0

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?

Eric Wofsey
  • 330,363
jeff
  • 13
  • 1
    What do you mean by equivalent? If you mean homeomorphic, then I think you need to tell us what base point you are using in $I$. If you mean homotopy equivalent, then the answer is yes (because both $S^1 \wedge I$ and the join of two disks are contractible). – Rob Arthan Jul 13 '19 at 10:33
  • Note that $S^1\wedge I$ is also known as $\Sigma I$, the reduced suspension of $I$. As Rob said, the homotopy type will not depend on the basepoint of $I$, but the homeomorphism type might – Maxime Ramzi Jul 13 '19 at 11:35
  • Thanks for the responses. Suppose I take the compact interval $[-1,1]$ with basepoint ${0}$ (and the circle's basepoint to be $s_0$). Will the smash between these two pointed spaces be homeomorphic to two disks joined at the appropriate basepoint? – jeff Jul 13 '19 at 11:39
  • "Join" has a different meaning; it sounds like you instead mean wedge sum. – Eric Wofsey Jul 13 '19 at 14:58
  • Oh yes. Sorry about that. Wedge sum ("bouquet" of two disks) is what I should have said. Thank you for editing that! – jeff Jul 13 '19 at 15:43

1 Answers1

0

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, $D^2\vee D^2\cong S^1\wedge [-1,1]$.)

Eric Wofsey
  • 330,363