What is homology/cohomology of $S^n \smallsetminus S^k$ and $\mathbb{R}P^n \smallsetminus \mathbb{R}P^k$?

Question

For $k=0, S^n \smallsetminus S^0$ is homotopy equivalent to $S^{n-1}$.

But, how to determine when $k$ is not equal to $0$?

Answer 1

$S^n \setminus S^k$ is homotopy equivalent to $S^{n-k-1}$. Further, $\Bbb{RP}^n \setminus \Bbb{RP}^k$ is homotopy equiavlent to $\Bbb{RP}^{n-k-1}$. These statements are straightforward to show and I will leave them to you as exercises.

Answer 2

No, obviously I'm not using that decomposition, William. Define the deformation retraction

$$f_t(x_0, \cdots, x_n) = (t x_0, \cdots, tx_k, x_{k+1}, \cdots, x_n) + (0, \cdots, 0, \sqrt{\frac{1-t^2}{n-k}(x_0^2 + \cdots x_k^2)+x_{k+1}^2}, \cdots, \sqrt{\frac{1-t^2}{n-k}(x_0^2 + \cdots x_k^2)+x_n^2}).$$ I'm sure you can at least compute that this is a deformation retraction of $S^n \setminus S^k$ onto $S^{n-k-1}$.

The definition of this homotopy is made completely clear by thinking of literally any example you can picture. You contract by a 'straight line' to the complementary sphere to $S^k$, the one point compactification of the radial deformation retraction of $\Bbb R^n \setminus \Bbb R^k$ onto $S^{n-k-1}$. It is also obviously Z/2 invariant, so descends to a deformation retraction downstairs. If you have difficulty with this I don't know what to tell you.