Any reflection is homotopic to the antipodal map on $\ S^{2n}$.

By the degree argument we see that any reflection on $\ S^{2n}$ is homotopic to the antipodal map.

But that seems a big theorem. I am looking for a straightforward argument.

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Hint: Consider a suitable path in $\mathrm{O}(2 n + 1)$ from $\textrm{diag}(1, \ldots, 1, -1)$ to $\textrm{diag}(-1, \ldots, -1, -1)$. – Zhen Lin Dec 9 '12 at 17:59
I think you mean a reflection specifically in a hyperplane, right? For example, reflecting $S^2$ in any axis in $\mathbb{R}^3$ is a half-turn about that axis, which is homotopic to the identity map. – Hew Wolff Dec 9 '12 at 22:46