Compute the integral homology groups \(H_k(X; \mathbb{Z})\) of the space \(X = Y \cup Z \subset \mathbb{R}^3\), where \(Y\) is the sphere \[ Y = \{ (x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 = 1 \} \] and \(Z\) is the ellipsoid \[ Z = \left\{ (x, y, z) \in \mathbb{R}^3 \;\middle|\; x^2 + y^2 + \frac{z^2}{4} = 1 \right\}. \]
Solution
1. The two surfaces are both homeomorphic to \(S^2\), and their intersection is the common equatorial circle.
The unit sphere \[ Y=\{x^2+y^2+z^2=1\} \] is \(S^2\). The linear map \[ (x,y,z)\longmapsto(x,y,2z) \] maps the unit sphere homeomorphically onto \[ Z=\left\{x^2+y^2+\frac{z^2}{4}=1\right\}, \] so \(Z\cong S^2\) as well.
If a point lies in \(Y\cap Z\), subtracting the two defining equations gives \[ z^2-\frac{z^2}{4}=0, \] hence \[ z=0. \] Substitution then gives \[ x^2+y^2=1. \] Therefore \[ A:=Y\cap Z = \{(x,y,0):x^2+y^2=1\} \cong S^1. \]
2. Mayer–Vietoris applies to the decomposition\[ X=Y\cup Z \]with intersection \(A\cong S^1\).
Give the common equator \(A\) its standard CW structure with one \(0\)-cell and one \(1\)-cell. Each of the upper and lower hemispheres of \(Y\) is then a \(2\)-cell attached along \(A\), and the same is true for the upper and lower halves of \(Z\). Thus \(X\) is a CW complex for which \(Y\), \(Z\), and \(A=Y\cap Z\) are subcomplexes. The Mayer–Vietoris sequence for a union of CW subcomplexes therefore gives \[ \cdots \longrightarrow \widetilde H_k(A) \longrightarrow \widetilde H_k(Y)\oplus\widetilde H_k(Z) \longrightarrow \widetilde H_k(X) \longrightarrow \widetilde H_{k-1}(A) \longrightarrow\cdots. \] By 1, \[ \widetilde H_k(A) \cong \begin{cases} \mathbb Z,&k=1,\\ 0,&k\ne1, \end{cases} \] and \[ \widetilde H_k(Y)\oplus\widetilde H_k(Z) \cong \begin{cases} \mathbb Z^2,&k=2,\\ 0,&k\ne2. \end{cases} \]
3. One has\[ H_k(X;\mathbb Z)=0 \qquad(k\ge3). \]
For \(k\ge3\), the relevant Mayer–Vietoris terms from 2 are \[ 0 \longrightarrow \widetilde H_k(X) \longrightarrow 0. \] Hence \[ \widetilde H_k(X)=0 \] for every \(k\ge3\).
4. The second homology group is\[ H_2(X;\mathbb Z)\cong\mathbb Z^3. \]
The relevant part of the reduced Mayer–Vietoris sequence is \[ 0 \longrightarrow \mathbb Z^2 \longrightarrow \widetilde H_2(X) \xrightarrow{\partial} \mathbb Z \longrightarrow 0. \] Exactness gives a short exact sequence \[ 0 \longrightarrow \mathbb Z^2 \longrightarrow \widetilde H_2(X) \longrightarrow \mathbb Z \longrightarrow 0. \] Because the quotient \(\mathbb Z\) is free abelian, this sequence splits. Therefore \[ \widetilde H_2(X) \cong \mathbb Z^2\oplus\mathbb Z \cong \mathbb Z^3. \] Since degree \(2\) is positive, reduced and unreduced homology agree there.
5. The first homology group is\[ H_1(X;\mathbb Z)=0. \]
Continuing the same exact sequence from 4 gives \[ \mathbb Z \longrightarrow 0 \longrightarrow \widetilde H_1(X) \longrightarrow 0. \] Exactness at \(\widetilde H_1(X)\) says \[ \operatorname{im}\bigl(0\to\widetilde H_1(X)\bigr) = \ker\bigl(\widetilde H_1(X)\to0\bigr). \] The left-hand side is \(0\), while the right-hand side is all of \(\widetilde H_1(X)\). Hence \[ \widetilde H_1(X)=0. \]
6. Finally,\[ H_0(X;\mathbb Z)\cong\mathbb Z. \]
Both \(Y\) and \(Z\) are path connected, and their intersection \(A\cong S^1\) is nonempty. Therefore their union \(X\) is path connected, so \[ H_0(X;\mathbb Z)\cong\mathbb Z. \]
7. Thus the integral homology groups are\[ \boxed{ H_k(X;\mathbb Z) \cong \begin{cases} \mathbb Z,&k=0,\\ 0,&k=1,\\ \mathbb Z^3,&k=2,\\ 0,&k\ge3. \end{cases}} \]
Combine <1>3–<1>6.