Understanding Contractible Spaces: Path Connected vs Simply Connected

  • Thread starter madness
  • Start date
In summary, a contractible space is homotopy equivalent to a single-point space {0}. As long as the space is path connected, it is contractible. The punctured disc is not contractible because every point is path connected to every other point, making it a single-point space.
  • #1
madness
815
70
I'm a bit confused about the idea of contractible spaces. These spaces are homotopy equivalent to a single-point space {0}. As far as I can see, a space is contractible if and only if it is path connected. However, wikipedia states that a contractible space must also be simply connected. Isn't the punctured disc contractible? Every point is path connected to every other point, so it should be simple to create a homotopy equivalence.
 
Physics news on Phys.org
  • #2
Then try it. :smile:

You might try a circle first; IMO they're easier to work with. (and they're homotopy equivalent to punctured discs anyways)
 
  • #3
I'm not sure about constructing an explicit homotopy, but I can show my general reasoning:

X is contractible if it is homotopy equivalent to the space {0}, i.e. if there exist maps f:X->{0} and g:{0}->X and homotopies h:gf (equiv) 1_X, k:fg (equiv) 1_{0}. f(x)=0 for all x, g(0)=x_0 for some x_0 in X. Hence gf(x)=x_0 for all x and fg(0)=0. So k is given by the constant homotopy and for any fixed x, h(x,t) is a path from x_0 to x. If such paths exist for all x in X then h is a homotopy equivalence and X is contractible. Hence a path connected space is contractible.
 
  • #4
Surely you can take your idea and write down some explicit functions for f, g, h, and k in the case where X is a circle?
 
  • #5
Well obviously I can't because the unit circle isn't contractible. But I seem to have proved that any path connected space is contractible, so I need to work out where my understanding has gone wrong. I think the problem is that h may not be continuous even if each path is individually.

Ok here's a quick attempt that I haven't really checked fully:

the points on the circle are given by x=(cos2pit,sin2pit) with t in [0,1), so they can be described in terms of t. As I showed in the proof (I'm not sure if you actually read it), f(t)=0 for all t and g(0) = x_0, which I choose to be (1,0). gf(t)=0 is already the identity. Let h(t,t') = (cos2pit't, sin2pit't), for t' in [0,1]. So h(t,t')=x_0 and h(t,t')=(cos2pit,sin2pit)=x(t). So my homotopy takes a path by traveling from the initial point x_0 around an arc until it reaches the point x.
 
  • #6
madness said:
But I seem to have proved that any path connected space is contractible, so I need to work out where my understanding has gone wrong. I think the problem is that h may not be continuous even if each path is individually.
Bingo! Your idea does successfully define functions, but they aren't continuous functions.

When you try to define continuous functions on the circle by instead defining them on [0,1), you need the following two things to be true:
  • f is a continuous function on [0,1)
  • [itex]\lim_{x \rightarrow 1^-} f(x)[/itex] = [itex]f(0)[/itex]

A similar thing is true for functions with domain S1xX for some other space X.
 
  • #7
Thanks for the help. I was pretty satisfied with your response until now when I was working through a past paper that disagrees. So what we really need is that

1) f is continuous on [0,1)
2) lim x->1 f(x)-f(0)=N for some integer N

Do you agree?
 
  • #8
Hurkyl wrote , in part:

" A similar thing is true for functions with domain S1xX for some other space X. "

In a way, but isn't it much hairier when you have a product space as a domain.?

If so, the only general way of shoing continuity is showing that the inverse image

is open in the product, which is often a hassle; sequential continuity does not

always help, and if your inverse image is not a basic or subasic set, then it is

pretty difficult to determine continuity of the map. I have the impression that

a lot of alg. topologists do not fully address some of these point-set issues in

continuity.


Hurkyl: I am a bit confused about your use of the term path; AFAIK, a path is,

by definition, a continuous map


Madness: I think if you work with free loops, then your group of paths

is trivial in a path-connected space. It is the fixed endpoints that do you in.

A common trouble, as in the case of the circle, is that, if you want to leave your

endpoints fixed, you need to tear the circle open before homotoping it to a point,

and , when the endpoints are fixed, tearing is discontinuous.
 
  • #9
madness said:
Thanks for the help. I was pretty satisfied with your response until now when I was working through a past paper that disagrees. So what we really need is that

1) f is continuous on [0,1)
2) lim x->1 f(x)-f(0)=N for some integer N

Do you agree?
This is an equally valid way to denote a continuous function from the circle to itself. Notice that any f from [0,1) to R that satisfies the condition allows us to define a function g from [0,1) to S1 with the property that g(0) is the limit of g(x) as x approaches 1.

It's true, but nontrivial, that to any function g from [0,1) to the circle that satisfies my condition, there exists a corresponding function f from [0,1) to R that satisfy the condition you stated.
 
  • #10
Ok but the function f:[0,1)->S1 is essentially the angle function for the loop right? So wouldn't your function necessarily have degree 0, whereas the integer N from my condition is the degree for an arbitrary loop?
 
Last edited:
  • #11
madness said:
Ok but the function f:[0,1)->S1 is essentially the angle function for the loop right? So wouldn't your function necessarily have degree 0, whereas the integer N from my condition is the degree for an arbitrary loop?
Nope. Pay attention to the codomain -- my function is actually mapping to the circle, whereas your function is mapping to the reals.

Let me clarify in two different ways, by writing an explicit degree 1 map to two different realizations of S1.

The first realization is the quotient R/Z. Consider the function
[tex]f(x) = x \pmod{\mathbf{Z}}[/tex]​
Does this satisfy my condition? Yes, because
[tex]\lim_{x \rightarrow 1^-} f(x) = 1 = 0 = f(0) \pmod{\mathbf{Z}}[/tex]​


The second realization is as the unit circle in R2. Consider the function
[tex]f(x) = \left( \cos x, \sin x \right)[/tex]​
Does this satisfy my condition? Yes, because
[tex]\lim_{x \rightarrow 1^-} f(x) = (1, 0) = f(0)[/tex]​



Notice that neither of us are talking about functions from the circle to itself. I'm talking about functions from [0,1) to S1. You are talking about functions from [0,1) to R. The reason we are doing this is because we are using a clever naming scheme! Rather than work with functions from the circle to itself directly, we are instead naming by simpler kinds of functions -- and then reinterpret topological properties of the circle in terms of the names.

Since we are using different naming schemes, the result of translation is different. That is why "f is continuous" looks slightly different between your scheme and mine.
 
  • #12
madness said:
I'm a bit confused about the idea of contractible spaces. These spaces are homotopy equivalent to a single-point space {0}. As far as I can see, a space is contractible if and only if it is path connected. However, wikipedia states that a contractible space must also be simply connected. Isn't the punctured disc contractible? Every point is path connected to every other point, so it should be simple to create a homotopy equivalence.
First, I hope that the "if and only if" is a typo. Second, it's fairly easy to see that a contractible space is simply connected, right? Since you can "contract" any loop to a point?
 

1. What is the difference between path connected and simply connected spaces?

The main difference between path connected and simply connected spaces is that while both types of spaces allow for continuous paths from one point to another, simply connected spaces also require every loop to be contractible. In other words, simply connected spaces do not have any "holes" or "voids" that cannot be continuously shrunk to a point, whereas path connected spaces may have such holes.

2. Can a space be both path connected and simply connected?

Yes, a space can be both path connected and simply connected. This type of space is known as a "contractible space", meaning that every loop in the space can be continuously shrunk to a point. All simply connected spaces are also path connected, but not all path connected spaces are simply connected.

3. How do contractible spaces relate to homotopy equivalence?

Contractible spaces are homotopy equivalent to a single point. This means that any two points in a contractible space can be connected by a continuous path, and any loop in the space can be continuously shrunk to a point. Homotopy equivalence is a relation between spaces that allows for continuous transformations between them.

4. Are all contractible spaces path connected?

Yes, all contractible spaces are path connected. This is because contractible spaces are defined as spaces where every loop is contractible, and path connected spaces allow for continuous paths from one point to another. Therefore, a contractible space must also be path connected.

5. What are some examples of contractible spaces?

Some examples of contractible spaces include the unit disk, the unit ball, and the n-dimensional Euclidean space. These spaces are contractible because any loop in these spaces can be continuously shrunk to a point. Other examples include the real line, the unit square, and the unit sphere.

Similar threads

Replies
40
Views
6K
  • Beyond the Standard Models
Replies
7
Views
1K
Replies
9
Views
3K
  • Differential Geometry
2
Replies
63
Views
8K
  • Differential Geometry
Replies
9
Views
5K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Special and General Relativity
Replies
11
Views
1K
Replies
6
Views
635
Replies
4
Views
893
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
Back
Top