Homeomorphism $\phi : T^2/A \to X/B$. What are $ T^2/A$ and $X/B$? The question I am working on asks me to construct a homeomorphism $\phi : T^2/A \to X/B$ where $T^2$, $A$, $X$ and $B$ are given as follows:


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*$T^2=S^1 \times S^1$ and $A \subset T^2$ is given by $A=S^1 \times\{1\}$.

*$X=S^1 \times [-1, 1]$ and $B = S^1 \times\{-1, 1\}$.


But first of all, I do not know what the spaces $ T^2/A $ and $X/B$ 'look like', or what they are mathematically. 
I can see that $A \subset T^2$ and $X \subset B$.
Can someone explain to me what they mean please? Then I can try and find an explicit homeomorphism between them.
 A: $T^2 = S^1\times S^1$ is a torus and $A = S^1\times\{1\}$ is one of the highlighted circles on torus below.
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$X = S^1\times[-1, 1]$ is a cylinder and $B = S^1\times\{-1, 1\}$ is the union of the boundary circles at each end.
The quotient $T^2/A$ is a torus with the circle $A$ collapsed to a point; depending on whether you collapse the purple circle or the red circle, you get two seemingly different spaces.
To visualise $X/B$, first collapse $S^1\times\{-1\}$ to a point and $S^1\times\{1\}$ to a point to obtain a sphere; these two points are the north and south pole. Then identify the south and north poles. Depending on whether you identify these two points on the inside or the outside of the sphere, you get two seemingly different spaces.


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*If you collapse the purple circle in the torus, you get the same space as you would get by identifying the north and south poles of a sphere by 'pushing them' inside to the centre.

*If you collapse the red circle in the torus, you get the same space as you would be identifying the north and south poles of a sphere by 'pulling them' towards each other on the outside of the sphere. 
Although these two spaces appear to be very different, they are in fact homeomorphic.
