Is T2 modulo permutation by S2 a familiar surface? If I have a 2d Torus $S_1 \times S_1$ but I identify $(x,y)\equiv(y,x)$, is that a familiar surface?
context: I am an engineering graduate student with a vanilla background in undergraduate topology pursuing a research question. I tried the usual gluing of edges but had difficulty embedding in 3 dimensions (which is my best chance at recognizing it). I felt the answer should be widely known on the internet but couldn't find it.
 A: Take the square $I \times I$ and form a torus, we have to take the quotient $(0,a) \sim (1,a)$, and $(x,0) \sim (x,1)$. Now, to make our identification $(x,y) \sim (y,x)$ we see that we can "fold the square" over diagonally (along the the line from $(0,0)$ to $(1,1)$.
But the order in which we take our quotients does not matter.
So, first make the "diagonal fold." Following the usual identication for the torus, we see that there are some orientation problems. To fix this, we can take another diagonal cut from $(1,0)$ to $(.5,.5)$. Remove both mini triangles and glue $(x,0)$ to $(y,1)$ together. Reiattaching the cut we made earlier, we see that the orientation has reversed,so this is just the mobius band.
This is difficult without a picture. Thankfully, 3blue1brown has an excellent video on just this around time 12:00 should give a very nice geometric intuition for why the answer is a mobius band (with beautiful animations)
A: Many thanks to David Hartley for pointing out that an earlier version of this answer was wrong.
Have a look at the nets or fundamental polygons of the torus $T^2$ and the real projective plane $RP^2$ that you can find on this Wikipedia page. What your quotient construction does to the net of $T^2$ comprising a square whose "word" is $aba^{-1}b^{-1}$ is to fold it in half along the diagonal $d$ and identify $a^{-1}$ with $b^{-1}$ (and hence $a$ with $b$). So you get a triangle with word $aad$. That gives you a surface with boundary. The boundary is given by the edge $d$ and is homeomorphic to a circle. If you glue in a disc whose net is a triangle with word $dcc^{-1}$, you get a closed surface with net given by a square with word $aacc^{-1}$. Collapsing the disc you glued in to a point (as you may because the the triangle with word $dcc^{-1}$ is homeomorphic to a disc), you get the digon with word $aa$ that is the fundamental polygon of $RP^2$. So your space is $RP^2$ with a disc-shaped puncture, which is homeomorphic to a Möbius strip.
A: It is homeomorphic to a disk with some points on the boundary glued together.
Perhaps the easiest way to see this is to start with the torus as the square $[0,1] \times [0,1]$ with opposite sides identified (glued), i.e., the top and bottom edges glued, and the left and right edges glued. The identification of $(x,y)$ with $(y,x)$ is just reflection across the diagonal line $y=x$. So the quotient space is (homeomorphic to) one of the triangles, say the lower triangle. The vertices of the triangle are identified because they were identified in the torus.
