# The Join of Two Copies of $S^1$

So I know the fact that the join of $S^1$ and $S^1$ is homeomorphic to the 3-sphere, but I'm having trouble "seeing" this. I'd prefer something that appeals to geometric intuition, but more formal abstract arguments are welcome as well! I want to invoke the unit quaternions here, but I can't think of a particularly simple way to do it.

If this question is too "soft" I sincerely apologize.

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@AgustíRoig Methinks in general, S^m joined to S^n is homeomorphic to S^(m+n+1)... – AsinglePANCAKE Sep 20 '12 at 18:39
The join in question is not $\vee$ but $*$: see this (One calls $\vee$ a wedge in this context, and dies a little every time one types \wedge to get a smash...) – Mariano Suárez-Alvarez Sep 20 '12 at 18:45
@AgustíRoig Methinks you're wedging and not joining... – AsinglePANCAKE Sep 20 '12 at 18:46
Ok, I mistook the "join" for the "wedge". Sorry. – a.r. Sep 20 '12 at 18:51
Thanks for your explanations, Asingle and Mariano. – a.r. Sep 20 '12 at 18:52

We discussed here a little while ago the fact that $S^3$ can be described as the result of gluing two solid tori along their boundaries.
To get a geometric reason for the homeomorphism $S^1*S^1\cong S^3$ you can look at how the two constructions are related: notice that two two tori have two central $S^1$s inside them...
(Alternatively: try to see what $S^1*[0,1]$ is, notice that you can describe $S^1$ as two copies of $[0,1]$ identified along the boundaries, and see how you can use this description on one of the two factors of $S^1*S^1$.)
(Another alternative: if you construct $S^1*S^1$ by doing identifications on $S^1\times[0,1]\times S^1$, cut the latter space in two parts $S^1\times[0,1/2]\times S^1$ and $S^1\times[1/2,1]\times S^1$, look at what the identifications do on each half, and then glue back the two parts)