# Fixed points of continuous involutions of the plane

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!

• Read this paper matwbn.icm.edu.pl/ksiazki/fm/fm43/fm43124.pdf – Petya Mar 14 '10 at 18:53
• This is the easy special case of a question asked here recently: mathoverflow.net/questions/17707/…. One of my answers contains a solution to the problem precisely under the additional condition of the continuity of the involution, so it answers this question. – Pete L. Clark Mar 14 '10 at 20:35
• In fact, the unknown who asked this question is the same unknown who asked the previous question. Strange... – Pete L. Clark Mar 14 '10 at 20:50

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.
• If X is a CW-complex and is doubly covered by a contractible space, then it is an Eilenberg-MacLaine space K(Z/2Z,1) and therefore is homotopy equivalent to $RP^\infty$. But the latter has nontrivial comology groups in infinitely many dimensions, thus X must be infinite dimensional. Maybe one can work around the assumption that X is a CW-complex, but I don't know for sure. – Sergei Ivanov Mar 15 '10 at 9:02
• @SergeiIvanov, please, what if we replace $\mathbb{R}$ by $\mathbb{C}$? – user237522 May 6 '17 at 22:05