# compute $\pi_{1}(X)$ with cell complex structure.

algebraic topology Hatcher exercise 1.2.7 Let $$X$$ be the quotient space of $$S^{2}$$ obtained by identifying the north and south poles to a single point. Put a cell complex structure on $$X$$ and use this to compute $$\pi_{1}(X)$$

solution :with Example $$0.8$$ on page $$11 .$$ Let $$0 -cell$$ be the image of the quotient of $$N$$ and $$S$$. Let arc $$B$$ connecting the north pole $$N$$ and the south pole $$S$$ be the $$1 -cell$$, and the rest of $$S^{2}$$ $$2-cell$$. That is, we attach the $$2 -cell$$ to $$1 -cell$$ by the map $$\varphi: S^{1}=\partial D^{2} \rightarrow B$$ Let $$U$$ be the arc $$B$$ together with its neighborhood on $$S^{2},$$ and $$V$$ be $$S^{2} \backslash B$$ or the interior of $$D^{2}$$ Then $$\pi_{1}(U)=\mathbb{Z}$$ and let $$\alpha$$ be the generator. It is clear that $$V$$ is simply connected so $$\pi_{1}(V)=0$$ and if we show $$U \cap V$$ is path connected then we can use Van Kampen theorem . Let $$\gamma \in \pi_{1}(U \cap V)$$ be a generator, then $$\varphi_{1}(\gamma)=\alpha \alpha^{-1}=0 .$$ By Van Kampen, $$\pi_{1}(X)=\frac{\pi_{1}(U) * \pi_{1}(V)}{N}=\frac{\mathbb{Z} * 0}{0}=\mathbb{Z}$$

how we can show $$U \cap V$$ is path connected (with draw shape)? is my proof true ?

• Hint: $U \cap V =U \setminus B$ by your definition. If you think about it, it is homeomorphic to a punctured disk... Jun 9 '20 at 16:40

If you like to be pedantic: $$S^2= \{ x=(x_0, x_1, x_2) \in \mathbb{R}^3 : \|x\|_2=1\} \quad N=(0,0,1) \quad S=(0,0-1)$$
and $$X=S/\{N \sim S\}$$. Set
$$\tilde{B} = \{x \in S^2 : x_0=0,\; x_1 \ge 0\} \qquad \tilde{U}= \{x \in S^2 : |x_0| <\varepsilon, x_{1} > -\varepsilon \} \qquad \tilde{V}= S^2 \setminus \tilde{B}$$
Both $$\tilde{B}$$ and $$\tilde{U}$$ contain $$N$$ and $$S$$, whereas $$\tilde{V}$$ does not contain any. You define $$B$$, $$U$$ and $$V$$ to be the images of $$\tilde{B}$$, $$\tilde{U}$$, $$\tilde{V}$$ respectively, by the quotient map $$S^2 \to X$$. Note that the tilded ones are all saturated sets with respect to such map. Your $$U \cap V = U \setminus B$$ is the image of $$\tilde{U} \setminus \tilde{B}$$ via the quotient map. Since $$N, S \notin \tilde{U} \setminus \tilde{B}$$ you get $$\tilde{U} \setminus \tilde{B} \simeq U \setminus B$$. Similarly you get $$V\simeq \tilde{V}$$ which is a disk as you already pointed out.
If you do stereographic projection from the point $$(0, -1, 0)$$ to the plane $$\{x_1=1\}$$ you get that $$\tilde{U} \setminus \tilde{B}$$ is mapped to something isomorphic to a an open disk subtracted by a stright segment contained in it, hence $$U \cap V$$ is path connected and you can use Van Kampen.