- Show that if \(X = S^2 \disjoint_{\id} S^2\) is a pushout along the equators, then \(H_n(X) = [\ZZ, 0, \ZZ^3, 0, \cdots]\).
Solution
1. \(X\) is the union of two copies of \(S^2\) glued along a common equator \(S^1\).
the pushout of two \(S^2\)’s along their equators (identified by the identity map).
2. Give the equator and both spheres compatible CW structures, so that the common equator \(S^1\) is a subcomplex of each sphere. Let \(A\) and \(B\) denote the two sphere subcomplexes of \(X\). Then \(A\cap B=S^1\) and \(A\cup B=X\).
Take the usual CW structure on \(S^2\) obtained from an equatorial \(S^1\) by attaching one \(2\)-cell along each side. After identifying the two equators, the images of the two spheres and their common equator are CW subcomplexes of the pushout.
3. The cellular chain complexes fit into a short exact sequence\[ 0\longrightarrow C_*(S^1)\xrightarrow{(i_*,-j_*)}C_*(S^2)\oplus C_*(S^2)\longrightarrow C_*(X)\longrightarrow0, \]and hence give the Mayer–Vietoris long exact sequence\[\cdots \to H_n(S^1) \to H_n(S^2) \oplus H_n(S^2) \to H_n(X) \to H_{n-1}(S^1) \to \cdots.\]
For CW subcomplexes \(A,B\) with \(X=A\cup B\), the cellular chain groups satisfy \[0\to C_*(A\cap B)\to C_*(A)\oplus C_*(B)\to C_*(X)\to0,\] because each cell of \(X\) lies in \(A\) or \(B\), and a cell occurring in both lies in \(A\cap B\). The associated long exact sequence in homology is the cellular Mayer–Vietoris sequence displayed above.
4. \(H_0(X) = \ZZ\).
\(X\) is connected.
5. \(H_1(X) = 0\).
5.1. The relevant part is \(H_1(S^1) \to H_1(S^2) \oplus H_1(S^2) \to H_1(X) \to H_0(S^1) \to H_0(S^2) \oplus H_0(S^2)\).
Mayer–Vietoris in low degrees.
5.2. \(H_1(S^2) = 0\), and the map \(H_0(S^1) \to H_0(S^2) \oplus H_0(S^2)\) is injective (both inclusions induce the identity on \(H_0\)).
\(H_0(S^1) = \ZZ\) and \(H_0(S^2) = \ZZ\), with the map \(\ZZ \to \ZZ \oplus \ZZ\) given by \(1 \mapsto (1,1)\), which is injective.
5.3. Hence \(H_1(X) = 0\).
exactness: \(H_1(X) \to H_0(S^1)\) is injective (kernel is image of \(H_1(S^2) \oplus H_1(S^2) = 0\)), and its image is the kernel of \(H_0(S^1) \to H_0(S^2) \oplus H_0(S^2)\), which is \(0\) (since that map is injective); so \(H_1(X) = 0\).
6. \(H_2(X) = \ZZ^3\).
6.1. The relevant part is \(H_2(S^1) \to H_2(S^2) \oplus H_2(S^2) \to H_2(X) \to H_1(S^1) \to H_1(S^2) \oplus H_1(S^2)\).
Mayer–Vietoris.
6.2. \(H_2(S^1) = 0\) and \(H_1(S^2) = 0\).
standard.
6.3. Hence \(0 \to \ZZ \oplus \ZZ \to H_2(X) \to \ZZ \to 0\) is exact.
\(H_2(S^2) \oplus H_2(S^2) = \ZZ \oplus \ZZ\) and \(H_1(S^1) = \ZZ\).
6.4. Therefore \(H_2(X) \cong \ZZ^3\).
the short exact sequence \(0 \to \ZZ^2 \to H_2(X) \to \ZZ \to 0\) splits (the last term is free), so \(H_2(X) \cong \ZZ^2 \oplus \ZZ = \ZZ^3\).
7. \(H_n(X) = 0\) for \(n \ge 3\).
\(H_n(S^1) = H_n(S^2) = 0\) for \(n \ge 3\), so the Mayer–Vietoris sequence gives \(H_n(X) = 0\).
8. Q.E.D.
\(H_0 = \ZZ\), \(H_1 = 0\), \(H_2 = \ZZ^3\), \(H_n = 0\) for \(n \ge 3\) (<1>4–<1>7).