1

Reduced cone of $S^n$. I could prove that the cone of $S^n$ is isomorphic to $D^{n+1}$. I can sense that that the reduced cone of $\tilde{C}(S^n)$ is $D^{n+1}$. But I cannot give an exact homeomorphism from $\tilde{C}(S^n)$ to $D^{n+1}$.

epsilon_delta
  • 688
  • Do you want an explicit homeomorphism? You get the reduced cone from the usual cone by contracting ${x_0}\times [0,1]$ where $x_0$ is the base point. Since you have got a homeomorphism from $C(S^n)$ to $D^{n+1}$, to get the reduced cone you have to quotient out a simple arc joining an interior point to a the boundary of $D^{n+1}$. It's not hard to see that this remains homeomorphic to $D^{n+1}$. – Baidehi Jul 01 '20 at 06:55
  • @User2592 But I cannot construct a homeomorphism. But I can see that the it is homeomorphic but it would be great if I could to construct a homeomorphism. – epsilon_delta Jul 01 '20 at 07:09

1 Answers1

2

Note, the basepoint of $D^{n+1}$ is not the origin. Rather it is the unit vector $e_1=(1,0,\dots,0)$, which is also the basepoint of its subspace $S^n$. Using the convexity of $D^{n+1}$ we see that the map $D^{n+1}\times I\rightarrow D^{n+1}$ given by $$F_t(x)=(1-t)\cdot x+t\cdot e_1$$ is a well-defined pointed contraction. Now restrict $F$ to $S^n\times I$ so that it becomes a pointed null-homotopy of the inclusion $S^n\hookrightarrow D^{n+1}$. Then for all $z\in S^n$ and $t\in I$ we have $$F_1(z)=e_1,\qquad F_t(e_1)=e_1.$$ In particular $F$ factors over the quotient $$CS^n=(S^n\times I)/(S^n\times\{1\}\cup\{e_1\}\times I)$$ to give a map $$\widetilde F:CS^n\rightarrow D^{n+1}.$$ We check that $\widetilde F$ is one-to-one and onto. Since $CS^n$ is compact and $D^{n+1}$ is Hausdorff, $\widetilde F$ is a homeomorphism.

Tyrone
  • 16,116