Attach a 2-disc \(D^2\) to a torus \(S^1\times S^1\) by the attaching map \(e^{2\pi it}\mapsto(e^{2\pi it},e^{2\pi it})\), thinking of the boundary \(\partial D^2\) and each \(S^1\) factor as the unit circle in the complex plane, and let \(X\) be the resulting space. Compute a presentation for the fundamental group of \(X\).
Solution
Let \[ T=S^1\times S^1 \] and take the basepoint \((1,1)\). Let \(a\) be the loop around the first factor and \(b\) the loop around the second.
1. The torus has presentation\[ \pi_1(T,(1,1)) \cong \langle a,b\mid aba^{-1}b^{-1}=1\rangle. \]
Use the standard CW structure on \(T\) with one \(0\)-cell, two oriented \(1\)-cells \(a,b\), and one \(2\)-cell whose boundary is attached along the commutator word \[ aba^{-1}b^{-1}. \] The Seifert–van Kampen theorem for this cell attachment gives the displayed presentation.
2. The attaching map\[ \gamma(t)=\bigl(e^{2\pi it},e^{2\pi it}\bigr) \]represents the element \(ab\in\pi_1(T,(1,1))\).
Under the standard identification \[ \pi_1(S^1\times S^1,(1,1)) \cong \pi_1(S^1,1)\times\pi_1(S^1,1) \cong \mathbb Z\times\mathbb Z, \] the loop \(\gamma\) projects to the positive generator in each circle factor. Hence its class is \((1,1)\). The classes of \(a\) and \(b\) are \((1,0)\) and \((0,1)\) respectively, so \[ [\gamma]=[a][b]=ab. \]
3. Attaching the disk gives\[ \pi_1(X) \cong \langle a,b\mid aba^{-1}b^{-1}=1,\ ab=1\rangle. \]
Attaching a \(2\)-cell along a loop quotients the fundamental group by the normal closure of that loop class. Applying the Seifert–van Kampen theorem to the attachment and using <1>1–<1>2 therefore adds the single relator \[ ab=1 \] to the torus presentation.
4. The group in 3 is infinite cyclic.
The relation \(ab=1\) gives \[ b=a^{-1}. \] After substituting this into the commutator relation, that relation becomes \[ aa^{-1}a^{-1}a=1, \] which is automatic. Thus every element is represented by a power of \(a\).
More explicitly, define \[ \Phi:\mathbb Z\longrightarrow \langle a,b\mid [a,b]=1,ab=1\rangle, \qquad n\longmapsto a^n. \] It is surjective because \(b=a^{-1}\). Conversely the assignments \[ a\longmapsto1, \qquad b\longmapsto-1 \] satisfy both relators and therefore induce a homomorphism to \(\mathbb Z\) inverse to \(\Phi\). Hence \[ \pi_1(X)\cong\mathbb Z. \]
5. A presentation answering the question is therefore\[ \boxed{ \pi_1(X) \cong \langle a,b\mid [a,b]=1,\ ab=1\rangle \cong\mathbb Z. } \]
This is exactly <1>3–<1>4, with the commutator notation \[ [a,b]=aba^{-1}b^{-1}. \]