Let \(X = S^n\cross I\) with its ends glued together by a map \(S^n \circlearrowleft\) of degree \(d\), calculate \(H_*(X)\).
Solution
Let \(M_f\) denote the mapping torus of \(f:S^n\to S^n\).
1. For \(n\ge1\), the Wang exact sequence of the mapping torus contains\[ \cdots\to H_k(S^n)\xrightarrow{1-f_*}H_k(S^n) \to H_k(M_f)\to H_{k-1}(S^n)\xrightarrow{1-f_*}H_{k-1}(S^n)\to\cdots. \]
This is the Mayer–Vietoris sequence for a cover of the base circle by two arcs, after identifying the two components of the overlap. The difference between the two gluing maps is \(1-f_*\), yielding the displayed Wang sequence.
2. On \(H_n(S^n)\cong\mathbb Z\), the map \(1-f_*\) is multiplication by \(1-d\), while on \(H_0(S^n)\cong\mathbb Z\) it is zero.
By definition of degree, \(f_*\) acts by multiplication by \(d\) on \(H_n(S^n)\). Since \(S^n\) is connected for \(n\ge1\), every self-map induces the identity on \(H_0\).
3. If \(n\ge2\), then\[ \boxed{ H_k(M_f;\mathbb Z)\cong \begin{cases} \mathbb Z,&k=0,1,\\ \mathbb Z/(1-d),&k=n,\\ \mathbb Z,&k=n+1\text{ and }d=1,\\ 0,&\text{otherwise}, \end{cases}} \]where \(\mathbb Z/(0)=\mathbb Z\).
For degrees \(n,n+1\), exactness gives \[ 0\to H_{n+1}(M_f)\to\mathbb Z\xrightarrow{1-d}\mathbb Z\to H_n(M_f)\to0. \] Thus \(H_{n+1}=\ker(1-d)\) and \(H_n=\operatorname{coker}(1-d)\). Around degree \(1\), the zero map \(1-f_*:H_0(S^n)\to H_0(S^n)\) gives \(H_1(M_f)\cong\mathbb Z\). All intermediate groups vanish because \(S^n\) has no homology in degrees \(1,\dots,n-1\).
4. If \(n=1\), then\[ \boxed{ H_0(M_f)=\mathbb Z,\qquad H_1(M_f)\cong\mathbb Z\oplus\mathbb Z/(1-d),\qquad H_2(M_f)=\ker(1-d), } \]and higher homology vanishes.
The Wang sequence gives \[ 0\to H_2(M_f)\to\mathbb Z\xrightarrow{1-d}\mathbb Z \to H_1(M_f)\to\mathbb Z\xrightarrow{0}\mathbb Z\to H_0(M_f)\to0. \] Hence \(H_2=\ker(1-d)\) and there is a short exact sequence \[ 0\to\mathbb Z/(1-d)\to H_1(M_f)\to\mathbb Z\to0. \] It splits because \(\mathbb Z\) is free.
5. If one also allows \(n=0\), the four self-maps of \(S^0\) have reduced degrees \(1,-1,0,0\). For degree \(1\) the mapping torus is \(S^1\sqcup S^1\); for degree \(-1\) it is \(S^1\); and for either degree-\(0\) constant map it is a circle with a contractible interval attached. Thus\[ (d=1):\quad H_0=H_1=\mathbb Z^2, \]while\[ (d=-1\text{ or }0):\quad H_0=H_1=\mathbb Z, \]with all higher groups zero.
For \(S^0=\{a,b\}\), the mapping torus is obtained from two intervals by identifying each top endpoint with the image under \(f\) at the bottom. The identity closes both intervals separately; the transposition joins them into one circle; a constant map closes one interval into a circle and attaches the other interval as a tree edge.