Prove these 3 spaces are homotopy equivalent The image is below. 
(a) $S^2$ with a diameter.
(b) $T^2$ with a disk in the middle hole.
(c) $S^2$ tangent with $S^1$ .
I think they may the deformation retract of the same space. But I can't found them exactly.

 A: Consider the folowing theorm :

If (X,A) is a CWpair consisting of a CWcomplex X and a contractible subcomplex A,
  then the quotient map X→X/A is a homotopy equivalence

Thing about what should be your CW pair for achiving these ?
A: Eli's answer is really good, as you'll understand when you're a little further along. 
But for now, consider deforming the sphere-with-diameter by shrinking the diameter to a single point, so that the north and south poles become one point in the new space $X$. 
Then consider shrinking the disk in the middle of the torus in part (b) so that it becomes a single point. 
The resulting two spaces look a LOT alike. Indeed, you could probably put polar coordinates on  them in such a way that the two spaces are naturally homeomorphic by a map like $S(\theta, \phi) \mapsto T(\theta, \phi)$, where $S$ and $T$ are the parameterizations of the squeezed sphere and the squeezed torus. 
As for showing that "c" is the same as "a" and "b", I'll leave that to you (in part because I don't instantly see how to do it!). [Or you can read on, now that I've thought for a moment or two more.]
Take drawing "a" and make the line from the north to south pole go OUTSIDE the sphere instead of inside. Clearly the resulting space is homeomorphic to the one shown in drawing "a". Now move the two attachment points for this arc  --- initially the N and S pole --- towards the equator, until they become a single point. That looks a lot like figure "c". 
