How do you show an $m$-simplex is homeomorphic to a cone? 
Show that, for $0 \le i \le m$,  $[p_0, ..., p_m]$ is homeomorphic to the cone $C[p_0, ..., \bar p_i, ..., p_m]$ with vertex $p_i$, where $\bar p_i$ means delete $p_i$.

It absolutely makes sense intuitively: $C[p_0, ..., \bar p_i, ..., p_m]$ is just $[p_0, ..., p_m]$ with a point taken out and then creating a new point via. the vertex which can be deformed to be homeomorphic to $[p_0, ..., p_m]$.
But I'm a little confused as to how to show this analytically through a homeomorphism.
Anyone have any ideas?
 A: You haven't stated your definition of the simplex $[p_0,\ldots,p_m]$, but I'll use the standard $$[p_0,\ldots,p_m] = \left\{ \sum_k a_k p_k : \sum_k a_k = 1 \text{ and } a_k \geq 0\right\}.$$
Recall that the cone on a space $X$ is $CX = X \times I/X \times 1$. Define a map $$[p_0,\ldots,p_m] \to C[p_0,\ldots,\overline{p_i},\ldots,p_m], \quad \sum_k a_kp_k\mapsto \begin{cases}\displaystyle\left(\sum_{k \neq i} \frac{a_k}{1-a_i}p_k,a_i\right) & \text{ if } a_i \neq 1 \\ (p_i,1) & \text{ if } a_i = 1\end{cases}.$$
Intuitively, this map "stretches out" each layer $\{ \sum_k a_kp_k  : a_i = c\}$ onto the corresponding cone layer $[p_0,\ldots,\overline{p_i},\ldots,p_m] \times c$ for all $c \in [0,1)$ and maps the tip $p_i$ of the simplex to the tip of the cone. Continuity of this map is rather clear except perhaps at the special tip $p_i$ of the simplex. But note that images of sets of the form $$[p_0,\ldots,\overline{p_i},\ldots,p_m] \times (1-\epsilon,1] \subset [p_0,\ldots,\overline{p_i},\ldots,p_m] \times I,$$ $\epsilon > 0$, form a neighborhood basis around the tip of the cone. And the preimage by our map of any such neighborhood is just $\{ \sum_k a_kp_k  : a_i > 1 - \epsilon\} \subset [p_0,\ldots,p_m]$, which is a neighborhood of the tip $p_i$ of the simplex.
Finally, it is easy to see that our map is a bijection (each layer $\{ \sum_k a_kp_k  : a_i = c\}$ of our simplex corresponds bijectively to the layer $[p_0,\ldots,\overline{p_i},\ldots,p_m] \times c$ of the cone). As the simplex is compact and the cone is Hausdorff, our map is a homeomorphism.
