Homological groups of the $T^2 = S^1 \times S^1$ and quotient out the circle $S^1 \times \lbrace x \rbrace$ for some point $x \in S^1$ Homological groups of the $T^2 = S^1 \times S^1$ and quotient out the circle $S^1 \times \lbrace x \rbrace$ for some point $x \in S^1$
I am trying to calculate the homology groups of this space, but first, I would like to know what this space looks like, that is, what topological space it looks like and what the difference if we made the quotient not for $S^1 \times \lbrace x \rbrace$  but for $ \lbrace x \rbrace\times S^1$ , thank you.
 A: Presumably you're quotienting out the co-ordinate circles here (that is, the ones which go "horizontally" and "vertically" around the torus.
I should point out that what I'm about to say is all going to be rather imprecise, but since you're asking about what stuff "looks like" I think that's okay.
If you quotient out the "horizontal" circle, you get a $T^2$ without the hole in the middle - you've closed up that hole because you've identified the $S^1$ which went around it to a point.
If you quotient out the "vertical" $S^1$, you get something which still looks like $T^2$, but is a sort of stretched out cresent moon. That is, if you imagine a cresent moon and then you take the two tips and stretch them around to join them together. You have this hole in the middle as you started with the $T^2$, but now you've taken a circle which sits around the outsi-... Ugh. I realise now this takes an effort to explain in words. Here's a picture of this second quotient:

I can't find a good picture of the first one online, but that's easier to visualise anyway.
The homologies of these two spaces are of course the same because they're homeomorphic - but this is not obvious when you have pictures of these spaces in your head! Thanks, Algebraic Topology :)
