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In Hatcher there is a proof that the antipodal map of a sphere $S^n$ has degree $(-1)^{n+1}$ that uses a $\Delta$-complex structure (e.g. see this question).

I wonder if there is another proof, that doesn't use the $\Delta$-complex structure. My intuition is that I only need to replace $\Delta_0-\Delta_1$ with the generator of $H_n(U\cap V)$ that appears in the standard Mayer-Vietoris sequence for $S^n$.

Does this work?

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Is your idea the following? If so, then yes this works.

Take the usual cover of the sphere $S^n$ by open neighborhoods of the upper and lower hemispheres, $U$ and $V$. Let $\sigma$ denote the antipodal map. WLOG we can assume that $\sigma U = V$. We also assume that $U\cap V$ deformation retracts onto the equator, $S^{n-1}$.

Then $\sigma$ induces a map of triples $(S^n,U,V)\to (S^n,V,U)$. Thus since $H^n(U)=H^n(V)=H^{n-1}(U)=H^{n-1}(V)=0$ for $n>1$, we get a map of Mayer-Vietoris sequences: $$ \require{AMScd} \begin{CD} 0 @>>>H_n (S^n) @>\partial_{S^n,U,V}>> H_{n-1}(S^{n-1}) @>>> 0 \\ @. @V\sigma_* VV @V\sigma_* VV @. \\ 0 @>>>H^n (S^n) @>\partial_{S^n,V,U}>> H_{n-1}(S^{n-1}) @>>> 0. \\ \end{CD} $$

Now the trick is to recall the definition of the boundary map, $\partial_{S^n,U,V}$. Given a cycle $z \in C_n(S^n)$, find an equivalent cycle of the form $x-y$, with $x$ a cycle in $C_n(U)$, $y$ a cycle of $C_n(V)$, apply the boundary map to the pair $(x,y)$ to get the pair $(\partial x,\partial y)$, then pullback along the inclusion of $C_{n-1}(S^{n-1})\hookrightarrow C_{n-1}(U)\oplus C_{n-1}(V)$. I.e., take the chain $\partial x = \partial y$, which is a chain in $S^{n-1}$.

The only difference when computing the boundary map $\partial_{S^n,V,U}$ is that now we need to have the cycle from $C_n(U)$ be negative. So we write $z=(-y)-(-x)$, where $x$ and $y$ are the same cycles as before. Then we apply the boundary map to get the pair $(-\partial y, -\partial x)$, then the final chain is $-\partial x = -\partial y$ regarded as an element of $C_{n-1}(S^{n-1})$.

Thus $\partial_{S^n,U,V} = -\partial_{S^n,V,U}$. This says that for $n>1$ $$\deg_{S^n}\sigma = -\deg_{S^{n-1}}\sigma.$$ Moreover, since $\deg_{S^1}\sigma=1$, this yields that $\deg_{S^n}\sigma = (-1)^{n+1}$, as desired.

Not sure if this is what you meant, but do let me know.

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  • $\begingroup$ This is so clear and beautiful! Thank you very much, this helped me a lot. $\endgroup$ – Paweł Czyż Feb 14 '20 at 19:27

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