Homology of the closed topologist sine curve The closed topologist sine curve, $X$, is the subspace of $R^2$ consisting of all the points $(x,\sin(1/x))$ for $x \in (0,1]$, and all points $(0,y)$ for $y \in [-1,1]$ and an arc from $(0,-1)$ to $(1,\sin(1))$ Compute the singular homology of $X$, using a suitable Mayer-Vietoris sequence.
So to utilize the Mayer-Vietoris sequence I need an excisive couple, one condition that will gaurentee that a pair is an excisive pair is if the union of their interiors cover the space. 
$X_1$ = $(x,\sin(1/x))$ for $x \in (0,1]$ and the arc from $(0,-1)$ to $(1,\sin(1))$ 
$X_2$ = all points $(0,y)$ for $y \in [-1,1]$ and the arc from $(0,-1)$ to $(1,\sin(1))$
I don't know i'm just taking a shot in the dark really I guess. Can anyone offer some insight for me? Thanks!!
 A: You do not consider the closed topologist's sine curve, but the Warsaw circle. See To show that Warsaw circle is simply connected.
Let $S = \{(x,\sin\frac{1}{x}):0<x \le 1 \}$, $L = \{0\} \times [-1,1]$ and $T = L  \cup S$. This is the closed topologist's sine curve. Your space $X$ is obtained by joining $(0,-1), (1,\sin(1)) \in T$ by an arc $A$ such that $A$ does not meet other points of $T$.
Your subspaces $X_1, X_2$ do not form an excisive couple. The points of $L \subset X_2$ are no interior points of $X_1$ and $X_2$.
Choose a homeomorphism $h : [0,1] \to A$. Let $X_1 = h((0,1))$ and $X_2 = X \setminus \{h(1/2)\}$. Then you have an excisive couple. $X_1$ is contractible, thus
$$H_n(X_1) = \begin{cases} 0 & n > 0 \\ \mathbb Z & n = 0\end{cases}$$
The space $X_2$ contains $T$ as a strong deformation retract and $T$ has two path components ($L$ and $S$) which are both contractible. Hence
$$H_n(X_2) \approx H_n(T) \approx H_n(L) \oplus H_n(S) = \begin{cases} 0 & n > 0 \\ \mathbb Z \oplus \mathbb Z & n = 0\end{cases}$$
Moreover, $X_1 \cap X_2$ is the disjoint union of two copies $I_1, I_2$ of an open interval. Thus
$$H_n(X_1 \cap X_2) \approx H_n(I_1) \oplus H_n(I_2) = \begin{cases} 0 & n > 0 \\ \mathbb Z \oplus \mathbb Z & n = 0\end{cases}$$
Finally, since $X$ is path-connected
$$H_0(X) = \mathbb  Z .$$
By Mayer-Vietoris we get for $n > 0$ an exact sequence
$$0 \to H_n(X) \to H_{n-1}(X_1 \cap X_2) \to H_{n-1}(X_1) \oplus H_{n-1}(X_2) \to H_{n-1}(X) \to \ldots$$
For $n > 1$ we know that $H_{n-1}(X_1 \cap X_2) = 0$, thus $H_n(X) = 0$.
Let us come to $n = 1$. We get the exact sequence
$$0 \to H_1(X) \to H_0(X_1 \cap X_2) \to H_0(X_1) \oplus H_0(X_2) \to H_0(X) \to 0$$
If $K$ denotes the kernel of $H_0(X_1) \oplus H_0(X_2) \to H_0(X)$, we get a short exact sequence
$$0 \to K \to H_0(X_1) \oplus H_0(X_2) \to H_0(X) \to 0$$
Since $H_0(X) = \mathbb Z$, this sequence splits and we get
$$H_0(X_1) \oplus H_0(X_2) \approx K \oplus H_0(X) .$$
We already know $H_0(X_2) = \mathbb Z \oplus \mathbb Z$ and $H_0(X_1) = H_0(X) = \mathbb Z$, thus
$$\mathbb Z \oplus \mathbb Z \oplus \mathbb Z \approx K \oplus \mathbb Z .$$
This shows that $K \approx \mathbb Z \oplus \mathbb Z$. But $K$ agrees with the image of $H_0(X_1 \cap X_2) = \mathbb Z \oplus \mathbb Z \to H_0(X_1) \oplus H_0(X_2)$. This is only possible if this map is injective. Thus we must have $H_1(X) = 0$.
A: We do not need the Mayer-Vietoris sequence and what we use is primary.
Computing the homology group of the part $X_2=\lbrace(0, y)\ |\ -1 \leq y \leq 1\rbrace$ is easy because it is bounded and convex. Hence $n=0$，$H_0(X_2)=Z$ and $n \geq 1$，$H_n(X_2)=0$
The part of $X_1=\lbrace(x, \sin(\frac{1}{x})\ |\ 0 < x \leq 1\rbrace$ is a bit complicated. Define $f \in S_n(X_1)$ and $f(t)=(x(t), y(t))$ where $t$ is a point of $n$-simplex. For the $n$-simplex is closed, $\inf x(t) > u > 0$. Hence we can define $K=\lbrace(x, \sin(\frac{1}{x}))\ |\ u \leq x \leq 1\rbrace$, then $im(f) \subseteq K$.
If $cls\ y \in H_n(X_1)$, we have $supp\ y \subseteq \text{some}\ K$. And assume $j$ is the inclusion between $K$ and $X_1$. Then $cls\ y \in im\ j^*$ where $j^*$ is the inclusion between $H_n(K)$ and $H_n(X_1)$. Because $K$ is homeomorphic to $I=[0, 1]$, $n \geq 1$, $H_n(K)=0$. Hence $cls\ y = 0$ in $H_n(X_1)$ and $n \geq 1$, $H_n(X_1)=0$.
Then finally $n=0$, $H_0(X)=Z \oplus Z$ and $n \geq 1$, $H_n(X)=0$.
