TITLE: The De Rham Cohomology of $\mathbb{R}^n - \mathbb{S}^k$
QUESTION [11 upvotes]: I'm reading Madsen and Tornehave's "From Calculus to Cohomology" and tried to solve this interesting problem regarding knots. 
Let $\Sigma\subset \mathbb{R}^n$ be homeomorphic to $\mathbb{S}^k$, show that $H^p(\mathbb{R}^n - \Sigma)$ equals $\mathbb{R}$ for $p=0,n-k-1, n-1$ and $0$ for all other $p$. Here $1\leq k \leq n-2$. 
Now the case $p=0$ is obvious from connectedness and the two other cases are easily solved by applying the fact that
$H^{p+1}(\mathbb{R}^{n+1} - A) \simeq H^p(\mathbb{R}^n - A),~~~~p\geq 1$
and
$H^1(\mathbb{R}^n - A) \simeq H^0(\mathbb{R}^n - A)/\mathbb{R}\cdot 1$
So what is my problem, really?
Now instead let's look at this directly from Mayer-Vietoris. If $\hat{D}^k$ is the open unit disk and $\bar{D}^k$ the closed. Then $\mathbb{R}^n - \mathbb{S}^k = (\mathbb{R}^n - \bar{D}^k)\cup (\hat{D}^k)$ and $(\mathbb{R}^n - \bar{D}^k)\cap (\hat{D}^k) = \emptyset$ 
Now
$H^p(\mathbb{R}^n - \bar{D}^k) \simeq H^p(\mathbb{R}^n - \{ 0 \})$ since $\bar{D}^k$ is contractible. And $H^p(\mathbb{R}^n - \{ 0 \})$ is $\mathbb{R}$ if $p=0,n-1$ and $0$ else. Since $\hat{D}^k$ is open star shaped we find it's cohomology to be $\mathbb{R}$ for $p=0$ and $0$ for all other $p$.
This yields and exact sequence
$\cdots\rightarrow 0\overset{I^{\ast}}\rightarrow H^{n-1}(\mathbb{R}^n - \mathbb{S}^k) \overset{J^{\ast}}\rightarrow \mathbb{R} \rightarrow 0\cdots$
So due to exactness I find that $\ker(J^*) = \text{Im}(I^*) = 0$ and that $J^*$ is surjective, hence 
$H^{n-1}(\mathbb{R}^n - \mathbb{S}^k) \simeq \mathbb{R}$. 
But ... If I apply the exact same approach to $p = n-k-1$ my answer would be $0$ for $H^{n-k-1}(\mathbb{R}^n - \mathbb{S}^k)$. 
Where does this last approach fail?

REPLY [3 votes]: Your question has been answered by Tom.  But I am also not sure if you are aware that the point of the problem was to show that the cohomology of a sphere embedded in euclidean space is independent of the embedding. You seem to think only of the standard embedding.  As Tom mentioned, duality is one way to prove this independence.
(This might have been better as a comment, but I am not yet allowed to comment.)