TITLE: Quotient of solid torus by swapping coordinates on boundary
QUESTION [14 upvotes]: Let $T$ be the solid 2-torus and let $\sim$ be the equivalence relation on $T$ generated by the relation $\{(\alpha,\beta) \sim (\beta,\alpha) \mid \alpha, \beta \in S^1\}$ on the boundary $\partial T=S^1\times S^1$. What kind of space is $T/{\sim}$?
I believe that it is $S^3$, but I can't prove it.
I posted this question already here:
https://math.stackexchange.com/questions/4339019/identifying-boundary-of-solid-torus-with-itself-by-swapping-coordinates
Unfortunately, I didn't get a satisfying answer. Not sure if it's because it is harder than I thought or actually nobody cares. Maybe both :)

REPLY [9 votes]: As yet another proof: let $M$ be the quotient of $\partial T$. Let $U$ be a small closed regular neighbourhood of $M$. Thus $U$ is a solid torus. Let $V$ be the closure of the complement of $U$. So $V$ is $T$, minus an open regular neighbourhood of its boundary. So $V$ is another solid torus.  Let $S$ be the intersection of $U$ and $V$. So $S$ is a (genus one) Heegaard surface for the three-manifold.  It is an “exercise” to show that the boundaries of the meridian disk in $U$ and $V$, respectively, meet once. Thus $S$ is the standard Heegaard splitting of the three-sphere.