# The degree of the antipodal map without $\Delta$-complex

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?

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.

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