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- Thread starter madness
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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.f

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

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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.

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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.

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Bingo! Your idea does successfully define functions, but they aren't continuous functions.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.

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 S

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1) f is continuous on [0,1)

2) lim x->1 f(x)-f(0)=N for some integer N

Do you agree?

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" 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.

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This is an equally valid way to denote a continuous function from the circle to itself. Notice that any

1) f is continuous on [0,1)

2) lim x->1 f(x)-f(0)=N for some integer N

Do you agree?

It's true, but nontrivial, that to any function

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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?

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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 S

The first realization is the quotient

[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

[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 S

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.

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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?

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