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From a homotopy-equivalence point of view you can take a (solid) straw and make it shorter and shorter until it is two dimensional (an annulus). Note that this

From a homotopy-equivalence point of view you can take a (solid) straw and make it shorter and shorter until it is two dimensional (an annulus). Note that this is a deformation retract (ie it is a “continuous deformation” but not the kind commonly used to try to explain a homeomorphism: a straw is like an annulus differently to how a donut is like a mug). The process can be done again, making the ring thinner and thinner until it is a circle. From here it seems reasonable to say that there is one hole in a circle so there is one hole in a straw.

Grustaf 8y ago
What do you mean by “until it is two-dimensional”? You can’t continuously change it’s dimension. It’s a two dimensional surface embedded in three dimensions, at least that’s how he modelled it in the beginning of the video.
- danharaj 8y ago
  Homotopy equivalence is even coarser than homeomorphism. A solid ball is homotopy equivalent to a point.
- dan-robertson 8y ago
  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)
wool_gather 8y ago
> it seems reasonable to say that there is one hole in a circle I'm not sure this is reasonable. There's only a hole in a circle if you are interacting with it from a higher dimension. If you're restricted to the plane, you can't access the area on the other side of the circle's border. Does a ping pong ball have a hole? It's empty inside just like the circle.
- dan-robertson 8y ago
  A ping pong ball does have a hole but slightly differently from a circle in that the hole in a ping pong ball is 3-dimensional whereas the hole in the circle is 2-dimensional. One needn’t embed the circle in a higher dimension to count holes. In the following sense a circle has one hole: the fundamental group of the circle is Z. That is, if you pick a point on the circle then the ways to go on a walk and get back to where you started are, modulo boring things like turning back on yourself or how fast you walk, to go round the hole in the middle of the circle clockwise an integer number of times (so 0 is staying still and -1 is once anti-clockwise). If you look at another shape, say a line the fundamental group is 1 as there is no nontrivial way to return to your starting point. On the surface of a torus (think inner tube) it turns out that you can go round the central hole (eg the whole bike wheel) some number of times and go round the small hole (eg the cross section of a bit of the tube) and it doesn’t matter which order you do them in. On the other hand if you take the shape made from two circles attached at a single point this also has two holes but unlike the torus it does matter which order you go round the holes. A path is described by some sequence of abcd where one cannot consecutively write ab,ba,cd, or dc: let a/b mean go round the first circle forwards/backwards, and c/d go round the other circle forwards/backwards then the sequence describes what to do. I’m not quite sure what you mean when you say restricted to a plane. In some sense a circle does not exist embedded in any space and it’s holefulness (being topological) shouldn’t depend on the space it’s embedded in. If you do embed it into a plane one can note that the circle partitions the plane into two regions (call this one hole) and two circles would partition the plane into three regions (if the circles intersect more than once then this is really equivalent to multiple circles). If you put a circle into R3 (or S3), one can count holes by looking at the shape of the space minus the circle again (in the way talked about in the video)

Grustaf 8y ago

What do you mean by “until it is two-dimensional”? You can’t continuously change it’s dimension. It’s a two dimensional surface embedded in three dimensions, at least that’s how he modelled it in the beginning of the video.

danharaj 8y ago
Homotopy equivalence is even coarser than homeomorphism. A solid ball is homotopy equivalent to a point.
dan-robertson 8y ago
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)

wool_gather 8y ago

> it seems reasonable to say that there is one hole in a circle I'm not sure this is reasonable. There's only a hole in a circle if you are interacting with it from a higher dimension. If you're restricted to the plane, you can't access the area on the other side of the circle's border. Does a ping pong ball have a hole? It's empty inside just like the circle.

dan-robertson 8y ago
A ping pong ball does have a hole but slightly differently from a circle in that the hole in a ping pong ball is 3-dimensional whereas the hole in the circle is 2-dimensional. One needn’t embed the circle in a higher dimension to count holes. In the following sense a circle has one hole: the fundamental group of the circle is Z. That is, if you pick a point on the circle then the ways to go on a walk and get back to where you started are, modulo boring things like turning back on yourself or how fast you walk, to go round the hole in the middle of the circle clockwise an integer number of times (so 0 is staying still and -1 is once anti-clockwise). If you look at another shape, say a line the fundamental group is 1 as there is no nontrivial way to return to your starting point. On the surface of a torus (think inner tube) it turns out that you can go round the central hole (eg the whole bike wheel) some number of times and go round the small hole (eg the cross section of a bit of the tube) and it doesn’t matter which order you do them in. On the other hand if you take the shape made from two circles attached at a single point this also has two holes but unlike the torus it does matter which order you go round the holes. A path is described by some sequence of abcd where one cannot consecutively write ab,ba,cd, or dc: let a/b mean go round the first circle forwards/backwards, and c/d go round the other circle forwards/backwards then the sequence describes what to do. I’m not quite sure what you mean when you say restricted to a plane. In some sense a circle does not exist embedded in any space and it’s holefulness (being topological) shouldn’t depend on the space it’s embedded in. If you do embed it into a plane one can note that the circle partitions the plane into two regions (call this one hole) and two circles would partition the plane into three regions (if the circles intersect more than once then this is really equivalent to multiple circles). If you put a circle into R3 (or S3), one can count holes by looking at the shape of the space minus the circle again (in the way talked about in the video)