Let \(X\) be a space obtained by attaching two \(2\)-cells to the torus \(S^1\times S^1\), one along a simple closed curve \(1\times S^1\) and the other along \(S^1\times 1\). Calculate the fundamental group and homology groups of \(X\).
Solution
1. Give the torus its standard CW structure with one \(0\)-cell, two oriented \(1\)-cells \(a,b\), and one \(2\)-cell \(c\) attached by the commutator\[ aba^{-1}b^{-1}. \]Take \(a\) to represent \(S^1\times 1\) and \(b\) to represent \(1\times S^1\).
The usual square model of the torus has one vertex after identifying all four corners, two \(1\)-cells from the two pairs of opposite sides, and one \(2\)-cell whose boundary word is the commutator. Thus \[ \pi_1(T^2) \cong \langle a,b\mid aba^{-1}b^{-1}=1\rangle \cong \mathbb Z^2. \]
2. Attaching the two additional \(2\)-cells kills both generators of the fundamental group, so\[ \boxed{\pi_1(X)=1}. \]
Let \(e_a^2\) be the new \(2\)-cell attached along \(S^1\times1\), and let \(e_b^2\) be the one attached along \(1\times S^1\). By Seifert–van Kampen, attaching a \(2\)-cell along a loop adds the relation represented by that loop. Hence \[ \pi_1(X) \cong \langle a,b\mid [a,b],a,b\rangle =1. \]
3. The cellular chain complex of \(X\) in positive degrees is\[ 0 \longrightarrow C_2(X)\cong\mathbb Z^3 \xrightarrow{\partial_2} C_1(X)\cong\mathbb Z^2 \xrightarrow{\partial_1} C_0(X)\cong\mathbb Z \longrightarrow0. \]
There is one \(0\)-cell, the two original \(1\)-cells \(a,b\), and three \(2\)-cells: the original torus cell \(c\) and the two newly attached cells \(e_a^2,e_b^2\). There are no cells in dimensions at least \(3\). Since both ends of each \(1\)-cell are the unique \(0\)-cell, \[ \partial_1=0. \]
4. With ordered bases\[ C_2(X)=\mathbb Z\langle c,e_a^2,e_b^2\rangle, \qquad C_1(X)=\mathbb Z\langle a,b\rangle, \]the cellular boundary map is\[ \partial_2(c)=0, \qquad \partial_2(e_a^2)=a, \qquad \partial_2(e_b^2)=b. \]Equivalently,\[ [\partial_2] = \begin{pmatrix} 0&1&0\\ 0&0&1 \end{pmatrix}. \]
For a \(2\)-cell in a CW complex with one vertex, its cellular boundary records the total degree with which its attaching map traverses each \(1\)-cell.
The original torus cell has attaching word \[ aba^{-1}b^{-1}, \] whose exponent sum in both \(a\) and \(b\) is zero, so its cellular boundary is zero. The cell \(e_a^2\) is attached once around \(a\), giving boundary \(a\), and \(e_b^2\) is attached once around \(b\), giving boundary \(b\).
5. Therefore\[ \boxed{ H_i(X;\mathbb Z) \cong \begin{cases} \mathbb Z,&i=0,2,\\ 0,&i\ne0,2. \end{cases}} \]
By 4, \(\partial_2\) is surjective, so \[ H_1(X;\mathbb Z) = \ker\partial_1/\operatorname{im}\partial_2 = \mathbb Z^2/\mathbb Z^2 =0. \] Also \[ \ker\partial_2 = \mathbb Z\langle c\rangle, \] and there is no \(C_3\), so \[ H_2(X;\mathbb Z) = \ker\partial_2 \cong \mathbb Z. \] The space \(X\) is connected, hence \(H_0(X;\mathbb Z)\cong\mathbb Z\), and there are no cells above dimension \(2\), so \(H_i(X;\mathbb Z)=0\) for \(i\ge3\).