What is $\displaystyle \pi_1(S^1 \vee S^1 \vee S^2)$?
Is it $\displaystyle \mathbb{Z} * \mathbb{Z} * \{1\}$ by Van Kampen's Theorem?
Thanks, I just wanted to check this.
Yes.
Since $\displaystyle S^2$ is simply connected, $\displaystyle \pi_1(S^1 \vee S^1 \vee S^2) \cong \mathbb{Z} * \mathbb{Z}$.
The free product of a free group $\displaystyle \mathbb{Z}$ and a free group $\displaystyle \mathbb{Z}$ is a free group having a presentation $\displaystyle \{\alpha, \beta\}$, where $\displaystyle \alpha$ = 1 or -1 in the first $\displaystyle \mathbb{Z}$ and $\displaystyle \beta$ = 1 or -1 in the second $\displaystyle \mathbb{Z}$.
For instance,
$\displaystyle 1_1 1_1 1_1 1_2 1_2$ (followed by a member in the first group) .... ,
reduces to
$\displaystyle 3_1 2_2 ...$ ,
where $\displaystyle x_i$ denotes the element x in the i-th $\displaystyle \mathbb{Z}$, i=1 or 2, and the group operation between elements in the same group is an addition.
As you know, $\displaystyle \mathbb{Z} * \mathbb{Z}$ and $\displaystyle \mathbb{Z} \times \mathbb{Z}$ are two different things. The former is non-abelian because a free group on a nonempty set S is abelian iff S has exactly one element. The latter group is abelian, which is isomorphic to the fundamental group of a torus having a presentation $\displaystyle \{\alpha, \beta\ | \alpha\beta\alpha^{-1}\beta^{-1}=1\}$.