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 ?

  • $\begingroup$ Hint: $U \cap V =U \setminus B$ by your definition. If you think about it, it is homeomorphic to a punctured disk... $\endgroup$
    – Dunnò000
    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.


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