0

The book I'm using states the following version of Van-Kampen Theorem:

Van-Kampen Suppose that linearly connected cell complexes $X$, $Y$ intersect along their common connected subcomplex $Z = X \cap Y$. Then a presentation of the group $\pi_1(X \cup Y)$ can be obtained as follows:

  1. Write down some generators of $\pi_1(X)$ and $\pi_1(Y)$.
  2. Write down relations for $\pi_1(X)$ and $\pi_1(Y)$.
  3. Write down one more series of relations, one for each generator $c$ of the group $\pi_1(Z)$. This relation has the form $\varphi_1(c) = \varphi_2(c)$, where $\varphi_1(c)$ is an expression of $c$ in the generators of $\pi_1(Y)$.

By linear connected I think he mean path connected.

After this he asks to calculate the fundamental group of the figure eight. So, we have to find the fundamental group of $S^1 \vee S^1$. By the definition of wedge sum, we take a disjoint union modulo a relation identifying to base points. So, it makes sense that $$\pi_1(S^1 \vee S^1) \simeq \Bbb{Z} \ast \Bbb{Z}.$$ But to apply this version of Van-Kampen we need to see $S^1 \vee S^1$ as a cell complex with subcomplexes $X,Y$ under the hypothesis of the theorem such that $|X \cap Y|=1$. How can I do that?

Greg
  • 466
  • 1
    Have you looked carefully at the definition of a cell complex and some examples? To be clear, the following is meant in a positive and not negative way: Figuring out a CW complex structure for a figure 8 is an exercise you should practice and figure out if you're going to work with CW complexes at all. You need to start by defining how many points are in the 0-skeleton, how many line segments are in the 1-skeleton, how many discs are in the 2-skeleton... – user176372 Jun 05 '24 at 18:12
  • @user176372 I'm not sure if it is correct, but if $X$ and $Y$ are cell complex, then it is the disjoint union $X \cup Y$. For the wedge sum, we choose two point $x \in X$ and $y \in Y$ (which are $0$-cells) and we collapse the subcomplex formed by $x,y$. This collpase is a cell complex and is the wedge sum – Greg Jun 05 '24 at 19:01
  • @user176372 but which are the $X$ and $Y$ used to apply Van Kampen? I this $X$ is a $1$-cell of $S^1 \cup S^1 - {x,y}$ with the new $0$-cell and the same for $Y$. Both intersecting on the new $0$-cell. But I'm not sure if it is correct – Greg Jun 05 '24 at 19:06
  • @Greg You only have 3 cells, among which there is a unique pair that covers the space. What other choice could you possibly make? :) – Ben Steffan Jun 05 '24 at 21:36
  • @BenSteffan I'm not sure if I understand. We take two copies of $S_1^1$ and $S_2^1$. Then we take their disjoint union $S_1^1 \cup S_2^1$. After this we choose two points $x_1 \in S_1^1$ and $x_2 \in S_2^1$ and collpase the subcomplex ${x_1,x_2}$ to a point $x_0$. So, $x_0$ is a new $0$-cell. The cells of $S_1^1 \vee S_2^1$ are the cells of $S_1^1 \cup S_2^1 - {x_1,x_2}$ join with $x_0$. So I think we have two arcs connected by a single point. Then one arc plus $x_0$ join with other arc plus $x_0$ covers the space and they intersect in ${x_0}$. – Greg Jun 05 '24 at 22:17
  • @BenSteffan I think the intuition may be right but I'm not sure if this is formally correct. – Greg Jun 05 '24 at 22:18
  • 1
    @Greg I have trouble following the issue. You have correctly identified that you have one 0-cell and two 1-cells (the arcs), so you apply van Kampen to $X$ one arc and $Y$ the other, what else could you do? At some point you need to jump :) Or is the issue that about closure? The 1-cells here are "closed," in the sense that they contain whatever they're attached to, in this case the $x_0$, whence each arc identifies with a copy of $S^1$. – Ben Steffan Jun 05 '24 at 22:26
  • @BenSteffan yes, the point is about the $1$-cells to be closed. When I consider $X$, for example, I have to take one arc in $S_1^1$ minus one point plus the new point $x_0$, is it? – Greg Jun 05 '24 at 22:40
  • Just the arc and $x_0$, there is no need to take out points anywhere. In $S^1 \vee S^1$ the arc is already attached to $x_0$. – Ben Steffan Jun 05 '24 at 22:49
  • 1
    You do not need a CW-structure on $S^1 \vee S^1$. See https://math.stackexchange.com/q/4625240. – Paul Frost Jun 06 '24 at 14:24

0 Answers0