# Find the fundamental group of the torus with an open disc removed

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.

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

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)
• 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. May 26, 2013 at 10:58