# Rotman Introduction to Algebraic Topology Question 0.5

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!

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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 $a,b\in I$.

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Thanks user32240! Works perfectly! – Mmhmm Dec 28 '12 at 15:13
You are welcome! – Bombyx mori Dec 28 '12 at 15:37