# Prove that $C(D^n) \cong D^{n+1}.$

Prove that $$C(D^n) \cong D^{n+1},$$ where $$C(D^n)$$ represents the cone of $$D^n.$$

What I know is that $$C(D^n) = \frac {D^n \times I} {D^n \times \{1\}}.$$ I want to find an onto map $$f : D^n \times I \longrightarrow D^{n+1}$$ whose set of fibres is precisely $$\frac {D^n \times I} {D^n \times \{1\}}.$$ But I couldn't able to find such a function. Can anybody help me finding such a function?

Thanks!

EDIT $$:$$ I thought that the map

$$(x,t) \longmapsto \left ((1-t)x, (1-t) \sqrt {1 - \|x\|^2} \right )$$

would work but the only problem is that the map is not onto.

## 2 Answers

This is intuitively clear, but to write down an explicit homeomorphism is not trivial. We shall do it in two steps.

1. Let $$D^{n+1}_+ = \{(x_1,\ldots, x_{n+1}) \in D^{n+1} \mid x_{n+1} \ge 0\}$$ be the closed upper half of $$D^{n+1}$$. We define a homeomorphism $$h : C(D^n) \to D^{n+1}_+$$ by identifying the disk $$D^n \times \{t\}$$ with the disk $$D^n_t = \{(x_1,\ldots, x_{n+1}) \in D^{n+1} \mid x_{n+1} = t\}$$ and the tip of $$C(D^n)$$ with the north pole $$N = (0,\ldots,0,1)$$ of $$D^{n+1}$$.
More precisely we define $$\phi : D^n \times I \to D^{n+1}_+, \phi(x,t) = (\sqrt{1 - t^2} x,t).$$ This is well-defined because $$\lVert (\sqrt{1 - t^2} x,t) \rVert^2 = (1-t^2)\lVert x \rVert^2 + t^2 \le 1- t^2 + t^2 = 1$$. The fibers are $$\phi^{-1}(N) = D^n \times \{1\}$$ and $$\phi^{-1}(y,t) = \{(\frac{y}{\sqrt{1 - t^2}},t)\}$$ for $$(y,t) \ne N$$. Note that $$\lVert y \rVert^2 + t^2 = \lVert (y,t) \rVert^2 \le 1$$, thus $$\lVert \frac{y}{\sqrt{1 - t^2}}\rVert \le 1$$. i.e. $$\frac{y}{\sqrt{1 - t^2}} \in D^n$$.
This shows that $$\phi$$ is onto and induces a homeomorphism $$h : C(D^n) \to D^{n+1}_+$$.

2. Next we identify $$D^{n+1}_+$$ with $$D^{n+1}$$ by stretching the line segments $$L_x$$ connecting $$(x,0) \in D^n \times \{0\}$$ with $$(x,\sqrt{1- \lVert x \rVert^2}) \in S^n$$ linearly to the line segments $$L'_x$$ connecting $$(x,-\sqrt{1- \lVert x \rVert^2})$$ with $$(x,\sqrt{1- \lVert x \rVert^2})$$.
More precisely we define $$g : D^{n+1}_+ \to D^{n+1}, g(x, t) = (x,2t -\sqrt{1- \lVert x \rVert^2}) .$$ This is well-defined because for $$(x,t) \in D^{n+1}_+$$ we have $$0 \le t$$ and $$\lVert x \rVert^2 + t^2 \le 1$$, hence $$0 \le t \le \sqrt{1- \lVert x \rVert^2}$$ which implies that $$-\sqrt{1- \lVert x \rVert^2} \le 2t - \sqrt{1- \lVert x \rVert^2} \le \sqrt{1- \lVert x \rVert^2}$$ and therefore $$\lVert g(x,t) \rVert^2 = \lVert x \rVert^2 + (2t -\sqrt{1- \lVert x \rVert^2})^2 \le \lVert x \rVert^2 + 1 - \lVert x \rVert^2 = 1 .$$ $$g$$ is injective: If $$g(x,t) = g(x',t')$$, we get $$x = x'$$ and $$2t -\sqrt{1- \lVert x \rVert^2} = 2t' -\sqrt{1- \lVert x' \rVert^2} = 2t' -\sqrt{1- \lVert x \rVert^2}$$, i.e. $$t = t'$$.
$$g$$ is surjective: Given $$(x,s) \in D^{n+1}$$, we have $$\lVert x \rVert^2 + s^2 \le 1$$, thus $$-\sqrt{1- \lVert x \rVert^2} \le s \le \sqrt{1- \lVert x \rVert^2}$$. Defining $$t = \frac{s + \sqrt{1- \lVert x \rVert^2}}{2}$$ we get $$0 \le t \le \sqrt{1- \lVert x \rVert^2}$$, thus $$(x,t) \in D^{n+1}_+$$ and $$g(x,t)= (x,s)$$.
Therefore $$g$$ is a homeomorphism.

If you want, you can explicitly write down $$g \circ \phi : D^n \times I \to D^{n+1}$$ which induces the desired homeomorphism $$C(D^n) \to D^{n+1}$$.

• Great answer. Thank you so much. By the time you posted this answer I was asleep. That's why there is a delay of accepting this nicely written answer. Sorry for that. Thanks again for your kind help. – Fanatics Jun 11 at 4:27

If you want a direct homeomorphism instead, think of $$C(D^n)$$ as inscribed in $$D^{n + 1}$$ and let $$f: C(D^n) \to D^{n + 1}$$ be a map that stretches a radial segment from the centre of $$D^{n + 1}$$ to $$\partial C(D^n)$$ onto the radial segment to $$S^n$$ in the same direction.