For topological spaces \(X,Y\), the mapping cone \(C(f)\) of a map \(f:X\to Y\) is defined to be the quotient space \[ (X\times[0,1])\amalg Y/\sim, \] where \((x,0)\sim(x',0)\) for all \(x,x'\in X\) and \((x,1)\sim f(x)\) for all \(x\in X\). Let \[ \phi_k:S^n\longrightarrow S^n \] be a degree-\(k\) map for some integer \(k\). Find \(H_i(C(\phi_k))\) for all \(i\).
Solution
Write \[ C_k=C(\phi_k). \] Unless stated otherwise, homology has coefficients in \(\ZZ\).
1. The mapping cone is the cofiber of \(\phi_k\), so its reduced homology fits into the exact sequence\[ \cdots\longrightarrow \widetilde H_i(S^n) \xrightarrow{(\phi_k)_*} \widetilde H_i(S^n) \longrightarrow \widetilde H_i(C_k) \longrightarrow \widetilde H_{i-1}(S^n) \longrightarrow\cdots. \]
The cone on \(S^n\) is attached to the target \(S^n\) along its base by \(\phi_k\). Thus \(C_k\) is the mapping cofiber of \(\phi_k\), and the displayed sequence is the reduced homology exact sequence of a cofiber. Equivalently, one may identify \(C_k\) with \[ S^n\cup_{\phi_k}D^{n+1} \] and use cellular homology.
2. In the only potentially nonzero reduced degrees, exactness gives\[ 0\longrightarrow \widetilde H_{n+1}(C_k) \longrightarrow \ZZ \xrightarrow{\cdot k} \ZZ \longrightarrow \widetilde H_n(C_k) \longrightarrow0. \]
The reduced homology of \(S^n\) is \(\ZZ\) in degree \(n\) and zero in all other degrees. By definition of degree, \[ (\phi_k)_*:\widetilde H_n(S^n)\longrightarrow\widetilde H_n(S^n) \] is multiplication by \(k\). Substituting these facts into 1 gives the displayed exact sequence, and all other reduced homology groups of \(C_k\) vanish.
3. If \(k\neq0\), then\[ \widetilde H_{n+1}(C_k)=0, \qquad \widetilde H_n(C_k)\cong\ZZ/k\ZZ. \]
Multiplication by a nonzero integer \(k\) on \(\ZZ\) is injective, with cokernel \(\ZZ/k\ZZ\). Apply exactness in 2.
4. If \(k=0\), then\[ \widetilde H_{n+1}(C_0)\cong\ZZ, \qquad \widetilde H_n(C_0)\cong\ZZ. \]
For \(k=0\), the middle map in 2 is the zero homomorphism. Its kernel and cokernel are both \(\ZZ\).
5. For the standard positive-dimensional interpretation \(n\ge1\), the answer is\[ H_i(C_k)\cong \begin{cases} \ZZ,&i=0,\\ \ZZ/k\ZZ,&i=n\text{ and }k\neq0,\\ \ZZ,&i=n\text{ and }k=0,\\ \ZZ,&i=n+1\text{ and }k=0,\\ 0,&\text{otherwise}. \end{cases} \]
4. For \(k=\pm1\), the notation \(\ZZ/k\ZZ\) denotes the zero group, as required.
6. If the source convention permits \(n=0\), then necessarily \(k\in\{-1,0,1\}\), and\[ H_i(C_k)\cong \begin{cases} \ZZ,&i=0, \quad k=\pm1,\\ \ZZ^2,&i=0, \quad k=0,\\ \ZZ,&i=1, \quad k=0,\\ 0,&\text{otherwise}. \end{cases} \]
The group \[ \widetilde H_0(S^0)\cong\ZZ \] is generated by the difference of the two points. A map \(S^0\to S^0\) either fixes the two points, swaps them, or identifies them, so its induced map on reduced \(H_0\) is multiplication by \(1,-1\), or \(0\), respectively. Thus only \(k=1,-1,0\) can occur.
The exact sequence in 2 remains valid with \(n=0\). For \(k=\pm1\), both its kernel and cokernel vanish, so \(C_k\) is connected with no positive reduced homology and therefore has only \(H_0\cong\ZZ\). For \(k=0\), \[ \widetilde H_0(C_0)\cong\ZZ, \qquad \widetilde H_1(C_0)\cong\ZZ. \] The first equality means that \(C_0\) has two path components, so \[ H_0(C_0)\cong\ZZ^2, \] and the second gives \(H_1(C_0)\cong\ZZ\).