Let \(X\) be the space obtained by attaching a torus \(T=S^1\times S^1\) to a cylinder \(C=S^1\times I\) via a homeomorphism of the circle \(S^1\times\{(1,0)\}\) of \(T\) with the circle \(S^1\times\{0\}\) of \(C\). Compute the homology groups of \(X\).
Solution
Let \[ A=S^1\times\{(1,0)\}\subseteq T, \qquad B=S^1\times\{0\}\subseteq C, \] and let \(h:A\to B\) be the homeomorphism used in the attachment.
1. The torus \(T\subseteq X\) is a strong deformation retract of \(X\).
On the cylinder define \[ H_C:C\times I\longrightarrow C, \qquad H_C((z,t),s)=(z,(1-s)t). \] At \(s=0\) this is the identity, at \(s=1\) its image is \(B=S^1\times\{0\}\), and every point of \(B\) is fixed for every \(s\).
Extend this homotopy over the attached space \(X\) by keeping every point of \(T\) fixed. This is well-defined on the gluing locus: if \(a\in A\) is identified with \(h(a)\in B\), then the torus-side homotopy fixes \(a\) and the cylinder-side homotopy fixes \(h(a)\), so the two representatives remain identified for all \(s\). By the quotient property this gives a continuous homotopy \[ H:X\times I\longrightarrow X. \] It starts at the identity, fixes \(T\) pointwise throughout, and at time \(1\) sends the whole cylinder into its attached boundary \(B\), hence into the copy of \(A\) already lying in \(T\). Thus \(H\) is a strong deformation retraction of \(X\) onto \(T\).
2. For every \(n\ge0\),\[ H_n(X;\mathbb Z)\cong H_n(T;\mathbb Z). \]
A deformation retraction is a homotopy equivalence, and singular homology is invariant under homotopy equivalence. Apply 1.
3. The cellular chain complex of the standard CW structure on\[ T=S^1\times S^1 \]is\[ 0\longrightarrow\mathbb Z \overset{0}{\longrightarrow} \mathbb Z^2 \overset{0}{\longrightarrow} \mathbb Z \longrightarrow0. \]
Use the usual CW structure with one \(0\)-cell, two loop \(1\)-cells \(a,b\), and one \(2\)-cell attached by the word \[ aba^{-1}b^{-1}. \] Both \(1\)-cells begin and end at the unique \(0\)-cell, so the cellular boundary \[ \partial_1:\mathbb Z^2\to\mathbb Z \] is zero.
For \(\partial_2\), the coefficient of each oriented \(1\)-cell is the total signed number of times the attaching map traverses it. The commutator word traverses \(a\) once positively and once negatively, and likewise for \(b\). Hence both coefficients vanish and \[ \partial_2=0. \] There are no cells in dimensions greater than \(2\).
4. Therefore\[ H_n(X;\mathbb Z) \cong \begin{cases} \mathbb Z, & n=0,\\ \mathbb Z^2, & n=1,\\ \mathbb Z, & n=2,\\ 0, & n\ge3. \end{cases} \]
By 3, the homology of the torus is the homology of a chain complex whose boundary maps are zero, so \[ H_0(T)\cong\mathbb Z, \qquad H_1(T)\cong\mathbb Z^2, \qquad H_2(T)\cong\mathbb Z, \] with all higher groups zero. Transfer these groups to \(X\) using 2.