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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.

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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 $C(D^n) \to D^{n+1}$.

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  • $\begingroup$ 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. $\endgroup$ – Fanatics Jun 11 at 4:27
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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.

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