# Contractible spaces

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.

Hurkyl
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Then try it.

You might try a circle first; IMO they're easier to work with. (and they're homotopy equivalent to punctured discs anyways)

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.

Hurkyl
Staff Emeritus
Gold Member
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?

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 travelling from the initial point x_0 around an arc until it reaches the point x.

Hurkyl
Staff Emeritus
Gold Member
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)
• $\lim_{x \rightarrow 1^-} f(x)$ = $f(0)$

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

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?

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.

Hurkyl
Staff Emeritus
Gold Member
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.

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:
Hurkyl
Staff Emeritus
Gold Member
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
$$f(x) = x \pmod{\mathbf{Z}}$$​
Does this satisfy my condition? Yes, because
$$\lim_{x \rightarrow 1^-} f(x) = 1 = 0 = f(0) \pmod{\mathbf{Z}}$$​

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

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.

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?