In the course on algebraic topology I came across a result called Brouwer's Fixed Point Theorem which states that any map (continuous function) $f : D^n \longrightarrow D^n$ has a fixed point for $n \geq 2.$ The proof uses No Retraction Theorem which states that there doesn't exist any retraction from $D^n$ onto $S^{n-1},$ for $n \geq 2.$ Now I have a question here.
Question $:$ Can we impose some further condition on $f$ so that it has a fixed point on $\partial D^n\ $?
Our instructor suggested that we may take a map which takes the upper hemisphere to the upper hemisphere then such maps will have a fixed point on the boundary. But I don't get his point. Could anyone suggest me some idea on it?
Thanks for reading.
Best Answer
What your instructor means is this:
The upper hemisphere of $S^{n-1} = \partial D^n$ is the space $$S^{n-1}_+ = \{(x_1,\ldots, x_n) \in S^{n-1} \mid x_n \ge 0 \} .$$ The map $h : S^{n-1}_+ \to D^{n-1}, h((x_1,\ldots, x_n)) = (x_1,\ldots, x_{n-1})$, is clearly a homeomorphism.
If $f(S^{n-1}_+) \subset S^{n-1}_+$, then consider the map $$g : D^{n-1} \stackrel{h^{-1}}{\longrightarrow} S^{n-1}_+ \stackrel{f}{\longrightarrow} S^{n-1}_+ \stackrel{h}{\longrightarrow} D^{n-1} .$$ Using the Brouwer Fixed Point Theorem, $g$ has a fixed point $y$. Thus $x = h^{-1}(y) \in S^{n-1}_+$ is a fixed point of $f$.