2

How to visualise the reduced suspension of $I=[0,1]?$

Here is an answer, Smash product of $S^1$ with the interval $I$ but I don't get it. So, it will be helpful for me if you can give a pictorial view.Thanks

SOUL
  • 1,042
  • Do you understand how to visualize the unreduced suspension? – Eric Wofsey Oct 14 '20 at 16:00
  • Yes, that's a double cone.Right? – SOUL Oct 14 '20 at 16:01
  • 3
    This depends on your choice of basepoint for $I$. For instance $\Sigma(I,0)\cong S^1\wedge(I,0)\cong D^2$, whilst $\Sigma(I,\frac{1}{2})\cong D^2\vee D^2$. ($D^2$ is the unit disc in $\mathbb{R}^2$ based at $(1,0)$). If $I$ has the basepoint $0$ (or $1$), then you should visualise its reduced suspension as a disc. – Tyrone Oct 14 '20 at 17:54
  • Thanks. I can see now. Now it remains to prove it. I Will try it. – SOUL Oct 14 '20 at 18:34
  • Did you manage to work things out Tom? – Tyrone Oct 15 '20 at 13:14
  • @Tyrone, will you please check my solution? – SOUL Nov 02 '20 at 01:21

1 Answers1

0

We claim that, $\Sigma (I,0)\cong D^2/A$ where, $A=\{(r,0):r\in [0,1]\}$ Consider the map,

$F:I\times I\to D^2/A$ given by,

$F(t,r)=[re^{2πit}]$ then $F$ is continuous.

Next, by universal mapping property and noting that , $I\times I $ is compact and $D^2/A$ is Hausdorff, we conclude that, $I×I/\left(\{0,1\}×I\bigcup I×\{0\}\right)\cong D^2/A$ and thus our claim is proved.

SOUL
  • 1,042
  • Your solution is correct, but you don't need to collapse out anything (i.e. $D^2$ is homeomorphic to $D^2/A$). Note that $\Sigma(I,0)=(S^1\times I)/(S^1\times 0\cup1\times I)$ is the same thing as the reduced cone on $S^1$, and this is just $D^2$. See my answer here for instance. – Tyrone Nov 02 '20 at 16:15