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.

I've been stuck on a problem from Rotman's Introduction to Algebraic Topology for a while. I'm doing the exercises outside of class right now so it's difficult to ask for help. I'm hoping someone here can help out.

The question is:

Let $f,g:I \to I \times I$ be continuous; let $f(0)=(a,0)$ and $f(1) = (b,1)$, and let $g(0) = (0,c)$ and $g(1)=(1,d)$ for some $a,b,c,d \in I$. Show that $f(s)=g(t)$ for some $s,t \in I$; that is, the paths intersect. (Hint: Use Theorem 0.3 (Brouwer's Fixed Point Theorem for the disc $D^n$) for a suitable map $I \times I \to I \times I$ (There is a proof in [Maehara]; this paper also shows how to derive the Jordan curve theorem from the Brouwer theorem.)

Here, $I=[0,1]$. I have shown in a previous problem if $X$ is homeomorphic to $D^n$ then any continuous function from $X$ to $X$ has a fixed point so we can use this on $I \times I$. I tried reading the paper by Maehara (http://www.maths.ed.ac.uk/~aar/jordan/maehara.pdf) but it seems that the problem in the paper is very different from this problem:

First, the domains of their functions are $[-1,1]$ so they are able to define the function $F: [-1,1] \times [-1,1] \to [-1,1] \times [-1,1]$ in a way to get a contradiction.

Second, their functions $f,g$ are paths on a rectangle with end points a,b,c,d. Ours are just paths on $I \times I$ so they are able to conclude that the difference in the first coordinate of $F(-1,t_0)$ is nonnegative to get a contradiction.

I've toyed with a similar method as in the paper by defining an $F: I \times I \to I \times I$ similarly but with absolute values in the numerators of both coordinates but I can't get a contradiction from this since I only have that the image of $F$ must have as one of its coordinates the value 1.

Any help is much appreciated!

share|improve this question
add comment

1 Answer

All we need is a point $(x,y)$ where $f(x)=g(y)$. Since $f_{x}$'s range include the $y$-axis and $g_{y}$'s range include $x$-axis by intermediate value theorem, we can construct $F$ to be $$F(g(b)_{x},f(a)_{y})=(f(a)_{x},g(b)_{y})$$ where for $[c,d]\in I\times I$ we define $b=\inf g_{x}^{-1}(d)$ as $g_{x}^{-1}(d)$ could have multiple values, and similarly for $a=\inf f_{y}^{-1}(c)$. Then $F$ is well defined on the whole $I\times I$. Further $F$ is a continuous map from $I\times I$ to itself. By Brower fixed point theorem you have a fix point of the form $$(c,d)=(g(b)_{x},f(a)_{y})=(f(a)_{x},g(b)_{y})$$ which implies $$f(a)=g(b)$$ for some $a,b\in I$.

share|improve this answer
    
Thanks user32240! Works perfectly! –  Mmhmm Dec 28 '12 at 15:13
    
You are welcome! –  Bombyx mori Dec 28 '12 at 15:37
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.