Homology of $\mathbb R^n-f(S^r)$, where $f$ is embedding

Question

Let $f:S^r\rightarrow\mathbb R^n$ be an embedding. What can we say about the homology of its complement $X:=\mathbb R^n-f(S^r)$?

So my first thought was viewing $\mathbb R^n\subset S^n$ and using the Jordan curve theorem.

Jordan curve Theorem. If $f:S^r\rightarrow S^n$ is an embedding, then $\tilde H_*(S^n-f(S^r))=\mathbb Z[n-r-1]$, that is, the reduced homology vanishes in all degrees except for $n-r-1$.

Now applying this to the composition $\tilde f:S^r\xrightarrow f \mathbb R^n\hookrightarrow S^n$, we get an inclusion $$\tilde H_k(\mathbb R^n-f(S^r))\rightarrow\tilde H_k(S^n-\tilde f(S^r)),$$ right? Can we conclude that $\tilde H_k(\mathbb R^n-f(S^r))=0$ for $k\neq n-r-1$?

Will it be $\mathbb Z$ for $k=n-r-1$?

Answer 1

There is a long exact exact sequence of (reduced) homology groups containing the relative groups $H_k(S^n \setminus f(S^r), \mathbb R^n \setminus f(S^r))$. Excision shows that they are isomorphic to $H_k(S^n, \mathbb R^n)$. Since $\mathbb R^n$ is contractible, we have $H_k(S^n, \mathbb R^n) \approx H_k(S^n,*) \approx \tilde H_k(S^n)$.

For $k \ne n-1$ we get $\tilde H_{k+1}(S^n) = 0$ and for $k \ne n-r-1$ w get $H_k(S^n \setminus\tilde f(S^r)) = 0$.

Thus for $k \ne n-1, n-r-1$ the sequence $0 \to \tilde H_k(\mathbb R^n \setminus f(S^r))\rightarrow 0$ is exact and therefore $\tilde H_k(\mathbb R^n \setminus f(S^r)) = 0$.

Now let $r > 0$. Then the sequence $0 \to \mathbb Z \to \tilde H_{n-1}(\mathbb R^n \setminus f(S^r))\rightarrow 0$ is exact, thus $\tilde H_{n-1}(\mathbb R^n \setminus f(S^r)) = \mathbb Z$. Moreover $0 \to \tilde H_{n-r-1}(\mathbb R^n \setminus f(S^r))\rightarrow \tilde H_{n-r-1}(S^n \setminus f(S^r)) \to 0$ is exact, thus $\tilde H_{n-r-1}(\mathbb R^n \setminus f(S^r)) = \mathbb Z$.

Let $r = 0$. Then $0 \to \mathbb Z \to \tilde H_{n-1}(\mathbb R^n \setminus f(S^r)) \to \mathbb Z\rightarrow 0$ is exact. The sequence splits, thus $\tilde H_{n-1}(\mathbb R^n \setminus f(S^r)) = \mathbb Z \oplus \mathbb Z$.

Summarzing we have

$$\tilde H_{n-1}(\mathbb R^n \setminus f(S^r)) = \mathbb Z[n-r-1] \oplus \mathbb Z [n-1] .$$