Show that the cone $c(S^n)$ is homeomorphic to $\overline{B}^{n+1}$.
The cone is defined as $c(S^n) = (S^n \times I)/(S^n \times\{1\})$. The book I'm reading says that this quotient comes with a projection $p:S^n \times I \to S^n \times\{1\}$ such that $$p(x)= \begin{cases} S^n \times\{1\}, &x \in S^n \times\{1\} \\ \{x\}, &x \in S^n \times I \setminus (S^n \times\{1\}) \end{cases}$$
Pictorially I see that the cone formed for example from $S^1$ is just the regular solid we have on $\Bbb R^3$, but I don't understand how does this mapping $p$ work? It doesn't seem to map the cylinder $S^1 \times I$ to the cone? All the points on top of the cylinder lying on $S^1 \times\{1\}$ are just mapped to the set itself and all other points to the singletons $\{x\}$?
How should I interpret this $p$?
First of all, the target space is a quotient by an equivalence relation on $S^n\times I$ (namely that given by $(x, s)\sim(y, t)$ if $(x, s)=(y, t)$ or $s=1=t$), so the points in the cone are actually equivalence classes of the points in the cylinder instead of the points in the cylinder. This is why it looks like it is sending a point to itself, but it actually does not. It sends a point $(x, t)$ with $0\le t<1$ to the equivalence class $[(x, t)]=\{(x, t)\}\neq(x, t)$. The quotient topology is also defined in a way that corresponds to the equivalence classes of $(x, t)$ getting closer and closer to each other as $t$ gets closer to 1. Also, note that even though $S^n\times\{1\}$ is a set by itself, it is actually also the case that $[(x, 1)]=S^n\times\{1\}$ for any $x\in S^n$. So, what this means is that the set that is being collapsed to a point actually does become an element in the quotient space (or its underlying set) instead of being a subset like it originally was. What $p$ then does is that it takes any $(x, t)\in S^n\times I$ and sends it to $[(x, t)]$. If $t=1$, then this does not depend on $x$ but is always sent to the same element, $S^n\times\{1\}\in c(S^n)$.
Hope this helps! $(\ ^\circ\smile^\circ)$