# Special case for Alexander's Duality

Let $C_1$ and $C_2$ be homeomorphic closed sets in $S^n$.Prove that: $$H^{r}_{dR}(S^n\setminus C_1 )\cong H^{r}_{dR}(S^n\setminus C_2 )$$

Here's what I tried:

Let $p \in S^n\setminus C_1$ , $U_1=B_\epsilon (p)\cap S^n$ such that $U_1\cap (S^n\setminus C_1)=\phi$ and $V_1=(S^n\setminus C_1)\setminus \left \{ p \right \}$.Using Mayer Vietoris I get a long exact sequence where $H^{r}_{dR}((S^n\setminus C_1)\setminus \left \{ p \right \})\cong H^{r}_{dR}(\mathbb{R}^n \setminus \bar{C_1})$ where $C_1$ is homeomorphic to $\bar{C_1}$ by the stereographic projection , $H^{r}_{dR}(U_1 \cap V_1) \cong H^{r}_{dR}(S^{n-1})$ and $U_1$ is contractible so $H^r_{dR}(U_1) \cong 0$ when $r\neq 0$ and $H^0_{dR}(U_1) \cong \mathbb{R}$. The long sequence looks like this:

$$...\rightarrow H^{r-1}_{dR}(U_1\cap V_1)\rightarrow H^{r}_{dR}(S^n\setminus C_1)\rightarrow H^{r}_{dR}(U_1) \oplus H^{r}_{dR}(V_1) \rightarrow H^{r}_{dR}(U_1\cap V_1) \rightarrow...$$

or

$$...\rightarrow H^{r-1}_{dR}(S^{n-1})\rightarrow H^{r}_{dR}(S^n\setminus C_1)\rightarrow H^{r}_{dR}(U_1) \oplus H^{r}_{dR}(\mathbb{R}^n \setminus \bar{C_1}) \rightarrow H^{r}_{dR}(S^{n-1}) \rightarrow...$$

By Alexander's Duality theorem : $H^{r}_{dR}(\mathbb{R}^n \setminus \bar{C_1}) \cong H^{r}_{dR}(\mathbb{R}^n \setminus \bar{C_2})$ where $\bar{C_2}$ is homeomorphic to $C_2$ by the stereographic projection, then we have similar sequences:

$$...\rightarrow H^{r-1}_{dR}(U_2\cap V_2)\rightarrow H^{r}_{dR}(S^n\setminus C_2)\rightarrow H^{r}_{dR}(U_2) \oplus H^{r}_{dR}(\mathbb{R}^n \setminus \bar{C_2})\rightarrow H^{r}_{dR}(S^{n-1}) \rightarrow...$$

or

$$...\rightarrow H^{r-1}_{dR}(S^{n-1})\rightarrow H^{r}_{dR}(S^n\setminus C_2)\rightarrow H^{r}_{dR}(U_2) \oplus H^{r}_{dR}(\mathbb{R}^n \setminus \bar{C_2}) \rightarrow H^{r}_{dR}(S^{n-1}) \rightarrow...$$

where $U_2$, $V_2$ are similarly defined as $U_1$, $V_1$.

I don't know if I can apply some dimension argument here because the only terms differing in the sequences are the ones involving $S^n\setminus C_1$ and $S^n\setminus C_2$.

Any idea would be appreciated.

• The statement is false, no sense in trying to prove it. – Ryan Budney Nov 5 '12 at 4:27
• @RyanBudney -I'm so sorry, missed the most important part :$C_1$ and $C_2$ are homeomorphic.I just edited it. – jandrew Nov 5 '12 at 4:32