Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is a homework problem in an undergraduate topology class that I am ludicrously (and probably stupidly) stuck on. Any guidance would be appreciated:

Let X be a connected space, $f:X\to X$ a continuous involution (i.e. f is its own inverse), $g:X\to\mathbb{R}$ continuous. Prove that $\exists$ $x\in X$ such that $g(x)=g(f(x))$.

Here are the things I understand:

Some examples of $f$ include the identity map, $f(x)=-x$, $f(x)=\frac{1}{x}$ if $x\neq0$, etc. The result clearly holds if $f$ has a fixed point. I know any continuous involution in $\mathbb{R}^2$ has a fixed point, but we are on an arbitrary connected space, and I have no results I can use for such a thing. I can define a new continuous function $h:X\to\mathbb{R}$ so that $h(x)=g(f(x))-g(x)$. If this function has a zero, the result follows, but I don't know how to show it does. I could apply the intermediate value theorem since X is connected and $g:X\to\mathbb{R}$, but I again don't see how this would be helpful.

Thanks very much in advance!

share|improve this question
add comment

1 Answer

HINT: Let $h(x)=g\big(f(x)\big)-g(x)$. Suppose that $x\in X$ is such that $h(x)<0$. What can you say about $h\big(f(x)\big)$? (You’re actually on the right track towards the end of your question.)

share|improve this answer
    
Thank you! That's all the nudge I needed. –  N. A. Ericson Dec 14 '12 at 23:36
    
@N.A.Ericson: Thought it might be. :-) You’re welcome. –  Brian M. Scott Dec 14 '12 at 23:37
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.