2
$\begingroup$

Brouwer Fixed point theorem states that

Theorem: any continuous map from a unit two dimensional disc $D^2$ into itself has a fixed point.

Q: Is it possible to use this theorem for two homeomorph discs $D^2$ and $D'^2$ (with different boundary)? and in this case what is the meaninig of fixed point?

$\endgroup$
  • $\begingroup$ What do you mean by "use this theorem for two homeomorphic discs"? The theorem states that $D^2$ has FPP. And since homeomorphisms preserve FPP then any space homeomorphic to $D^2$ has FPP. Is that what you are looking for? $\endgroup$ – freakish Jan 12 '18 at 9:08
1
$\begingroup$

You cannot really formulate such a theorem at all.

Fixed points are for self-maps (i.e. $f:X \to X$ for some $X$).

The FPP (fixed point property) i.e. "every continuous self-map of $X$ has a fixed point" is a topological property. So any space $Y$ homeomeomorphic to $X$ has it when $X$ has it and vice versa.

This is proved easily by "map transportation": if $f: Y \to Y$ is a continuous self map and $h: X \to Y$ is a homeomorphism, then $h^{-1} \circ f \circ h$ is a continuous self map of $X$, so has a fixed point $p \in X$ and then $h(p)$ is one for $f$ (all this is simple to check). So your $D'^2$ does have the FPP. Maybe that is what you really want?

$\endgroup$
2
$\begingroup$

It doesn't make sense to talk about fixed points for a map $f : D^2 \to D'^2$ with differing domain and codomain. So no, you can't really use the theorem, because I don't even see how you could state it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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