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Homotopy and Continuous Functions

  1. Jan 4, 2014 #1

    WWGD

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    Hi, say X is a topological space with subspaces Y,Z , so that

    Y and Z are homotopic in X. Does it follow that there is a continuous

    map f:X→X with f(Y)=Z ? Do we need isotopy to guarantee the existence of

    a _homeomorphism_ h: X→X , taking Y to Z ?

    It seems like the chain of maps parameterized by t , for each x, in

    H: XxI→X with H(x,0)=Y and H(x,1)=Z would give us a path H(x,t)

    of continuous maps parameterized by t , by composing the maps at each stage t ; from

    t=0 to t=1 ; we get an infinite chain of maps, so that

    and that the composition H(x,s); s in [0,1]

    is a continuous map from X to X taking Y to Z. What if Y,Z are just curves in X ?
     
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  3. Jan 4, 2014 #2

    mathwonk

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    huh? this does not make sense to me.
     
  4. Jan 4, 2014 #3

    WWGD

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    What do you mean; if Y,Z are subspaces of X and Y,Z are homotopic to each other within X , i.e., there is
    H : X xI -->X with H(x,0)=Y; H(x,1)=Z , and H(x,t) continuous, does it follow that there exists a continuous map f: X-->X , so that f(Y)=Z ? What doesn't make sense about it? My idea was that the composition of the "deformations" parameterized by t in I=[0,1] , starting at t=0 , ending at t=1 , would allow us to construct a continuous function from X to X taking Y to Z .
     
  5. Jan 4, 2014 #4

    jgens

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    The answer is no. Take X = Z = R and Y = *. Then * is a deformation retract of X so we get a homotopy h:X x I→X such that h(x,0) = * and h(x,1) = x. However one can trivially verify that for every map f:X→X we have f(*) ≠ R.
     
  6. Jan 4, 2014 #5

    WWGD

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    O.K, good point. What if neither of the subspaces was contractible, and the subspaces were proper subspaces (i.e., neither subspace is the whole space) ?

    But still, jgens, you do have a continuous map in the opposite direction; the constant map f(x)=*.
     
    Last edited: Jan 4, 2014
  7. Jan 4, 2014 #6

    jgens

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    Essentially the same argument shows the answer is no again. Just choose X = R-{0} and Y = {-1,1} and Z = (-∞,-1]∪[1,∞) and argue as before.

    Edit: These are all seriously elementary examples. Before asking whether something is true on the forum, why not search for the easy counter-examples yourself?
     
  8. Jan 4, 2014 #7

    WWGD

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    Why not read my answer carefully before telling me my question is simple? I stated _neither subspace is contractible in the space_ . So why not actually read what I say before telling me what I'm saying is simple?

    Besides, I'm not aiming just for a "give me a counterexample". I'm aiming for conditions under which something holds or does not. Besides, I guess I'm not as brilliant as you are, and from time-to-time, I cannot
    find a clear answer.
     
  9. Jan 4, 2014 #8

    jgens

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    Neither of the subspaces are contractible. If you think they are, then you should check again.
     
  10. Jan 4, 2014 #9

    WWGD

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    What do you mean : ## (-\infty ,1] ## is not contractible ? It is homotopic to any interval ## (a,b]##.
     
  11. Jan 4, 2014 #10

    jgens

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    The interval (-∞,-1] is contractible. The disjoint union (-∞,-1]∪[1,∞) has the homotopy type of two points, however, and while homotopy type is crude it still captures the number of connected components.
     
  12. Jan 4, 2014 #11

    WWGD

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    Well, O.K, if you want to look for a "counterexamples in topology" approach using extreme cases, fine ; I'm looking for a general, comprehensive answer. Seriously: is your example significantly-different from having a contractible subspace? So don't preach to me on overly-simplistic.
     
  13. Jan 4, 2014 #12

    jgens

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    These cases are hardly extreme. They should be obvious after about 20 seconds of thought.

    Nope! But until you pose some reasonable restrictions on what you want, this is the best you're gonna get.

    I'm not claiming you're being overly simplistic. I'm claiming that you're lazy. When you stated the question those sorts of counter-examples should have been almost immediate. So then you can refine your question on your own and eventually pose the edited question to the PF community and get the sort of answer you want. I've no idea what you want the topologies on your subspaces to look like. So until you have a better idea what it is you want, the best I can do is tell you where things fail as they stand.
     
  14. Jan 4, 2014 #13

    WWGD

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    And still: your first counter was _not_ a counter. There is a continuous function taking R to * : the constant function. It is just not invertible, but I never asked for invertibility.

    And, yes, I did refine my question: No contractible subspaces; I thought it clear that if I want to avoid contractible subspaces, I also want to avoid subspaces with contractible components. Is it really very different to have contractible subspaces than to have subspaces in which the components are contractible? Let's also assume that both subspaces are connected. If not, we can argue connected-component-wise.

    Now, I understand if I regularly asked lazy questions. do you believe this is the case? Do you not allow for a lazy question from time-to-time? Do you live up to those standards all the time?
     
    Last edited: Jan 4, 2014
  15. Jan 4, 2014 #14

    jgens

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    Well in the first example we can never have f(Y) = Z. So answering the problem you asked literally it definitely works.

    The point was it should have started with that. Not took an example from me (or anyone else) to get you there.

    I'm no mind reader. Sorry if you hoped otherwise. This is not psychic forums :tongue:

    Often the answer here is no. You can occasionally work some magic using finite spaces and in that case it can be pretty important.

    It probably is not. But I do not have the time to continue helping you narrow down what you really want to ask here. Do everyone a favor and figure that out first. Then ask.

    If I feel a question is lazy, then yeah I will probably call the OP out. And lately I've not been asking questions on the forum. So for the last while I certainly have.
     
    Last edited: Jan 4, 2014
  16. Jan 4, 2014 #15

    WWGD

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    Well, it seems you believe you speak for all here. And, yes, you're being hyper-literal, when a small effort --that you're not willing to do -- would clear things up , and show it's no counterexample. But I'm the one who's lazy. And you're considering "occasionally using finite spaces", admitting this is a rare case, which you somehow choose to bring up. Maybe if you did a better, less lazy job of narrowing things down, I could have pinned things down . But I'm the one who's lazy. Cat. Kettle. Black. Right.
     
    Last edited: Jan 4, 2014
  17. Jan 4, 2014 #16

    jgens

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    Certainly not! But I do know other members appreciate it when the OP has put some thought into their questions.

    The difference is rather important. If you want a map f:X→X such that f(Y) = Z, then it's false. If all you want is either f(Y) = Z or f(Z) = Y, then my example fails. But you need to state which of these you do want. The difference is like asking for a homeomorphism and then turning around and saying: "Well I only really wanted either the function or its inverse to be continuous. I didn't mean both! Stop being so literal!"

    In my estimation yes :tongue:

    Well again keeping track of the connected components can be rather important. That it might not be most of the time is hardly reason for me to assume you want to exclude this case without your say-so.

    This is the point! It's not my job to narrow things down for you! It's your job to ask the question you mean to ask. If you put 20 seconds of thought into your question you could have found my counter-example or fixed the wording so that it said what you apparently really meant. That's why I'm calling you lazy.

    Edit: In any case, it is not like me calling you lazy carries any penalty, so do not get so worked up about it. If it irks you so much, then there is an obvious fix: figure out what you really mean to ask and check there are no any obvious counter-examples to your question.
     
  18. Jan 4, 2014 #17

    WWGD

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    Of course, there is no room in your view for the consideration that your observations/claims about my being lazy are subjective, and may just be wrong. Nor do you consider that _your_ views on what your role is or what is not, may be no more than your personal opinion.

    Feel free to ignore every single one of my posts from now on.
     
  19. Jan 4, 2014 #18

    jgens

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    Jesus Christ! Of course my opinion could be wrong. It is, after all, based solely off one thread.

    That is not the point. All you need to do is ask your question precisely and have given it some thought ahead of time. This is what I have been saying to you for post after post. It's not that unreasonable a thing to ask!
     
  20. Jan 4, 2014 #19

    WWGD

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    Ditto here. When I ask for non-contractible, give it a thought and don't come up with a space with contractible components, which is just-about the same thing. And using components? Why not simplify things, assume connectedness. Otherwise, the argument can be done identically arguing component-wise. Why bother needlessly complicating things by bringing-up cases that do not illuminate? How is that not lazy on _your_ part?
    How can you tell if I did think things thru, but still could not figure it out? Maybe I'm no expert in the area, or I'm not too sharp. How do you warrant the assumption that I did not think things thru? What makes you think that what's simple to you is also simple for me, and viceversa? You're making assumptions about me based on a single post I made. That seems unreasonable -- and lazy -- to me, and does not make it seem like you're interested in helping -- which you're not required to do, but then just don't reply if that is the case.

    It just seems like a poteito-potato thing; we seem to have different , incompatible views on how things should be done.
     
    Last edited: Jan 4, 2014
  21. Jan 4, 2014 #20

    jgens

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    The point I am making is that in some applications multiple contractible components can change the answer significantly. I cannot read minds. If you want to exclude this, then say so in your question statement.

    The disconnected spaces were to provide a counter-example. If the spaces were connected, then the new counter-example would have been identical to the first counter-example and would not have satisfied any of your new requirements. You asked a question. Based on what you gave me I answered the question in the negative. You might not like how I did it, but it works nonetheless.

    Because they were counter-examples to the question you asked? What you said was wrong and I provided you with an example why. Apparently this is not what you wanted?

    Again because it is not my job to narrow your question down. You asked a question and I provided you with reasons both times why it does not work. There is not much else I can do! I have no idea what you are looking for so that information needs to be filled in by you.
     
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