My question is in relation to the accepted answer here .
I have already proven that the fundamental group of the plane minus two points is just the same as wedge of two circles and hence is $\Bbb{Z}*\Bbb{Z}$ by considering the plane as an open disc and removing two points from the interior and then using Van Kampen's Theorem.
But I do not know how to compute the fundamental group of disc minus two points . As far as I can see there should be a few cases. For example if I remove two points from the interior then I should again be able to use Van-Kampen and conclude that it is the same as that of the wedge of two circles.(Although I am not very sure as to what the open sets should be. But I guess I'll need to use the subspace topology and modify the way I proved for the open disc a little).
But what if the two points are on the boundary or one point in the interior and the other on the boundary. In these cases I am having trouble as to apply Van Kampen as I cannot figure out what open sets should I choose.
Any help is appreciated .
As there are no answers, I will try to answer my own question based on the hints by @David Sheard.
I'll try and show it by means of a few pictures using paint.
Here $U\cup V= D^{2}\setminus\{a,b\}$ .
And $U$ deformation retracts to a circle and $V$ deformation retracts to a point.
So $\Pi_{1}(U)=\Bbb{Z}$ and $\Pi_{1}(V)=\{0\}$. And $U\cap V$ is simply connected
Hence By Van-Kampen Theorem we have that $\Pi_{1}(D^{2}\setminus\{a,b\})\cong \Pi_{1}(U)*\Pi_{1}(V)\cong \Pi_{1}(U)\cong \Bbb{Z},$ .
If two points were missing in the interior , I would use a similar sort of arguments to conclude that the fundamental group is $\Bbb{Z}*\Bbb{Z}=\langle\alpha,\beta\rangle$
If two points were missing from the boundary, then I would use a similar argument and conclude that the fundamental group is $\{0\}*\{0\}=\{0\}$
This allows me to conclude the result from the linked question that $\Bbb{Z}\cong\Pi_{1}(D^{2}\setminus\{a,b\})\cong\Pi_{1}(D^{2}\setminus\{a\})$ whereas $\Bbb{Z}*\Bbb{Z}\cong\Pi_{1}(\Bbb{R}^{2}\setminus\{a,b\})\not\cong\Pi_{1}(\Bbb{R}^{2}\setminus\{a\})\cong\Bbb{Z}$.
Where $b$ is a point on the boundary of $D^{2}$ and $a$ is a point interior to $D^{2}$.
Can anyone please verify my above conclusions?