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I do not see how using Van Kampen theorem we obtain:

(a) $\pi_1(S^1 \vee S^1)\cong \mathbb{Z} * \mathbb{Z}$ and

(b) $\pi_1(S^n) \cong \left\{1\right\}$ for $n\geq 2$.

In my lecture notes I have that Van Kampen is the main theorem that helps us compute the fundamental group of a space $X$ provided that $X$ is expressed $X=U_1 \cup U_2$ where $U_i \subset X$ for $i=1,2$ and $U_1 \cap U_2$ are open and path connected and that the fundamental groups $\pi_1(U_i)$ and $\pi_1(U_1 \cap U_2)$ are known. Moreover, the theorem as stated is:

Van Kampen Theorem: If $X = U_1 \cup U_2$ with $U_i$ open and path connected, and $U_1 \cap U_2$ path connected and simply connected, then the induced homomorphism $\Phi : \pi_1(U_1)*\pi_1(U_2)\to\pi_1(X)$ is an isomorphism; where $*$ is the free product of the groups $\pi_1(U_1)$ and $\pi_1(U_2)$.

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    $\begingroup$ So choose appropriate $U_1$ and $U_2$ that cover the space and have easy to compute $\pi_1$. $\endgroup$ – anomaly Dec 11 '17 at 21:11
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    $\begingroup$ Do you not have any examples in your lecture notes? Nor any book with examples? $\endgroup$ – Lee Mosher Dec 11 '17 at 21:13
  • $\begingroup$ You will (probably) need a more general version of the theorem for question $2$. $\endgroup$ – Andres Mejia Dec 12 '17 at 1:46
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For the first, choose open sets $U$ and $V$ such that $U$ contains one copy of $S^1$ and $V$ the other, such that $U\cap V$ is a cross at the basepoint, and such that $U$ and $V$ both deformation retract onto such copies of $S^1$. Then Van Kampen says $\pi_1(S^1\vee S^1)$ is the coproduct of $\pi_1(U)$ and $\pi_1(V)$ over $\pi_1(U\cap V)$. This last group is trivial, so you get the free product of $\pi_1(S^1)=\mathbb Z$ with itself.

For the second, take open sets $U$ and $V$ such that $U$ and $V$ are contractible and intersect at an open set that deformation retracts onto $S^{n-1}$. Since $n>1$, this set is connected, and Van Kampen says that $\pi_1(S^n)$ is the coproduct of two trivial groups over trivial maps from $\pi_1(S^{n-1})$. Obviously this is the trivial group.

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  • $\begingroup$ Note that the version of Van Kampen's theorem used in the OP cannot be used to argue for question $2$, since it assumes that $U \cap V$ is simply connected. $\endgroup$ – Andres Mejia Dec 12 '17 at 1:44

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