# 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.

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.