Smash product of $S^1$ with the interval $I$

Question

I'm trying to work through various examples of smash products of spaces. In order to check to see if what I'm doing is correct, is the smash product $S^1 \wedge I$ of the circle with the interval $[-1,1]$ with basepoint $0$ homeomorphic to the wedge sum of two disks?

Answer 1

Yes. By definition, $S^1\wedge [-1,1]$ is the quotient of the cylinder $C=S^1\times[-1,1]$ by the subspace $A=\{s_0\}\times[-1,1]\cup S^1\times\{0\}$. Now $C\setminus A$ has two connected components $(S^1\setminus\{s_0\})\times (0,1]$ and $(S^1\setminus\{s_0\})\times [-1,0)$. Thinking of $S^1\setminus\{s_0\}$ as an open interval, each of these connected components is an open square together with one of its boundary sides, with the rest of its boundary partially glued together to form $A$. When we collapse $A$ to a point, this means we are collapsing the other three boundary sides of each square to a point, which gives a space homeomorphic to a disk (the one boundary side becoming the whole boundary circle). So, $C/A$ is homeomorphic to a wedge sum of two disks (the basepoints of the disks being on their boundaries), the two disks coming from the two connected components of $C\setminus A$.

(To make all this rigorous, you can observe that $S^1\wedge[-1,1]$ without its basepoint is homeomorphic to $C\setminus A$ and thus $S^1\wedge[-1,1]$, being compact Hausdorff, is a 1-point compactification of $C\setminus A$. But $C\setminus A$ is homeomorphic to a disjoint union of two closed disks with a point on their boundaries removed, and $D^2\vee D^2$ is the 1-point compactification of that space. So, by the uniqueness of 1-point compactifications, $D^2\vee D^2\cong S^1\wedge [-1,1]$.)