# Fixed point of a cone over an $n$-point space

I'm trying to answer the following question:

Let $$CX$$ denote the cone over an $$n$$-point space $$X=\{1,\dots,n\}$$. Show that every continuous map $$f:CX\to CX$$ has a fixed point.

Any help is appreciated. The only fixed point theorem I've come across for this course is the Brouwer fixed point theorem for the disk, i.e., if $$f:B^2\to B^2$$ is continuous, then there exists a point $$x\in B^2$$ such that $$f(x)=x$$.

• Are you able to describe a cone geometrically? Do you know that union of two spaces $X, Y$ with fixed point property and $|X\cap Y| = 1$ also as a fixed point property? Commented Dec 29, 2023 at 13:39
• Are we in $\mathbb R ^2$, in $\mathbb R^3$, or other space? $X$ is not really the set of integers from $1$ to $n$? What do you mean by "the cone over" $X$ ? Commented Dec 29, 2023 at 13:42
• @Jean-ArmandMoroni Cone of a topological space Commented Dec 29, 2023 at 13:45
• @Jakobian I would describe it as lines from a common point to the $n$ points. It's possible that that property was mentioned but I cannot recall it honestly. How would you show such a thing?
– Seth
Commented Dec 29, 2023 at 14:00
• Sorry I'd write it sooner but my keyboard is malfunctioning Commented Dec 29, 2023 at 14:22

We can embed $$X$$ as a subset of $$S^1$$ by identifying $$k$$ with $$\zeta_k = e^{\frac{2k\pi i}{n}}$$. Let $$Z = \{ \zeta_1, \ldots, \zeta_n\}$$ and $$Z^* = \{t \zeta \mid t \in [0,1], \zeta \in Z\}$$. Clearly $$Z^*$$ is a homeomorphic copy of $$CX$$. It therefore suffices to show that each $$f : Z^* \to Z^*$$ has afixed point.

It is easy to see that there is a retraction $$r : B^2 \to Z^*$$ (drawing a picture helps, to derive an explicit formula is somewhat tedious).

Given $$f : Z^* \to Z^*$$, define $$F : B^2 \to B^2, F(z) = f(r(z)) .$$ This map has a fixed point $$z_0 \in B^2$$. We have $$z_0 = F(z_0) = f(r(z_0)) \in f(Z^*) \subset Z^*$$ and therefore $$r(z_0) = z_0$$. Hence $$f(z_0) = f(r(z_0)) = F(z_0) = z_0 .$$

• I think there is a typo: Is $f'=F$?
– Gerd
Commented Dec 30, 2023 at 8:38
• @Gerd Thank you! I corrected the typo. Commented Dec 30, 2023 at 9:23

Theorem. If $$X, Y$$ have fixed point property (in some ambient space $$Z$$), be closed, and $$X\cap Y = \{x_0\}$$, then $$X\cup Y$$ has fixed point property.

Proof: Let $$f:X\cup Y\to X\cup Y$$ be continuous, and $$g_Y(x) = \begin{cases} x_0, & x\in X \\ x,& x\in Y\end{cases},\ g_X(x) = \begin{cases} x,& x\in X\\ x_0, & x\in Y\end{cases}$$

Then $$f_Y = g_Y\circ f\restriction_Y$$ and $$f_X = g_X\circ f\restriction_X$$ have fixed points, so that there are $$x\in X, y\in Y$$ with $$f_X(x) = x$$ and $$f_Y(y) = y$$. If $$f(x)\in X$$ or $$f(y)\in Y$$ then we are done. Suppose on the other hand that $$f(x)\in Y$$ and $$f(y)\in X$$, then $$f_X(x) = x = x_0$$ and $$f_Y(y) = y = x_0$$ so that $$f(x) = f(y) = f(x_0)\in X\cap Y = \{x_0\}$$ so finally $$f(x_0) = x_0$$. $$\square$$

Corollary. Let $$X_1, ..., X_n$$ have fixed point property, be closed (in some space $$Z$$), and $$X_i\cap X_j = \{x_0\}$$ for $$i \neq j$$. Then $$X_1\cup ... \cup X_n$$ has fixed point property.

Proof: Induction. $$\square$$

In your case, $$CX$$ is a union of spaces $$I_i$$ homeomorphic to compact intervals such that the vertex $$x_0$$ is the common intersection of each two disjoint intervals. Hence $$CX$$ has fixed point property.