Hello. I would like to know how to prove that every continuous involution $F:\mathbb{R}^{2}\to\mathbb{R}^{2}$
(that is, $F(F(x))=x$ $\forall x \in \mathbb{R}^{2}$ ) has a fixed point?
Thank you very much in advance!
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Sign up to join this communityHello. I would like to know how to prove that every continuous involution $F:\mathbb{R}^{2}\to\mathbb{R}^{2}$
(that is, $F(F(x))=x$ $\forall x \in \mathbb{R}^{2}$ ) has a fixed point?
Thank you very much in advance!
Suppose $F$ has no fixed points. Define $x\sim y$ iff $x=y$ or $x=F(y)$. The the projection of $\mathbb R^2$ to the quotient $X=\mathbb R^2/\sim$ is a 2-fold covering map and hence $\pi_1(X)=\mathbb Z/2\mathbb Z$. Hence there are non-contractible loops in $X$, and it is easy to find one without self-intersections. The pre-image of such a loop is a Jordan curve in $\mathbb R^2$ invariant under $F$. The domain bounded by this curve is also invariant. Now apply Brouwer's fixed point theorem and that's it.