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

Question

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 $f(x)=x$.

Answer 1

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 $$f(z_0) = f(r(z_0)) = F(z_0) = z_0 .$$

Answer 2

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.