Undergrad Homotopy Definitions: Homeomorphisms, Homotopies & Retracts

[Silviu]

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!

 

[lavinia]

- 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 $X$.

 

[mathwonk]

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.

 

[lavinia]

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?

 

[Infrared]

@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 $\mathbb{R}^2$.

 

[mathwonk]

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.

 

[lavinia]

@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 $\mathbb{R}^2$.

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.

 

[Infrared]

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.

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.

 

[lavinia]

Let $\gamma_1,\gamma_2$ be your two paths agreeing at endpoints. My homotopy is $(1-t)\gamma_1+t\gamma_2$.

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

$F(t,s) = (1-t)\gamma_1(s)+t\gamma_2(s)$

 

[Infrared]

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.

 

[lavinia]

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 $\mathbb{R}^2$.

nice.

 

[mathwonk]

a piecewise geodesic path is contained in a finite union of great circles, hence not onto.

 

[Infrared]

Thanks, that does it.

 

[lavinia]

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.