Homotopy Equivalence: Definition & Examples

[Silviu]

Hello! I am a bit confused about the definition of homotopy for loops. So it looks like: Let $\alpha, \beta : I -> X$ be loops at $x_0$. They are said to be homotopic if there is a continuous map $F : I \times I -> X$ such that: $F(s,0)=\alpha (s), F(s,1)=\beta(s), F(0,t)=F(1,t)=x_0$ for all $s,t \in I$. As far as I understood this means that the 2 loops can be deformed from one to another. However I am not sure I understand why you can't find such a function if you can't deform them to one another. If we have let's say an annulus and a loop around the hole in it (which can't be reduced to a point) and another loop that can be reduced to a point they are not homotopic to each other. However if instead of the annulus we have a solid disk, they are homotopic. I can't seem to see why F can't be found in the first case, while it can in the second one.

 

[andrewkirk]

I can't seem to see why F can't be found in the first case, while it can in the second one.

The reason F cannot be found for the second one is that the loop is not homotopically equivalent to a point. Equivalently, a disc is not homeomorphic to an annulus. In more folksy terms, it's because the loop encircles a hole - it would have to go outside of the annulus at some stage in order to morph into a point.

It's easy to demonstrate a homotopy for the first case. Imagine the disk as embedded in Euclidean 2-space with a Cartesian coordinate system with origin at $x_0$. Then given a loop $\gamma$ based at $x_0$, the function
$$F:I\times I\to \mathbb R^2,\ \ F(u,v)=(1-u)\gamma(v)$$
is such a homotopy.

It is not easy to rigorously demonstrate that there is no homotopy in the second case. We either have to be content with our intuition that 'the hole gets in the way' or dive into algebraic topology.

 

[Silviu]

The reason F cannot be found for the second one is that the loop is not homotopically equivalent to a point. Equivalently, a disc is not homeomorphic to an annulus. In more folksy terms, it's because the loop encircles a hole - it would have to go outside of the annulus at some stage in order to morph into a point.

It's easy to demonstrate a homotopy for the first case. Imagine the disk as embedded in Euclidean 2-space with a Cartesian coordinate system with origin at $x_0$. Then given a loop $\gamma$ based at $x_0$, the function
$$F:I\times I\to \mathbb R^2,\ \ F(u,v)=(1-u)\gamma(v)$$
is such a homotopy.

It is not easy to rigorously demonstrate that there is no homotopy in the second case. We either have to be content with our intuition that 'the hole gets in the way' or dive into algebraic topology.

Thank you for your answer. Yes, intuitively it makes sense, but I was wondering if for more complicated shapes (in more dimensions I guess), where you can't use your intuition, is there a rigorous way to show that 2 loops are not part of the same homotopy class?

 

[lavinia]

Thank you for your answer. Yes, intuitively it makes sense, but I was wondering if for more complicated shapes (in more dimensions I guess), where you can't use your intuition, is there a rigorous way to show that 2 loops are not part of the same homotopy class?

Generally, distinguishing homotopy classes of loops is hard.

For instance on a sphere one can show that every closed loop is null homotopic. If the loop misses a point on the sphere one can remove the point and do the homotopy on the sphere minus a point (since the sphere minus a point is homeomorphic to a disk Andrewkirk's method can be used.). But how do you show this for a loop that is also a space filling curve that passes through every point on the sphere?

 

[lavinia]

For the case of an annulus this line of thought may help.

First suppose that $c(t)$ is a path in the closed interval $[0,1]$ with end points $c(0) = 0$ and $c(1) = 1$. It is clear that a homotopy $H$ that holds both end points fixed can not shrink the path to a single point.

But the closed interval can be made into a loop by pasting the end points together.

 

[Silviu]

For the case of an annulus this line of thought may help.

First suppose that $c(t)$ is a path in the closed interval $[0,1]$ with end points $c(0) = 0$ and $c(1) = 1$. It is clear that a homotopy $H$ that holds both end points fixed can not shrink the path to a single point.

But the closed interval can be made into a loop by pasting the end points together.

What I am confused about is where exactly in the deffiniton of F it is implied that it is forbidden to cut of a loop and paste it later (after you go around a hole for example)?

 

[lavinia]

What I am confused about is where exactly in the deffiniton of F it is implied that it is forbidden to cut of a loop and paste it later (after you go around a hole for example)?

The homotopy is required to keep $F(0,s) = F(1,s)$ for each $s$.

You can think of it as a map from $S^1×I$ , the circle cross the unit interval, where at time zero $F$ is the first loop and at time 1 it is the second loop.

 

[mathwonk]

If you are willing to consider smooth homotopies there is a method i like. I discovered it when I was teaching advanced calculus and looking for some application for the stokes and greens theorems. namely, a loop that is homotopic smoothly to a point, is the boundary of a parametrized disk, Hence by greens theorem, the integral of any differential form round that loop equals the integral of its differential over the disc. I particular a form with zero differential, i.e. a "closed" form, should have integral zero around the loop. But the closed "angle form" dtheta, has ± 2pi around a loop that goes once around the origin. More generally there are in every smooth manifold, closed forms that meaure precisely how many times a given cycle encloses a given "hole" in the space.

Another way to measure the winding number of a loop in the plane is use the exponential map and the lifting property for loops. I.e. only loops that shrink to a point in the punctured plane have closed lifts.