Let $X$ be a topological space. The cone $CX$ on $X$ is the cylinder $X \times I$ with the top $X \times 1$ identified to a point. Clearly for every $X$, $CX$ is contractible. Looking at the homeomorphism type is slightly less trivial: the cone on a point is an interval, the cone on an interval is a disk, the cone on the $n$-sphere is the $n+1$ ball etc.
My question: is there a nice way to see the cone on the cylinder $D^2 \times I$? It should be a $4$-dimensional manifold whose boundary is the cylinder...I have trouble seeing this.
$C(D^2 \times I)$ is homeomorphic to $C(D^3)$ which is homeomorphic to $D^4$ as k.stm suspected. More generally, you get a homeomorphism between $C(D^{n-1} \times I)$ and $D^{n+1}$ using the fact that if $X \cong Y$ then $CX \cong CY$ together with four homeomorphisms that I have tried to describe visually as follows:
1) $D^{n-1} \times I \simeq D^n$: the product $D^{n-1} \times I$ of the $(n-1)$-dimensional unit ball and the unit interval is homeomorphic to the $n$-dimensional unit ball $D^n$ by a radial dilation. (When $n = 2$, this says that a disk inscribed inside a square is homeomorphic to the square.)
2) $D^n \cong S^n_+$: the $n$-dimensional unit ball $D^n$ is homeomorphic to the upper hemisphere $S^n_+$ of the unit sphere $S^n \subseteq \mathbb{R}^{n+1}$ by a vertical projection.
3) $CS^n_+ \cong D^{n+1}_+$: the cone $CS^n_+$ is homeomorphic to the upper hemisphere $D^{n+1}_+$ of the $(n+1)$-dimensional unit ball $D^{n+1}$ by restriction of the obvious homeomorphism between $CS^n$ and $D^{n+1}$.
4) $D^{n+1}_+ \cong D^{n+1}$: $D^{n+1}_+$ is homeomorphic to $D^{n+1}$ by a vertical dilation.
Note that $C(D^{n-1}\times I)$ is a manifold with boundary but its boundary is not $D^{n-1}\times I$. Moreover, the cone on a manifold is not a manifold in general.