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

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.

• Do you know about the suspension isomorphism? Feb 28 '20 at 10:47
• @BenSteffan no, but it is the kind of thing I am searching! Feb 28 '20 at 10:51
• Suspension in this case is Mayer-Vietoris. Feb 28 '20 at 11:22
• @Tyrone is there a more general link between excision and Mayer Vietoris? I would be interested! Feb 28 '20 at 12:03

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