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 $f(x)=x$.
Best Answer
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 $$f(z_0) = f(r(z_0)) = F(z_0) = z_0 .$$