Which is the fundamental group of $\mathbb{R}^2-\{(0,0),(0,1)\}?$
Making a picture of this space with two closed curves with point basis on $(-1,0)$ and both disjoint each one involving $(0,0)$ and $(0,1)$, I can see that these curves are not homotopic. Even more, they are not contractible to a point. So my group has two classes of equivalence plus the constant curve, that are not homotopic of none of these curves. Then, can I afirme that this fundamental group is $\mathbb{Z}_3?$
Your space is homotopically equivalent to the double circle. Just like how $\mathbb{R}^2\setminus\{0\}$ is homotopically equivalent to the circle. As a result, the fundamental group is $\mathbb{Z}\coprod\mathbb{Z} $, the nonabelian coproduct of $\mathbb{Z}$ with itself.