Antipodal map not homotopic to identity without degree 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?
 A: 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
$F_t(x_0,x_1)=(cos((1-t)\pi)x_0-sin((1-t)\pi)x_1,sin((1-t)\pi)x_0+cos((1-t)\pi)x_1)$
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
$G^{odd}_t(\underline{x})=(F_t(x_0,x_1),F_t(x_2,x_3),\dots,F_t(x_n,x_{n+1}))$.
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
$G_t^{ev}(\underline{x})=(F_t(x_0,x_1),F_t(x_2,x_3),\dots,F_t(x_{n-1},x_{n}),-x_{n+1})$
Then end point of this homotopy is the map given by reflecting across the last coordinate
$\iota_n(x_0,\dots,x_n,x_{n+1})=(x_0,\dots,x_n,-x_{n+1})$
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.
