Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Homotopy homework help

  1. Jan 23, 2010 #1
    Prove every map [itex]e:X \rightarrow \mathbb{R}^n[/itex] is homotopic to a constant map.

    well i said that the constant map is [itex]c:X \rightarrow \mathbb{R}^n;x \mapsto c[/itex]
    since [itex]\{ c \} \subseteq \mathbb{R}^n[/itex] is a clealry a convex subspace and [itex]e(X)=\mathbb{R}^n[/itex] is a convex subspace of [itex]\mathbb{R}^n[/itex], e and c must be homotopic (using the fact that any two maps [itex]f,g: X \rightarrow Y[/itex] where Y is a convex subset of [itex]\mathbb{R}^n[/itex] are homotopic).

    however, i'm not sure if i can assume [itex]e(X) \subseteq \mathbb{R}^n[/itex] is a convex subset. probably not. any ideas?

    thanks.
     
  2. jcsd
  3. Jan 23, 2010 #2
    Re: homotopy

    You can't assume that, but you don't need to. The usual straight-path homotopy:
    [tex]H(s,t) = (1-t)e(s) + tc[/tex]
    still works since [itex]\mathbb{R}^n[/itex] is convex. You seem to be confusing the image and codomain. e has codomain [itex]\mathbb{R}^n[/itex], but its image e(X) is just a subset of [itex]\mathbb{R}^n[/itex]. You wrote:
    If you know this, then this is actually sufficient since in the case of e we have that Y is [itex]\mathbb{R}^n[/itex] and this is a convex subset of itself. In this quote e(X) is never mentioned and it doesn't matter that e isn't surjective.
     
  4. Jan 23, 2010 #3
    Re: homotopy

    so it's as easy as saying that since [itex]Y=\mathbb{R}^n[/itex] for both the map e and the map c and [itex]\mathbb{R}^n[/itex] is a convex subset of itself, e and c both map to convex subsets of [itex]\mathbb{R}^n[/itex] and are therefore homotopic by the homotopy [itex]H: X \times I \rightarrow \mathbb{R}^n ; (s,t) \mapsto (1-t)e(s)+tc[/itex]

    do i need to also show that this is in fact a homotopy?
    i.e. show [itex]h(s,0)=e(s),h(s,1)=c(s)=c[/itex]?
    or is the above enough

    thanks
     
  5. Jan 23, 2010 #4
    Re: homotopy

    Well that depends on how much detail you want to give. If you want to do it in great detail you would have to argue that h is continuous since it's the composition of continuous functions, and that it's in fact a homotopy from e to c. In my opinion both statement are so simple that I wouldn't bother to give more detail than you gave since you stated that h is from e to c and it's very simply to verify h(s,0) = (1-0)e(s)+0c=e(s) and h(s,1) = (1-1)e(s)+ 1c=c.
     
  6. Jan 24, 2010 #5
    Re: homotopy

    ok. thanks.

    the next bit asks me to show that if [itex]f:X \rightarrow S^n[/itex] is not surjective then f is homotopic to a constant map.

    well this means image is not codomain so f maps to f(X) but i can't find any useful theorems in my notes to prove there exists a homotopy. any suggestions?

    thanks.
     
  7. Jan 24, 2010 #6
    Re: homotopy

    We can find some [itex]N \in \mathbb{S}^n[/itex] such that N isn't in the image of f, so we can restrict its codomain to form a continuous map [itex]f' : X \to \mathbb{S}^n \setminus \{N\}[/itex] such that [itex]f = \iota \circ f'[/itex] where [itex]\iota : \mathbb{S}^n\setminus\{N\} \to \mathbb{S}^n[/itex] is the inclusion map.

    If [itex]N \in \mathbb{S}^n[/itex], then [itex]\mathbb{S}^n \setminus \{N\}[/itex] is homeomorphic to [itex]\mathbb{R}^{n}[/itex] by the stereographic projection [itex]\sigma : \mathbb{S}^n \setminus \{N\} \to \mathbb{R}^{n}[/itex]. You can use this fact to construct a homotopy from [itex]\sigma \circ f' : X \to \mathbb{R}^n[/itex] to a constant map, and you can then use the inverse stereographic projection [itex]\sigma^{-1}[/itex] to obtain your desired homotopy.
     
  8. Jan 24, 2010 #7
    Re: homotopy

    but how do you know that there is only one such [itex]N \in S^n[/itex] that isn't in the image of [itex]f[/itex]?

    also, couldn't i just say that from the argument above (i.e. in the earlier posts) that [itex]\sigma \circ f' : X \rightarrow \mathbb{R}^n[/itex] is homotopic to a constant map as it is an example of such a function [itex]e: X \rightarrow \mathbb{R}^n[/itex].
     
  9. Jan 24, 2010 #8
    Re: homotopy

    I don't. But I don't need to. We just need to remove point to get the homeomorphism, and we never state that f' is surjective (because we don't need that).

    Yes that was what I had in mind. Then you get a homotopy [itex]h : X\times I \to \mathbb{R}^n[/itex] You then compose it with [itex]\iota \circ \sigma^{-1} : \mathbb{R}^n \to \mathbb{S}^n[/itex] and show that [itex]\iota \circ \sigma^{-1} \circ h[/itex] is your desired homotopy.
     
  10. Jan 24, 2010 #9
    Re: homotopy

    so surely i can just write: and from part a) this works rather than ahving to write out the explicit homotopy.

    also how can the function [itex]\iota[/itex] exist? wouldn't it have to be one-to-many?
     
  11. Jan 24, 2010 #10
    Re: homotopy

    For the existence of h yes. But this still only shows them homotopic in [itex]\mathbb{R}^n[/itex] so you need to apply [itex]\iota \circ \sigma^{-1}[/itex] to get back to a homotopy in [itex]\mathbb{S}^n[/itex].


    [itex]\iota : \mathbb{S}^n\setminus\{N\} \to \mathbb{S}^n[/itex] is just the inclusion function, so it's simply the identity map of S^n with its domain restricted:
    [tex]\iota(x) = x \qquad \textrm{for }x \in \mathbb{S}^n \setminus \{N\}[/tex]
    It's not surjective, but we don't need it to be.
     
  12. Jan 24, 2010 #11
    Re: homotopy

    why is it important that [itex]\sigma[/itex] is a homeomorphism?

    also, we know h is a homotopy from the first part of the question.
    how do we go about showing that [itex]\iot \circ \sigma^{-1} \circ h[/itex] is also a homotopy? i can't find anything that says composition of functions with a homotopy leaves a homotopy. is that where the homeomorphism plays a role? even so, [itex]\iota[/itex] is just a function, not a homeomorphism.
     
  13. Jan 24, 2010 #12
    Re: homotopy

    Because a homeomorphism is precisely a continuous map with a continuous inverse. Thus it's exactly what is required for [itex]\sigma^{-1}[/itex] to exist and be continuous.

    A homotopy of maps from X to Y is simply a continuous map from X x I to Y. Since [itex]\sigma[/itex] is a homeomorphism [itex]\sigma^{-1}[/itex] is continuous. [itex]\iota = \textrm{id}|(\mathbb{S}^n \setminus\{N\})[/itex], so [itex]\iota[/itex] is the restriction of a continuous function and therefore [itex]\iota[/itex] is continuous. Since all three functions are continuous you know that [itex]\iota \circ \sigma^{-1} \circ h[/itex] is continuous and therefore a homotopy. You now only need to check that it's actually a homotopy from f to some constant map.
    [tex](\iota \circ \sigma^{-1} \circ h)(s,0) = (\iota \circ \sigma^{-1})(h(s,0)) = (\iota \circ \sigma^{-1})((\sigma \circ f')(s)) = (\iota \circ \sigma^{-1} \circ \sigma \circ f')(s) = (\iota \circ f')(s) = f(s)[/tex]
    [tex](\iota \circ \sigma^{-1} \circ h)(s,1) = (\iota \circ \sigma^{-1})(c(s)) =(\iota \circ \sigma^{-1})(c(0)) =(\iota \circ \sigma^{-1} \circ c)(0)[/tex]
     
  14. Jan 25, 2010 #13
    Re: homotopy

    thanks but i'm still not sure about a couple of things:

    (i) in your very last line shouldn't it read
    [itex](i \circ \sigma^{-1} \circ h)(s,1)=(i \circ \sigma^{-1})(c(s))[/itex]
    but then [itex]c(s)=c[/itex] and we don't know what [itex](i \circ \sigma^{-1})(c)[/itex] is so we just leave it as [itex](i \circ \sigma^{-1})(c)[/itex]. now the inverse stereographic projection just maps to some point in [itex]S^n - \{ N \}[/itex] and [itex]\iota[/itex] is just the identity on this space so we remain at this point [itex]\sigma^{-1}(c) \in S^n - \{ N \}[/itex]. since [itex]c[/itex] is constant, [itex]\sigma^{-1}(c)[/itex] will be constant and so we know we have a homotopy between f and the constant map. is that correct?

    (ii) to do with the reasoning:
    i accept that [itex]\iota, \sigma^{-1}[/itex] and [itex]h[/itex] are all continuous and that's grand. but only one of them is a homotopy so how can we deduce that the composition is a homotopy? surely this conclusion should come at the very end of your argument once we've verified the composition is a continuous map from f to a constant map (as this is the definition of a homotopy from f to c). is it not?

    (iii) i thought all homotopies mapped from something crossed with I (the unit interval). this composition of maps just maps from some space [itex]X[/itex] not [itex]X \times I[/itex] does it not?
     
  15. Jan 25, 2010 #14
    Re: homotopy

    This is correct and essentially the same as mine. Since c is constant c(0) in my post is the same as what you call c.

    A homotopy in general is nothing but a continuous map from I x X to Y for some space X,Y. We don't really know what this is a homotopy between, but if we have a map from I x X to S^n, then all we really need to check is that it's continuous. Thus it suffices to check that the functions are continuous. At this point we know that it's a homotopy, but we haven't yet verified what it's a homotopy from and to so that's what the last calculation is for.

    The composition is:
    [tex]X \times I \xrightarrow{h} \mathbb{R}^n \xrightarrow{\sigma^{-1}} \mathbb{S}^n \setminus \{N\} \xrightarrow{\iota} \mathbb{S}^n[/tex]
    so it's a map from X x I to S^n.


    Maybe the argument is easier to see in general. Let [itex]H : X \times I \to Y[/itex] be some homotopy from [itex]p : X \to Y[/itex] to [itex]q : X \to Y[/itex], and let [itex]g : Y \to Z[/itex] be a continuous map. Then [itex]g \circ H : X \times I \to Z[/itex] is a continuous map and therefore a homotopy from g(H(s,0)) = g(p(s)) to g(H(s,1))=g(q(s)). Thus we get a homotopy [itex]g \circ H : g \circ p \simeq g \circ q[/itex]. This establishes the following result:
    Thm: Let [itex]p,q : X \to Y[/itex] and [itex]g : Y \to Z[/itex] be continuous maps. Then [itex]p \simeq q[/itex] imply [itex]g \circ p \simeq g \circ q[/itex].
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook