Using homology and the notion of degree one can show that the antipodal map $S^n \to S^n$, $z \mapsto -z$ is not homotopic to the identity if $n$ is even. Do you know of a proof of this fact that does not rely on homology? Is there a proof that only uses homotopy theory?


We have $S^n=\{\underline{x}=(x_0,\dots,x_n)\in\mathbb{R}^{n+1}\mid \underline{x}^2=1\}$ with the antipodal map $A_n:S^n\rightarrow S^n$ defined by $A_n(\underline{x})=-\underline{x}$. Consider first $n=1$. Then $A_1\simeq id_{S^1}$ since it is just a rotation by $\pi$ degrees. Explicitly we have $F_t:A_1\simeq id_{S^1}$ given by


Now for $n>1$ we can use $F$ to get a homotopy of $A_n$ to some other map. If $n=2m+1$ is odd then $S^{2m+1}\subseteq\mathbb{R}^{2(m+1)}$ so there are an even number of coordinates and we can define $G^{odd}_t:A_n\simeq id_{S^n}$ by


Now if $n=2m$ is even then there are an odd number of coordinates so we define a homotopy $G^{ev}_t$ starting at $A_n$ by


Then end point of this homotopy is the map given by reflecting across the last coordinate


and this map is not homotopic to the identity. Indeed on $S^1$ the map $\iota_1$ is the degree $-1$ map. For $n>1$ choose a homeomorphism $S^n\cong S^{n-1}\wedge S^1$ (preferably using stereographic projection). Then we see explicitly that under this homeomorphism we have $\iota_n=id_{S^{n-1}}\wedge \iota_1$. That is, it is the $(n-1)$-fold suspension of $\iota_1$. Since suspension is an isomorphism $\Sigma:\pi_rS^r\xrightarrow{\cong}\pi_{r+1}S^{r+1}$ we get for $n$ even $A_n\simeq \iota_n=id_{S^{n-1}}\wedge \iota_1\simeq \Sigma^{n-1}\iota_1\simeq\Sigma^{n-1}(-1)\simeq -1$ is homotopic to the degree $-1$ map.


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