Let \(X\) be a path-connected Hausdorff space. The suspension \(SX\) is the quotient of the space \(X\times I\) by the smallest equivalence relation such that \((p,0)\sim(q,0)\) and \((p,1)\sim(q,1)\) for all \(p,q\in X\). Prove that the homology groups satisfy \[H_i(SX)=H_{i-1}(X)\] for all \(i>1\).
Solution
Let \[ q:X\times I\longrightarrow SX \] be the quotient map, and write \[ S=q(X\times\{0\}), \qquad N=q(X\times\{1\}) \] for the two suspension vertices. Define \[ U=q\bigl(X\times[0,2/3)\bigr), \qquad V=q\bigl(X\times(1/3,1]\bigr). \]
1. The subsets \(U\) and \(V\) are open in \(SX\) and satisfy\[ SX=U\cup V. \]
The only nontrivial equivalence classes for \(q\) are \(X\times\{0\}\) and \(X\times\{1\}\). Hence both \[ X\times[0,2/3) \qquad\text{and}\qquad X\times(1/3,1] \] are saturated for the quotient relation. Therefore \[ q^{-1}(U)=X\times[0,2/3), \qquad q^{-1}(V)=X\times(1/3,1]. \] The intervals \([0,2/3)\) and \((1/3,1]\) are open in the subspace topology of \(I=[0,1]\), so these inverse images are open in \(X\times I\). By the definition of the quotient topology, \(U\) and \(V\) are open.
Finally, \[ [0,2/3)\cup(1/3,1]=I, \] so their images cover \(SX\).
2. The open set \(U\) is contractible.
Define on the preimage of \(U\) the homotopy \[ F_U:\bigl(X\times[0,2/3)\bigr)\times I\longrightarrow U, \qquad F_U((x,t),s)=q(x,(1-s)t). \] It is continuous because \[ ((x,t),s)\longmapsto(x,(1-s)t) \] is continuous and its image remains in \(X\times[0,2/3)\).
The restricted map \[ q_U:X\times[0,2/3)\longrightarrow U \] is a quotient map: the domain is a saturated open subset of \(X\times I\), and restricting a quotient map to a saturated open subset again gives a quotient map. By the quotient-product fact from part A2 of this same exam, \(q_U\times\operatorname{id}_I\) is also a quotient map.
If two points have the same image under \(q_U\times\operatorname{id}_I\), then their \(s\)-coordinates agree and either their \((x,t)\) coordinates agree or both have \(t=0\). In the latter case \(F_U\) sends both points to the suspension vertex \(S\). Thus \(F_U\) is constant on the fibers of \(q_U\times\operatorname{id}_I\) and descends to a continuous homotopy \[ H_U:U\times I\longrightarrow U, \qquad H_U(q(x,t),s)=q(x,(1-s)t). \] At \(s=0\) this is the identity; at \(s=1\) it is the constant map to \(S\); and \(S\) is fixed throughout. Hence \(U\) is contractible.
3. The open set \(V\) is contractible.
Use the analogous homotopy toward the upper suspension vertex: \[ H_V(q(x,t),s) = q\bigl(x,(1-s)t+s\bigr). \] For \(t>1/3\) and \(0\le s\le1\), \[ (1-s)t+s=t+s(1-t)\ge t>1/3, \] so the homotopy remains in \(V\). As in 2, it descends continuously from \[ ((x,t),s)\longmapsto q\bigl(x,(1-s)t+s\bigr) \] through the quotient map on the saturated open subset \(X\times(1/3,1]\). At \(s=0\) it is the identity, at \(s=1\) it is constant at \(N\), and it fixes \(N\) throughout. Thus \(V\) is contractible.
4. The overlap \(U\cap V\) is homeomorphic to\[ X\times(1/3,2/3). \]
Since no point with height strictly between \(0\) and \(1\) is identified with any other point under the suspension relation, \[ q^{-1}(U\cap V)=X\times(1/3,2/3), \] and the restriction \[ q_0:X\times(1/3,2/3)\longrightarrow U\cap V \] is bijective.
2. A bijective quotient map is a homeomorphism: for any subset \(W\) of its target,\[ W\text{ is open} \quad\Longleftrightarrow\quad q_0^{-1}(W)\text{ is open}, \]which is precisely continuity of the inverse.Hence \(q_0\) is a homeomorphism.
5. The overlap \(U\cap V\) deformation retracts onto a copy of \(X\) and therefore\[ H_k(U\cap V)\cong H_k(X) \]for every \(k\ge0\).
Under the homeomorphism of 4, define \[ R_s(x,t)=\bigl(x,(1-s)t+s/2\bigr). \] Because \((1/3,2/3)\) is convex and contains \(1/2\), this remains in \[ X\times(1/3,2/3) \] for every \(s\). It is a deformation retraction onto \[ X\times\{1/2\}, \] which is homeomorphic to \(X\). Homotopy invariance of singular homology gives the stated isomorphisms.
6. For every \(i>1\), the Mayer–Vietoris connecting homomorphism gives an isomorphism\[ H_i(SX)\xrightarrow{\ \cong\ }H_{i-1}(U\cap V). \]
1. The relevant part is\[ H_i(U)\oplus H_i(V) \longrightarrow H_i(SX) \xrightarrow{\delta} H_{i-1}(U\cap V) \longrightarrow H_{i-1}(U)\oplus H_{i-1}(V). \]By <1>2–<1>3, \(U\) and \(V\) are contractible, so\[ H_j(U)=H_j(V)=0 \qquad(j>0). \]If \(i>1\), then both \(i\) and \(i-1\) are positive.Thus the two outer groups in the displayed exact segment vanish, and exactness forces \(\delta\) to be both injective and surjective.Hence it is an isomorphism.
7. For every \(i>1\),\[ \boxed{H_i(SX)\cong H_{i-1}(X).} \]
By 6, \[ H_i(SX)\cong H_{i-1}(U\cap V). \] By 5, \[ H_{i-1}(U\cap V)\cong H_{i-1}(X). \] Composing the two isomorphisms proves the required formula.