Re: Question on Homotopy Classes

On Dec 6, 2005 7:37 PM CT, Lee Rudolph wrote:

> Narcoleptic Insomniac
> <i_have_narcoleptic_insomnia@xxxxxxxxx> writes:
>
> >Okay, this is not a homework exercise, but it is an
> >exercise out of Munkres "Topology" text:
> >
> >9.52.2a) Given spaces X and Y, let [X, Y] denote the
> >set of homotopy classes of maps of X into Y. Let
> >I = [0, 1].
>
> Okay, I is the set of homotopy classes of 0 into 1.
> Now what?
> ...Oh, that isn't what you meant? Sorry!
>

: - P

>
> >Show that for any X, the set [X, I] has a single
> >element.
> >
> >We discussed the notation [X, Y] a bit in class and
> >from what I gather it seems to partition all the
> >continuous maps from X to Y by putting maps that are
> >homotopic to each other in the same class.
>
> That is indeed what "let [X, Y] denote the set of
> homotopy classes" wants to mean. Of course it can't
> really be said to *mean* that until it has been shown
> that "homotopy class" is well-defined, in other words,
> that "homotopy" (of maps) is an equivalence relation.
> This presumably precedes 9.52.2a).
>

Yes, this was done chapter 52 and we went over it in
class.

> >For instance, suppose there were only four
> >continuous maps a, b, c, d : X --> Y. Now suppose
> >that a was only homotopic to c, and that b was only
> >homotopic to d. The set [X, Y] would just have two
> >elements then correct? If so, how would I denote
> >those elements; [a] is in [X, Y] and [c] is in [X, Y]?
>
> That would be a fine notation, and I think it's one
> of the standard ones. Again, I presume that somewhere
> before 9.5.2a), Munkres sets out the notation *he*
> wants to use?
>

Yep, that was the notation given in the text. I suppose
these questions seem a bit obvious, but I was a bit
unsure because I had not seen many explicit examples.

>
> >Okay, so back to the exercise; show that for any X,
> >the set [X, I] has a single element. Intuitively I
> >interpret this to mean that all continuous maps from X
> >to I are homotopic to one another.
>
> I do not think "Intuitively" means what you
> apparently think it means. "By taking the given
> definitions at face value, I understand this to mean"
> (etc.) would be more on the mark.
>

Hahah, I think you're correct.

Intuition: The act or faculty of knowing or sensing
without the use of rational processes.

Can you tell I'm not a fan of English and grammar?

> >I know that two continuous maps f, g : X --> I are
> >homotopic if there exists a continuous mapping
> >F: X x I --> I such that F(s, 0) = f(s) and
> >F(s, 1) = g(s). How to show [X, I] consists of
> >only one element and what that element is (constant
> >mapping maybe?) alludes me.
>
> (*E*ludes.)
>

Heh, maybe I should take an English course next semester.

> Why "maybe"? If, indeed, [X, I] has only one
> element, then that single element is a homotopy class
> that contains every continuous map from X to I,
> including each of the constant maps. So, "(any choice
> that I like to make of a constant mapping, certainly!)"
> would be appropriate.
>
> At this point, I'm afraid that almost anything more I
> say will just be a dead giveaway. But here's something
> that may be closer to a mere hint. Consider the case
> when X is a one-point space, say X = {0}. Then a
> continuous map f from X to I is necessarily constant,
> and so may be identified with the element z = f(0) in
> I. Can you find a continuous mapping F from {0} x I to
> I such that F(s,0) = z and F(s,1) = 0; that is, a
> homotopy from your original constant map z to the
> constant map 0?
>

How about F(s, t) = f(s) - f(s) * t. Then at the 0-level
we have F(s, 0) = f(s), and since s can only be 0 if X =
{0} then F(0, 0) = f(0) = z. At the 1-level we have
F(s, 1) = f(s) - f(s) = 0. Moreover, the continuity of
F follows from the continuity of f.

So using the homotopy F above, I can show any continuous
function f: X --> I is homotopic to the constant map,
from which it follows that [X, Y] must consist of one
element.

Thanks again for the math (and English) help Lee.

>
> Draw what the graph of such a map (a subset of the
> product ({0}xI)xI, which is of course pretty much
> identical to the square IxI) would look like. Draw
> the *simplest* such graph. Think about it.
>
> Lee Rudolph
>
.

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