why $\pi_1(S^2)=0$ ?
I found the answer here but I didn't understand the answer.
My thinking :If $n=2$ then $\pi_1(S^2)\neq 0$ because $S^2$ is not contractible.
Now if I take $S^2-\{x\}$ then $\pi_1\left(S^2-\{x\}\right)=0$ because $S^2-\{x\}$ is contractible
Here $S^2-\{x\} \neq S^2$ .I am not getting why $\pi_1(S^2)=0?$
For $n \geq 2$. The $n$-sphere $S^n$ is simply connected.
Write the sphere as union of the two subsets obtained by removing the north pole and south pole, respectively. Using spherical coordinates we can write
\begin{align*} O_1 := S^n \setminus \lbrace (0, \ldots, 0, 1 \rbrace = \Phi([0,2\pi]^{n-1} \times (0, \pi]) \\ O_2 := S^n \setminus \lbrace (0, \ldots, 0, -1 \rbrace = \Phi([0,2\pi]^{n-1} \times [0, \pi)). \end{align*} Then \begin{equation*} O_1 \cap O_2 = S^n \setminus \lbrace (0, \ldots, 0, 1), (0, \ldots, 0,-1) \rbrace = \Phi([0,2\pi]^{n-1} \times (0,\pi)). \end{equation*} We see that $O_1,O_2$ and $O_1 \cap O_2$ are pathwise connected.
By means of the stereographic projection (taking as projection center the north- or south pole), $O_1$ and $O_2$ are homeomorphic to $\mathbb{R}^n$. Thus $\pi_1(O_1) = \pi_1(O_2) = \lbrace 1 \rbrace$.
Now for a open cover $\lbrace O_i : i \in I \rbrace$ of a topological space $(X, \mathcal{T})$ with \begin{equation} \bigcap_{i \in I}U_i \neq \emptyset \end{equation} and \begin{equation} \forall i,j \in I, U_i \cap U_j \mbox{ is pathwise connected} \end{equation} and $\iota_i : U_i \to X$ the set-theoretic inclusion map then $X$ is pathwise connected and the group $\pi_1(X)$ is generated by \begin{equation} \bigcup_{i \in I}{\pi_1(\iota_i)(\pi_1(U_i))}. \end{equation}
From this result we conclude $\pi_1(S^n) = \lbrace 1 \rbrace$.