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I'm trying to find a fundamental group of $\mathbb{T} \setminus \mathbb{D}$, the $2$-torus $\mathbb{T}$ with an open disc $\mathbb{D}$ removed. Any help would be really appreciated.

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    $\begingroup$ Try to see the torus as $\frac{\mathbb{R}^2}{\mathbb{Z}^2}$ $\endgroup$ May 26, 2013 at 9:37
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    $\begingroup$ Or even easier, see the torus as a square modulo some relation on its boundary. $\endgroup$
    – Abel
    May 26, 2013 at 9:40
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    $\begingroup$ The question seems to be about the fundamental group of $\mathbb{T}^2 \setminus \mathbb{D}$, not the torus itself. $\endgroup$
    – Martin
    May 26, 2013 at 9:42
  • $\begingroup$ @Martin yes, you are right $\endgroup$
    – xxxxx
    May 26, 2013 at 9:47
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    $\begingroup$ I edited your question to remove possible ambiguity. $\endgroup$
    – Martin
    May 26, 2013 at 9:49

1 Answer 1

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Big hint: View the torus as the square $I^2/\sim$ with the usual equivalence relation $\sim$ identifying opposite edges, and then removes a small disk from the interior of $I^2$. Can you see how this space is homotopy equivalent to a wedge of some number of circles? (I'll leave you to figure out how many)

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  • $\begingroup$ We will wide the hole so that there will be only one boundary. Then collapse two sides and obtain a wedge of two circles. Why they should not be collapsed too? $\endgroup$
    – xxxxx
    May 26, 2013 at 10:52
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    $\begingroup$ To put it precisely, you're performing a deformation retract on to the 1-skeleton, and the 1-skeleton is a wedge of two circles so you can not deform the space any more (similar to if you punctured a disk and then deformation retracted that on to the boundary circle). Another method for calculating the fundamental group of this space would be to use Van-Kampen's Theorem where $U$ is an open annulus around the removed disk, and $V$ is the entire space with a slightly thinner annulus removed from around the removed disk. $\endgroup$
    – Dan Rust
    May 26, 2013 at 10:58
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    $\begingroup$ Thank you for explanation. So the answer is Z*Z? $\endgroup$
    – xxxxx
    May 26, 2013 at 11:14
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    $\begingroup$ Yes it is. Now try and work out the fundamental group of the klein bottle with a disk removed, and the real projective plane with a disk removed :). $\endgroup$
    – Dan Rust
    May 26, 2013 at 11:17

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