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The reduced cone is given by: $$CX = （X \times I） /(X \times 1 ∪ x_0 × I),$$ where $I$ is the unit interval $[0,1], x_0$ is the base point. Let $x_0$ be any point in $S^1$. How can I show that $CS^1$ is homeomorphic to $D^2$? $S^1$ is the unit circle and $D^2$ is the unit disk.

I proved that the unreduced cone is homeomorphic to $D^2$, so I started this problem with showing a unit disk with equivalent relation on a radius is homeomorphic with $D^2$. But I cannot find a proper bijection between them. Anyone help?

Hamou
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Hui
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1 Answers1

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Hint:

Use the map $f:X\times I\to D^2$ defined by $f(x,t)=(1-t)x+tx_0$ (continuous and surjective).

You can show that $f(X\times I\cup x_0\times I)=\{x_0\}$.

also $f(x,t)=f(y,s)$ iff $(x,t),(y,s)\in X\times I\cup x_0\times I$.

The set $X\times I\cup x_0\times I$ is compact.

Hamou
  • 6,745