Excision theorem We consider $f:S^{n}\rightarrow S^{n}$ a continuous function. Then we consider $y\in S^{n}$ such as $f^{-1}(y)=\{x_{1},..., x_{p}\}$, $U_{1},..., U_{m}$ $x_{i}$-neighbourhoods, and $V$ y-neighbourhood such as $f(U_{i})\subset V$.

Prove that $H_{n}(S^{n},S^{n}\setminus x_{i})\simeq H_{n}(U_{i},U_{i}\setminus x_{i})$, $H_{n}(S^{n},S^{n}\setminus x_{i})\simeq H_{n}(S^{n})$ and $H_{n}(S^{n},S^{n}\setminus f^{-1}(y))\simeq \mathbb{Z}^{p}$.

For the first equivalence, I want to apply the excision theorem :

$H_{n}(S^{n}\setminus(S^{n}\setminus U_{i}),(S^{n}\setminus x_{i})\setminus (S^{n}\setminus U_{i}))\simeq H_{n}(S^{n},S^{n}\setminus x_{i})$, hence the result.

Is that true ? And for the others equivalence, excision ?
Thank you.
For the second one, I know that :

$H_{n+1}(S^{n},S^{n}\setminus x_{i})\rightarrow H_{n}(S^{n}\setminus x_{i})\rightarrow H_{n}(S^{n})\rightarrow H_{n}(S^{n},S^{n}\setminus x_{i})$ is exact.

 A: The first one is correct, since $S^n\setminus (S^n\setminus U_i)=U_i$ and so on. For the second one, you're definitely on the right track, but you need to look at another place in the sequence: $$H_n(S^n\setminus\{x_i\})\to H_n(S^n) \to H_n(S^n, S^n\setminus \{x_i\})\to H_{n-1}(S^n\setminus\{x_i\})$$
where the first and last terms are zero, and therefore the middle map is an isomorphism.
The last one, you need to look at the long exact sequence again, in the same dimensions, but add $\to H_{n-1}(S^n)$ to the right hand side. The sequence then becomes
$$
0\to \Bbb Z \to H_n(S^n, S^n\setminus f^{-1}(\{y\})) \to \Bbb Z^{p-1}\to 0
$$
and therefore the result must hold.
A: I'm super new to homology so take this answer with a grain of salt.
You definitely have the first part right. The third one looks like you apply excision, the fact that the homology of the disjoint union is a product of homologies, the second result, and the knowledge that $H_n(S^n)=\Bbb Z$.
For the second part, I would encourage you to use a different section of the exact sequence: remember that you want $H_n(S^n, S^n\smallsetminus x_i)$ and therefore having it as the first term of your exact sequence isn't doing you much good. If it were the second or third term and there were a bunch of zeros hanging around (hint: there are) then maybe you could get an isomorphism from the exactness.
