_2:V\to X$. These induce $\phi: U*V\to X$ defined by $g_1g_2\mapsto i_1(g_1)i_2(g_2)$ that can be extended to the homomorphism $f:\pi_1(U)*\pi_1(V)\to \pi_1(X)$ In this particular case the Theorem says that $f$ is surjective.Now note that $U$ is a retract of the only right circle and so its fundamental group is $\mathbb Z$.But now I have my problem: $\endgroup$ – Richard Jun 19 '16 at 10:39
  • $\begingroup$ We know that the generator of the right circle is the class of the closed path around $x_1$. Call it the class of $a$. How can I deduce that the class of $a$ is also the generator of the retraction of the circle, namely $U$? This seems a property of the functions of retraction and not of the isomorphism of Van Kampen. I hope it is clear now. $\endgroup$ – Richard Jun 19 '16 at 10:42
  • $\begingroup$ You're on the right track, but you must use the induced homomorphism of the retraction $U \to a$. You haven't drawn out all the conclusions that you need to regarding the behavior of this induced homomorphism. $\endgroup$ – Lee Mosher Jun 19 '16 at 14:22
  • $\begingroup$ First, to be clear, the retraction maps $U$ to $a$, not the other way around. Second, the retraction $r_1 : U \to a$ is more than just an ordinary retraction, it is a deformation retraction. Like any deformation retraction, the map $r_1$ is a homotopy equivalence, its homotopy inverse is the inclusion map $i_1 : a \to U$, and those two maps induce inverse isomorphisms $(r_1)_* : \pi_1(U,p) \to \pi_1(a,p)$ and $(i_1)_* : \pi_1(a,p) \to \pi_1(U,p)$ (where $p$ is the crossing point of the figure 8). $\endgroup$ – Lee Mosher Jun 19 '16 at 14:23
  • $\begingroup$ Since $i_1$ is the inclusion map, it follows that the isomorphism $(i_1)_* : \pi_1(a,p) \to \pi_1(U,p)$ takes the class of the loop $a$ in $\pi_1(a,p)$, which I'll denote as $[a]_a$, to the class of the loop $a$ in $\pi_1(U,p)$, which I'll denote as $[a]_U$. Since $[a]_a$ is a generator of the infinite cyclic group $\pi_1(a,p)$, it follows that $[a]_U$ is a generator of the cyclic group $\pi_1(U,p)$. $\endgroup$ – Lee Mosher Jun 19 '16 at 14:26

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