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I Homotopy Definitions

  1. Jul 21, 2017 #1
    Hello! I want to make sure I understand these definitions (mainly the difference between them), so please let me know if what I am saying is correct. So a ##\textbf{homeomorphism}## between 2 topological spaces, means that the 2 can be continuously deformed from one to another, while keeping a bijection between them (so a disk and a smaller disk inside it are homeomorphic, but a disk and a circle inside it are not). Then ##\textbf{homotopy}## means that 2 loops can be continuously deformed from one to another (not necessary bijectively - a circle and a point inside it are homotopic, for a simply connected space). Also, does homotopy applies just to loops, like 1 dimensional objects? And lastly the notion of ##\textbf{deformation retract}## means that 2 topologically spaces can be continuously transformed from one to another (not necessary in a bijective way - so a point is the deformation retract of a sphere). So a deformation retract is like midway between homotopy and homeomorphism (i.e. you can work not only with loops, but you don't need bijectivity)? Thank you!
     
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  3. Jul 21, 2017 #2

    lavinia

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    - A homeomorphism is a continuous bijection between two topological spaces whose inverse is also continuous. You can think of this as a continuous deformation but the language is unusual.

    - A homotopy is something that exists between two maps. if ##f## and ##g## map the topological space ##X## into the topological space ##Y## then a homotopy between them is a continuous map from ##X## cartesian product the unit interval ##F:X×I→Y## where ##F(x,0) = f(x)## and ##F(x,1) = g(x)##. ##f## and ##g## are said to be homotopic. Intuitively one imagines the image of ##f## continuously flowing into the image of ##g##.

    A loop is a map from the unit interval into a space whose values at 0 and 1 are equal or if you like a map from a circle into a space. Two loops ##l_1## and ##l_2## are homotopic if there is a homotopy ##F:S^1×I→X## with ##F(t,0) = l_1(t)## and ##F(t,1) = l_2(t)##. Usually these loops are chosen to have a common end point in which case the end point is kept fixed during the homotopy.

    - A subspace ##Y## of a space ##X## is called a deformation retract of ##X## if there is a homotopy ##F:X×I→X## that is the identity map at time zero, its image at time one is contained in ##Y## and ##F(y,1) = y## for all ##y∈Y##. So at time 1 ##X## is mapped into ##Y## while ##Y## itself is kept fixed.

    - There is also the idea of a retract(as opposed to a deformation retract).

    - It occurs to me that you are thinking of homotopy groups of a space. If so, then homotopic loops keeping endpoints fixed form a group called the fundamental group. There are higher homotopy groups which are made of homotopic maps of spheres into a space. These higher homotopy groups are all abelian.

    Here are a couple examples/exercises:

    - Any point in Euclidean space is a deformation retract of all of Euclidean space.
    - Any loop on a sphere is homotopic to the contant loop.(This isn't easy).
    - A circle is a deformation retract of an annulus.
    - a circle is no a retract of the disk. (Think of the circle as the boundary of the disk.)
    - Let X be ##R^3## minus the z-axis and the unit circle in the xy-plane. Let ##Y## be a small torus around the unit circle, small enough so that it does not touch the z-axis. Show that ##Y## is deformation retract of ##X##.
     
    Last edited: Jul 22, 2017
  4. Jul 23, 2017 #3

    mathwonk

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    any loop on the sphere that misses a point is homotopic to a constant, but by compactness, any loop can be decomposed into a finite sequence of loops, each segment of which lies in a small piece of the sphere, hemnce each of which can be straightened into a geodesic. thus any loop ios homotopic to a finite sequence of geodesics, hence one missing most points.
     
  5. Jul 24, 2017 #4

    lavinia

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    To restate your proof.

    The compactness argument subdivides the closed loop into a finite number of segments each lying in coordinate domain on the sphere. Each segment is homotopic to a geodesic arc so the union of these geodesic arcs is homotopic to the entire curve. But a piecewise geodesic is not space filling and so is null homotopic.

    How does one know that the geodesic arc is homotopic to the segment of the loop?
     
  6. Jul 24, 2017 #5
    @lavinia Perhaps I'm missing some subtlety, but any two paths in a coordinate patch of ##S^2## with the same endpoints must be endpoint-preserving homotopic since the same is true in ##\mathbb{R}^2##.
     
  7. Jul 24, 2017 #6

    mathwonk

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    yes, the local patches are homeomorphic to R^2. In fact a similar argument is then needed to show the piecewise geodesic is homotopic to a constant, since it lies in the complement of one point, which is thus also homeomorphic to R^2, hence contractible.

    for the first part, i guess you could retract the plane onto the rectangle above the interval [0,1], and then retract that rectangle onto that interval. This seems to leave the endpoints {0,1} fixed. i admit it is easier to say than to write down.
     
  8. Jul 24, 2017 #7

    lavinia

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    I agree but some detail is left out for the OP.

    If one has a space filling curve in ##R^2## how does one know that is is homotopic to say a straight line segment with the same endpoints? It would be nice to write the homotopy down.
     
  9. Jul 24, 2017 #8
    Let ##\gamma_1,\gamma_2## be your two paths agreeing at endpoints. My homotopy is ##(1-t)\gamma_1+t\gamma_2##.

    Edit: Of course, I should've also required each coordinate patch to actually contain a geodesic between any two points in it, but this is easy to arrange. Maybe to avoid some technical difficulty (like showing that a piecewise geodesic path is not onto) it would be best to to write my loop ##\gamma## as a finite sequence of loops which are either constant at my basepoint or hit my base point only at their endpoints. Each of the nontrivial parts can be homotoped into a fixed geodesic by basically the argument above (say ##\gamma:[a,b]\to\mathbb{R}^2## is such a part, corresponding to a path ##(a,b)\to\mathbb{R}^2##. The base point condition just becomes ##\lim_{t \to a,b}|\gamma(t)|=\infty##). Arguments are in any topology book, but it's fun to work out for oneself.
     
    Last edited: Jul 24, 2017
  10. Jul 24, 2017 #9

    lavinia

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    The OP might want to know why this map is continuous.

    ##F(t,s) = (1-t)\gamma_1(s)+t\gamma_2(s)##
     
    Last edited: Jul 24, 2017
  11. Jul 24, 2017 #10
    The map ##I^2\to\mathbb{R}^6## given by ##(t,s)\mapsto ((1-s),\gamma_1(t),s,\gamma_2(t))## is continuous when composed with projection onto any factor of the target (since this itself is the composition of a projection, which is continuous, and another continuous function) and thus is continuous by the universal property of product spaces. Then use that scalar multiplication and addition are continuous functions on ##\mathbb{R}^2##.

    This is hopefully enough to satisfy the OP.
     
  12. Jul 24, 2017 #11

    lavinia

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    nice.
     
  13. Jul 24, 2017 #12

    mathwonk

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    a piecewise geodesic path is contained in a finite union of great circles, hence not onto.
     
  14. Jul 24, 2017 #13
    Thanks, that does it.
     
  15. Jul 28, 2017 #14

    lavinia

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    Here is another proof that the sphere is simply connected which illustrates a general method that can be used for any space. This method calculates the fundamental group of a space in terms of two path connected open sets whose intersection is also path connected. The general theorem is called Van Kampen's Theorem. The sphere is a particularly simple case where the two path connected open sets are contractible.

    On the sphere choose ##U_1## and ##U_2## to be the sphere minus two antipodal points ##U_1 = S^2- x## and ##U_2 = S^2- (-x)##. Choose the antipodal points so that the base point ##p## of the loop ##γ## lies in the intersection ##U_1∩U_2##. The ##U##'s are both contractible since they are both homeomorphic to the Euclidean plane (Use stereographic projection to prove this.).

    Split the unit interval ##[0,1]## into finitely many closed intervals ##[t_{i},t_{i+1}]## such that ##γ([t_{i},t_{i+1}])## lies entirely in one of the two open sets ##U_1## and ##U_2## (maybe both). One can assume without loss of generality that at the end points ##t_{i}## ##γ(t_{i})## is not one of the antipodal points.

    For each ##t_{i}## (except for ##0## and ##1##) choose a path ##α_{i}## in ##U_1∩U_2##. from the base point ##p## to ##t_{i}## These paths ##α_{i}## allow one to create a new loop ##ϒ## consisting of a series of loops ##α_{i}γ([t_{i},t_{i+1}])α_{i+1}^{-1}## each of which lies entirely in ##U_1## or ##U_2##. It follows that each of these loops is null homotopic (since the ##U##'s are contractible) and therefore that ##ϒ## is also null homotopic. Further ##ϒ## is homotopic to the original loop ##γ## because ##α_{i}^{-1}α_{i}## is null homotopic and ##ϒ= γ([0,t_1])(α_1^{-1}α_1)γ([t_1,t_2])(α_2^{-1}α_2)...(α_{n-1}^{-1}α_{n-1})γ([t_{n-1},1])##.

    - Van Kampen's Theorem describes the fundamental group of a space in terms of the two path connected open sets and their path connected intersection. These sets may have non-trivial fundamental groups. The algebraic statement of Van Kampen's theorem is a little abstract so I won't go into it here unless you want me to or perhaps you can start a new thread.There is also a decent Wikipedia article on it.
     
    Last edited: Jul 31, 2017
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