I want to prove that the antipodal map from $S^n$ to $S^n$ is homotopic to the identity map if $n$ is odd. (I know it's actually true if and only if)
If I consider the map $ H(x,t) = (1-2t)x $, why doesn't this work? I don't really see how this is any different if $n$ is odd or even. Is it continuous? Again, I don't see why it isn't
Thanks
As Henning or Rasmus have explained to you, $H(x,t) = (1-2t)x$ fails to land within $S^n$ for $t\neq 0,1$. The usual trick to arrange that, namely redefine
$$ H(x,t) = \dfrac{(1-2t)x}{\| (1-2t)x \|} \ , $$
doesn't work either because this new $H$ is not continuous for $t= \frac{1}{2}$.
If you want an specific homotopy from $\mathrm{id}_{S^{2n-1}}$ to $-\mathrm{id}_{S^{2n-1}}$, here is one.
Let's start with $n=1$. Let's think $S^1$ as the set of complex numbers $z \in \mathbb{C}$ of modulus 1: $\vert z \vert = 1$. Consider the map:
$$ H: S^1 \times I \longrightarrow S^1 $$
defined by
$$ H(z,t) = e^{i\pi t} z \ . $$
This map is plainly continuous, its image lies on the sphere $\vert H(z,t) \vert = \vert e^{i\pi t}\vert \cdot \vert z \vert = \vert z \vert = 1$ and $H(z,0) = z$ and $H(z,1) = -z$. Hence, it may be the homotopy you were looking for, right?
For $n \geq 1$, exactly the same map works. Now you think your odd sphere $S^{2n-1}$ as being the subset of complex $n$-tuples $z = (z_1, \dots , z_n) \in \mathbb{C}^n$ of modulus 1, $\vert z \vert = \sqrt{\vert z_1\vert^2 + \dots + \vert z_n\vert^2 } = 1$. And you consider the same map:
$$ H: S^{2n-1} \times I \longrightarrow S^{2n-1} \ . $$
Namely,
$$ H(z,t) = e^{i\pi t} z = (e^{i\pi t} z_1 , \dots ,e^{i\pi t} z_n ) \ . $$
The same proof we did for $n=1$ applies now and shows that $H$ is the desired homotopy.
Notice that, since we had to rely on complex numbers, this does not work for even spheres.
Finally, if you prefer, another way to write the map $H$ for $n=1$, now thinking $S^1$ as the set of points $(x,y) \in \mathbb{R}^2$ such that $\| (x,y) \| = 1$, could be
$$ H(x,y,t) = \begin{pmatrix} \cos (\pi t) & -\sin (\pi t) \\ \sin (\pi t) & \cos (\pi t) \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \ . $$
That is, we are rotating $(x,y)$ more and more, as $t$ increases, till we arrive at $(-x,-y)$. Notice that you need to be in $\mathbb{R}^2$ to do this: you couldn't rotate anything if we were talking about the "even" sphere $S^0 = \left\{ -1, +1\right\} \subset \mathbb{R}$ and clearly there is no homotopy from $\mathrm{id}_{S^0}$ to $-\mathrm{id}_{S^0}$; that is, a path inside $S^0$, joining $-1$ and $+1$.
This follows easily from some algebraic topology, but you can:
1) prove that $GL(n,\mathbb{C})$ is path-connected (by hands)
2) $U(n)$ is path-connected, because they are homotopy-equivalent (equivalence being the Gram-Schmidt orthogonalisation.)
Every element of $U(2n)$ restricts to a map of $S^{2n-1}$. But both identity and antipodal maps lie in $U(n)$, so they are connected by a path. This will be your required homotopy.
Your $H(x,t) = (1-2t)x$ fails to land within $S^n$ if $t$ is strictly between $0$ and $1$.
Hint: $-I\in SO(k)$ iff $k$ is even.
Here is why your approach does not work:
If $t$ is not $0$ or $1$, then $(1-2t)$ will have modulus less than $1$. Hence $(1-2t)x$ does not belong to the sphere anymore. But if $H$ was a homotopy of maps to the sphere, it would take values on the sphere.
