Is $\mathbb S^2$ just a notation or $\mathbb S^2\cong \mathbb S\times \mathbb S$ (in a possible way)? I was wondering : $\mathbb S\times \mathbb S$ is a torus and $\mathbb S^2$ is a sphere. Is $\mathbb S^2$ just a notation or it's justified by the fact that $$\mathbb S^2\cong \mathbb S\times \mathbb S \ \ ?$$
(I wouldn't see how a torus would be a sphere since the fundamental group of the sphere is trivial whereas, the fundamental group of the torus is $\mathbb Z^2$, but maybe it could have a link between sphere and torus that would justify $\mathbb S^2$ and $\mathbb S\times \mathbb S$.
 A: Using the smash product, we do have $S^n=S^1\wedge S^1\wedge\cdots\wedge S^1$ (with $n$ copies of $S^1$). However, I think that that is only coincidental. Note that this way you do get $S^2$ as a quotient space of $S^1\times S^1$.
The main reason, to most people, that $S^n$ is denoted by $S^n$ is probably just that it is the $n$-dimensional sphere, and that's that. The $n$ denotes the dimension of the space, and doesn't explicitly refer to some kind of $n$-times  repeated operation.
A: As Ted Shifrin pointed out in comments, $S^n$ is not a product. But there is another way to interpret the notation $S^n$ that makes it seem a little systematic. For any topological space $X$, let $S(X)$ denote the suspension of $X$: This is the topological space obtained from the "cylinder" $X\times [0,1]$ by identifying $X\times \{0\}$ to a point and $X\times \{1\}$ to another point. Then if $S^0$ denotes the zero-sphere (i.e., the two-point discrete space), then it's not hard to show that $S(S^0)$ is homeomorphic to $S^1$, and in general $S^n(S^0) \approx S(S^{n-1})$ is homeomorphic to $S^n$.  
