Homology of spheres: how to show that $\tilde{H}_{n-1}(S^{n-1})\cong\tilde{H}_{n}(S^{n})$, without Mayer Vietoris

Question

I am searching for a proof of this fact, which is used for instance, in showing that the reflection of $S^n$ has deegre $-1$.
Rotman proves it through Mayer-Vietoris, but in my professors' notes this result is often exploited before introducing that tool.
It seems to me my notes suggest this homeomorphism comes from the "equatorial" inclusion $i:S^{n-1}\hookrightarrow S^n$ bbut I can't see why.
The same result should also hold for any suspension.

Answer 1

Consider $S^n$ to be the suspension $\Sigma S^{n - 1}$ for all $n \geq 1$. Let $C^n_+$ and $C^n_-$ denote the upper and lower cone of $\Sigma S^{n - 1}$, respectively, and let $S^{n - 1} \subseteq \Sigma S^{n - 1}$ be the equator. Let $U$ be a neighborhood of the north pole that can be excised from $C^n_+$.

We now look at the long exact sequence

$$ \begin{equation*} \cdots \rightarrow \tilde{H}_k(C^n_+) \rightarrow \tilde{H}_k(S^n) \rightarrow \tilde{H}_k(S^n, C^n_+) \rightarrow \tilde{H}_{k - 1}(C^n_+) \rightarrow \cdots \end{equation*} $$

Since $C^n_+$ is contractible, we have that $\tilde{H}_k(S^n) \cong \tilde{H}_k(S^n, C^n_+)$ for all $k$. Using our neighborhood $U$ from above, by excision, $\tilde{H}_k(S^n, C^n_+) \cong \tilde{H}_k(S^n \setminus U, C^n_+ \setminus U)$, which by (relative) deformation retraction is isomorphic to $\tilde{H}_k(C^n_-, S^{n - 1})$. Considering again the long exact sequence of the pair:

$$ \begin{equation*} \cdots \rightarrow \tilde{H}_k(S^{n - 1}) \rightarrow \tilde{H}_k(C^n_-) \rightarrow \tilde{H}_k(C^n_-, S^{n - 1}) \rightarrow \tilde{H}_{k - 1}(S^{n - 1}) \rightarrow \cdots \end{equation*} $$

we obtain that $\tilde{H}_k(C^n_-, S^{n - 1}) \cong \tilde{H}_{k - 1}(S^{n - 1})$ for all $k$, hence $\tilde{H}_k(S^n) \cong \tilde{H}_{k - 1}(S^{n - 1})$ for all $k$, and the claim follows.