Consider the torus $\mathbb{T}$ given by $aba^{-1}b^{-1}$ with vertex $p$ in the square of identifications. Now remove a disk $D$ from it (I based its boundary $c$ on $p$). Then, I get the space given by $aba^{-1}b^{-1}c$. Now, I consider two concentric circumferences and let $\mathbb{T}\setminus D = A\cup B$, where $A$ is the exterior of the smaller one and $B$ the interior of the bigger one. So, $A\cap B$ is an open annulus, which is path connected. Let $x_0\in A\cap B$.
Since $B$ is contractible, $\pi_1(B,x_0)$ is trivial. Since $A\cap B$ is an open annulus, $S^1$ is a deformation retract of it, so $\pi_i(A\cap B,x_0)=\langle\varepsilon_0 | \rangle\cong\mathbb{Z}$, where $\varepsilon_0$ is the class of the canonical turn to $S^1$.
Now, notice that $A$ can be retracted with deformation into the boundary of the identification polygon $aba^{-1}b^{-1}c$, which consists in the punctual union of three circumferences $a,b$ and $c$. Then, $\pi_1(A,p)=\langle\alpha_a,\alpha_b,\alpha_c|\rangle\cong F_3$, where $\alpha_a,\alpha_b,\alpha_c$ are the classes of the canonical turns to the circumferences $a,b$ and $c$. But we want to base the fundamental group of $A$ in $x_0$. Consider $\gamma$ and path between $x_0$ and $p$. Let $\varepsilon_1 = [\gamma a\bar{\gamma}]$,$\varepsilon_2 = [\gamma b\bar{\gamma}]$ and $\varepsilon_3 = [\gamma c\bar{\gamma}]$, where $\bar{\gamma}$ denote the path $\gamma$ walked in the other way. Then, $\pi_1(A,x_0)=\langle\varepsilon_1,\varepsilon_2,\varepsilon_3|\rangle$.
Then, by Seifert-Van Kampen theorem, $\pi_1(\mathbb{T}\setminus D)=\langle\varepsilon_1,\varepsilon_2,\varepsilon_3 | R_{12}\rangle$, where $R_{1,2}=\{i_{1*}(\varepsilon_0)i_{2*}(\varepsilon_0)^{-1}\}$. Here, $i_{1*}$ and $i_{2*}$ are the homomorphisms induced by the inclusions $i_1:A\cap B\hookrightarrow A$ and $i_2:A\cap B\hookrightarrow B$.
We have that $i_{1*}(\varepsilon_0)=[\gamma ab\bar{a}\bar{b}c\bar{\gamma}]=[\gamma a\bar{\gamma}\gamma b\bar{\gamma}\gamma\bar{a}\bar{\gamma}\gamma\bar{b}\bar{\gamma}\gamma c\bar{\gamma}]=\varepsilon_1\varepsilon_2\varepsilon_1^{-1}\varepsilon_2^{-1}\varepsilon_3$ and $i_{2*}=1_{\pi_1(B,x_0)}$.
So, with this, we conclude that $\pi_1(\mathbb{T}\setminus D,x_0)=\langle\varepsilon_1,\varepsilon_2,\varepsilon_3 | \varepsilon_1\varepsilon_2\varepsilon_1^{-1}\varepsilon_2^{-1}\varepsilon_3\rangle$
Is my approach correct? If it is, is there another way of presenting $\pi_1(\mathbb{T}\setminus D,x_0)$ in terms of abelian or free products?
Thank you very much!
Best Answer
A punctured torus deformation retracts to figure-eight (wedge sum of two circles).