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 :)



Quotient of solid torus by swapping coordinates on boundary
Your quotient manifold is homeomorphic to $S^3$. I know this because it is a closed 3-manifold and its fundamental group is trivial, so I'm quoting the Poincare conjecture/Perelmann's theorem. The fundamental group can be seen to be trivial because there is a CW-decomposition of the (2-manifold) torus with one 0-cell, three 1-cells and two 2-cells that is respected by your involution, where the 1-cells are the longitude, meridian and diagonal of the torus, and the two 2-cells are triangles. The involution swaps the longitude and meridian while fixing the diagonal pointwise, and swaps the two 2-cells. The quotient of the torus by the involution has one vertex, two 1-cells and one 2-cell. The fundamental group of this space is infinite cyclic, generated by the loop around the 1-cell that started life as both the longitude and meridian of the torus.  

This quotient space can be viewed as a copy of the Mobius strip, as in Taras Banakh's comment.


But now think about how the insides fits in: a CW-structure on the solid torus can be made by adding in one more 2-cell, attached around the meridian of the torus, and one 3-cell. The extra 2-cell wraps around the generator for the fundamental group of the piece discussed in the previous paragraph. So the fundamental group of the overall space is trivial.


