i need some help please my professor asked for
homology groups of $S^2\cup d$ which $d={\{ (0,0,t) \mid -1 \leq t \leq 1}\} $ .
We just find homology groups with some points and Mayor-veitoris and CW-complex structures. For solving this problemوi thought we can take $S^2/(0,0,0)$ as $u$ and $d/(0,0,0)$ as $v$ in M-V theorem.We know $\mathbb R^3$ is contractible and if we look at $d$ as a subset of $\mathbb R^3$ ,could i say that $v$ is contractible and is homotope with a point? please guide me for taking $u$ and $v$ or making right CW-Complex for this and let me know that is the intersection between them homotope with a point or not, please...
Thanks in Advance
$\newcommand{\Z}{\mathbb{Z}}$If you want to use Mayer-Vietoris, let $A = S^2$ and $B = d$. Set $X = A \cup B$; then $A \cap B$ is the disjoint union of two points $N$ and $S$ (the north and south poles). Then \begin{align*} H_k(A) = H_k(S^2) &= \begin{cases} \Z & k=0,2 \\ 0 & \text{otherwise} \end{cases}\\ H_k(B) = H_k(\text{pt}) &= \begin{cases} \Z & k=0 \\ 0 & \text{otherwise} \end{cases}\\ H_k(A \cap B) = H_k(\text{pt}) \oplus H_k(\text{pt}) &= \begin{cases} \Z \oplus \Z & k = 0\\ 0 & \text{otherwise} \end{cases} \end{align*} Since $X$ is connected, $H_0(X) = \Z$. One portion of the Mayer-Vietoris sequence is $$ H_2(A \cap B) \longrightarrow H_2(A) \oplus H_2(B) \longrightarrow H_2(X) \longrightarrow H_1(A \cap B) $$ or $$ 0 \longrightarrow \Z \oplus 0 \longrightarrow H_2(X) \longrightarrow 0 $$ So we know $H_2(X) = \Z$. Another portion is $$ H_1(A) \oplus H_1(B) \longrightarrow H_1(X) \stackrel{\partial_*}{\longrightarrow} H_0(A \cap B) \stackrel{(i_*,j_*)}{\longrightarrow} H_0(A) \oplus H_0(B) $$ Since $H_1(A) \oplus H_1(B)=0 \oplus 0 = 0$, $\partial_*$ is an injection. So $H_1(X)$ is isomorphic to $\operatorname{im} \partial_* = \ker (i_*,j_*)$.
Now $H_0(A \cap B)$ is a free abelian group on two generators: $[N]$ and $[S]$. And $H_0(A) \oplus H_0(B)$ is a free abelian group on two generators: $[A]$ and $[B]$. Since $i$ and $j$ are the inclusion maps from $A \cap B$ to $A$ and $B$, we have \begin{align*} i_*([N]) = i_*([S]) &= [A] \\ j_*([N]) = j_*([S]) &= [B] \\ \end{align*} So the kernel of $(i_*,j_*)$ is generated by $[N]-[S]$. This means $H_1(X) = \Z$.
We can find a generator of $H_1(X)$, too. Let $c$ be an arc running from $N$ to $S$ along the sphere. Then $d+c$ is a cycle but not a boundary. And if you follow the construction, you can see why $\partial_*([d+c]) = [N] - [S]$.