Let the straw be the following set: X = { (x,y,z) in R3 : |z| < 1, 1 < x^2 + y^2 < 2 } Change the numbers until this is straw-like enough for you. Define t

Let the straw be the following set: X = { (x,y,z) in R3 : |z| < 1, 1 < x^2 + y^2 < 2 } Change the numbers until this is straw-like enough for you. Define the deformation retract f : I * X -> X, (where I is the unit interval [0,1]): f_t(x,y,z) = (x,y,z*(1-t)) This is continuous and well-defined and retracts the straw onto the space f_1(X) = { (x,y,0) : 1<x^2+y^2<2 } Which is two dimensional. Now do something similar to make the annulus a circle. Now what it actually means for two spaces, X, Y to be homotopy equivalent is that there are continuous f : X -> Y and g : Y -> X such that fg is homotopic to the identity on Y and gf is homotopic to the identity on X. Now for maps a,b : A -> B to be homotopic, there must be some (continuous) map H : I * A -> B such that H_0 = and H_1 = b. So why does a deformation retract lead to a homotopy equivalence? Let R : I * X -> X be a deformation retract and r = R_1, Y = r(X) so r : X -> Y. Then we get a homotopy equivalence between X and Y by the functions r : X -> Y and i: Y -> X the inclusion function (ie the identity function). ri = id (definition of a retract) so done here ir is the same as R_1 : X -> X so just use the homotopy H_t = R_{1-t} to get the equivalence to id (R_0=id by definition of deformation retract)