\(S_1 \vee S_1\) with two discs attached via \((ab)^3\) and \((ab)^6\)
Solution
1. The fundamental group has presentation\[ \pi_1(X)=\langle a,b\mid (ab)^3=1,(ab)^6=1\rangle. \]
The \(1\)-skeleton is \(S^1\vee S^1\), with generators \(a,b\). Attaching a \(2\)-cell along a loop imposes the corresponding word as a relation.
2. The second relation is redundant, and\[ \boxed{\pi_1(X)\cong\mathbb Z*\mathbb Z/3}. \]
The relation \((ab)^3=1\) already implies \((ab)^6=1\). Set \(c=ab\), so \(b=a^{-1}c\). Then \[ \pi_1(X)\cong\langle a,c\mid c^3=1\rangle\cong\mathbb Z*\mathbb Z/3. \]
3. The cellular chain complex in dimensions \(2,1,0\) is\[ 0\longrightarrow\mathbb Z^2\xrightarrow{\partial_2}\mathbb Z^2\xrightarrow{0}\mathbb Z\longrightarrow0, \]with\[ \partial_2= \begin{pmatrix} 3&6\\ 3&6 \end{pmatrix}. \]
For a \(2\)-cell attached by a word in the wedge generators, the cellular boundary records the total exponent sums of the generators. The words \((ab)^3\) and \((ab)^6\) have exponent-sum vectors \((3,3)\) and \((6,6)\) respectively.
4. Hence\[ \boxed{H_1(X;\mathbb Z)\cong\mathbb Z\oplus\mathbb Z/3} \]and\[ \boxed{H_2(X;\mathbb Z)\cong\mathbb Z}. \]
The image of \(\partial_2\) is generated by \((3,3)\), so \[ H_1\cong\mathbb Z^2/\langle(3,3)\rangle\cong\mathbb Z\oplus\mathbb Z/3. \] The kernel of \(\partial_2\) consists of pairs \((x,y)\) satisfying \(3x+6y=0\), hence is infinite cyclic generated by \((-2,1)\). Since there are no \(3\)-cells, this kernel is \(H_2\).
5. Also \(H_0(X)\cong\mathbb Z\) and \(H_k(X)=0\) for \(k\ge3\).
The complex is connected and \(2\)-dimensional.