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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 .

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    $\begingroup$ Remember that the fundamental group is a homotopy invariant, not just a homeomorphism invariant. The closed disk is homotopic to the plane, so this should help give you intuition for what the answer should be in each case. Alternatively, recall that when applying van Kampen, your open sets need to be open in $D^2$, not open in $R^2$, so in particular your open sets may include (parts of) the boundary $\endgroup$
    – David Sheard
    Jun 11, 2022 at 13:46
  • $\begingroup$ @DavidSheard I have tried to arrive at an answer based on your hints. Can you please check it and tell me if it is correct? $\endgroup$
    – Dovahkiin
    Jun 13, 2022 at 17:37
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    $\begingroup$ Your answer looks correct, well done $\endgroup$
    – David Sheard
    Jun 13, 2022 at 20:42
  • $\begingroup$ @DavidSheard Thanks for your hints !! $\endgroup$
    – Dovahkiin
    Jun 14, 2022 at 7:17

1 Answer 1

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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.

enter image description here

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?

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    $\begingroup$ That looks fine, but I'll add that you can do all three cases with deformation retraction arguments; no need to do repeated Van-Kampen arguments. For instance two points are removed from the interior, you get a deformation retraction of the whole plane with those two points. $\endgroup$
    – Lee Mosher
    Jun 13, 2022 at 18:38
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    $\begingroup$ @LeeMosher I thought about that . for example when two points are removed from the interior, I can just use the fact that the open disc is homeomorphic to $\Bbb{R}^{2}$. But I could not do it for the boundary part. Can you please add an answer or an additional comment explaining it to me further. $\endgroup$
    – Dovahkiin
    Jun 13, 2022 at 19:12
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    $\begingroup$ If say two points are missing from the boundary, just deformation retract to a smaller, concentric closed disc with no points missing. $\endgroup$
    – Lee Mosher
    Jun 13, 2022 at 19:21
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    $\begingroup$ @LeeMosher Got it. So say my disc was of radius $2$ . Then if I just use $H(x,t)=\frac{x}{1-t+t||x||}$ then it would just deformation retract to the unit disc with no missing points right? $\endgroup$
    – Dovahkiin
    Jun 14, 2022 at 7:20
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    $\begingroup$ That's the right idea but the wrong formula, because when $t=1$ you get $H(x,1) = \frac{x}{\|x\|}$ which is undefined for $x=0$ and takes everything else to the unit circle. What you want is something like $H(x,t) = \left(1-\frac{t}{2} \right) x$. $\endgroup$
    – Lee Mosher
    Jun 14, 2022 at 13:06

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