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We first prove that fixed-point-free sphere maps are homotopic to the antipodal map. If $f:S^d\to S^d$ has no fixed point, defineThe denominator could vanish only if $f(x)$ is a positive multiple of $x$. Both are unit vectors, so this would require $f(x)=x$ and $t=1/2$, excluded by the hypothesis. Thus $H$ is a continuous homotopy from $f$ to $x\mapsto -x$.

On $S^{2r}$, with $r\ge 1$, the antipodal map has degree $(-1{)}^{2r+1}=-1$, since it is the boundary map of the ambient linear transformation $-I$ on ${\mathbb{R}}^{2r+1}$. Homotopy invariance therefore gives degree minus one for every fixed-point-free action map. The degree of each homeomorphism is $\pm 1$, and functoriality makes degree a group homomorphismEvery nonidentity element in a free action has degree minus one, so this homomorphism has trivial kernel. HenceFor $S^0$ the same bound follows immediately from its two points. This proves that free actions on even-dimensional spheres have order at most two, without an Euler-characteristic argument requiring a cell structure on the quotient.