# Is the two-torus $T^2$ equivalent to the 2-sphere $S^2$ minus two open disks with the boundaries identified (in opposite orientation)

I know that the pinched torus is equivalent to $$S^2$$ with two points identified. I'm wondering about the general torus now. It appears that taking two open disks from $$S^2$$ and identifying their boundaries (the circles) would then be general two-torus. Is this correct?

What about $$S^2 \times S^1$$ being the same as taking two open 3-balls from $$S^3$$ and identifying their $$S^2$$ boundaries???

I know usually the covering map is two-to-one from the 2-torus to the two-sphere (with 4 ramification points) Do we still have a double cover in the case I describe or are they actually the same topologically?

• The answers to your questions are: Yes; Yes; and Yes. Commented Jun 8, 2021 at 23:53
• @LeeMosher yes to the double cover also you meant??? Commented Jun 8, 2021 at 23:54
• Actually, since I see two questions in your last question, I suppose that the answer is yes and yes: yes (we still have a double cover) and yes (they are actually the same topologically). Commented Jun 8, 2021 at 23:56
• @LeeMosher Just one more thing.. do the boundaries that are identified then consist of branch points then where the double-cover fails? THank you!! Commented Jun 9, 2021 at 0:06
• You should think more carefully about what it is that you really want to ask. I was unsure whether you had done that already, since the answers to your questions were just repeating the same three letter word. Your most recent question in the comments does not really make sense, though. Commented Jun 9, 2021 at 2:07

To see that, you can consider the sphere minus two circles as a $$D$$ from which you removed a small disc inside, by identifying the boundaries it gives the $$2$$-torus, do the same for any dimension