From what I know, a continuous function can be intuitively thought of as a mapping that does not tear but only stretches or rotates, at least in the Euclidean space. Let $X$ be the one-point union of two circles in $\mathbb{R}^2$. Does the following map give a continuous function from $X$ to itself that does not have any fixed point? I'm trying to show that $X$ does not have the fixed-point property.

enter image description here Thinking of $X$ as having the subspace topology inherited from $\mathbb{R}^2$, is the operation of rotating (even if in the 'process' of rotating, the image fails to lie in $X$) still continuous?

Edit: I guess the second map isn't a continuous function from $X$ to itself. My intuition is then $X$ does have the fixed-point property because mapping the intersection point anywhere else other than to itself breaks continuity. How can I show this rigorously?

  • 2
    $\begingroup$ First reflect the second circle onto the first about the vertical, then rotate the image 90 degrees counterclockwise. It is a composition of two continuous (even linear) maps, hence continuous. It does not have fixed points. Your reasoning about the intersection point would only work if the map is also required to have a continuous inverse, i.e. to be a homeomorphism. $\endgroup$ – Conifold Mar 30 '19 at 6:33
  • $\begingroup$ Is the second map a continuous function from $X$ to itself? I'm worried about the fact that the composition of the two functions need to end up being a function $X\rightarrow X$. Or, can we just think of the first map as $X\rightarrow S^1$ and the second map as $S^1\rightarrow X$, in which case everything is ok? $\endgroup$ – user500144 Mar 30 '19 at 6:44
  • $\begingroup$ Also, in the case that we require it to be a homeomorphism, how can I rigorously prove that every homeomorphism from $X$ to itself has a fixed point? $\endgroup$ – user500144 Mar 30 '19 at 6:56
  • $\begingroup$ $X\rightarrow S^1$ interpretation works, or you can think of the map as defined on the whole of $\mathbb{R}^2$ but maps $X$ into itself. To prove that the intersection point must be mapped into itself by a homeomorphism you'll need a topological invariant that distinguishes it from regular points. That is not elementary. $\endgroup$ – Conifold Mar 30 '19 at 7:08
  • $\begingroup$ What kind of invariant are you talking about? I've only studied a few of the very basic concepts of algebraic topology. I'd appreciate a hint. $\endgroup$ – user500144 Mar 30 '19 at 7:16

Conifold already showed you a continuous $f: X \to X$ without a fixpoint: mirror the right hand : first apply $(x,y) \to (-x,y)$ (assuming we have two circles meeting in $(0,0)$ on both sides of the $y$-axis) and then rotate the left circle over some angle. This composition has no fixpoint, clearly.

But if $h: X \to X$ is a homeomorphism it must map the intersection point to itself, because that is the unique cut-point of $X$: If $h(0) = p \neq 0$ then $h\restriction_{X\setminus\{0\}}$ is a homeomorphism of $X\setminus \{0\}$ with $X\setminus \{p\}$ and the latter is connected while $X\setminus \{0\}$ is not.

So $X$ does not have the FFP, but it does have the FFPH: the fixed point property for homeomorphisms.


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