# Show that every orientation reversing homeomorphism of the real line has a fixed point.

A homeomorphism f is said to be orientation reversing if for any $$x we have $$f(z). Show that every orientation reversing homeomorphism of the real line has a fixed point.

This is a question on my assignment sheet (not for credit) that I've been thinking about for days but not made any progress on. I feel like this will be easy to answer once I find the trick, any hints would be great!

• IVT${}{}{}{}{}$? – Lord Shark the Unknown Oct 13 '18 at 18:21
• I was thinking that but I'm not quite sure how to apply it – SpaghettiMonsterMan Oct 13 '18 at 18:23

If $$f(0)=0$$, we are done.
If $$f(0)>0$$, then $$f(f(0)), and if $$f(0)<0$$, then $$f(f(0))>f(0)$$. In both cases, we have poitns $$x_1,x_2$$ with $$f(x_1) and $$f(x_2)>x_2$$. Then the Intermediate Value Theorem tells us that the continuous function $$x\mapsto f(x)-x$$ has a zero between $$x_1$$ and $$x_2$$, i.e., $$f$$ has a fixed point.
• @ihf I'm not sure what your comment means, but the hypothesis that it $f$ is an orientation reversing homeomorphism has been used in this proof, i.e. in the case $f(0)>0$ we may conclude $f(f(0))<f(0)$. This proof would break down if you only assumed $f$ was continuous. – Lee Mosher Oct 13 '18 at 20:32
• @LeeMosher, I meant, does $f$ need to be a bijection? With continuous inverse? Sure, orientation reversing is essential. – lhf Oct 14 '18 at 10:56
• Okay. The only assumption you need is that $f$ is strictly decreasing. In this case $f(\mathbb{R})$ is an open interval (bounded or unbounded), and $f$ is an orientation reversing homeomorphism from $\mathbb{R}$ to $f(\mathbb{R})$. – Paul Frost Oct 15 '18 at 14:58