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Inverse functions for f:R^m->R^m , or f:X^m->Y^m

  1. Mar 9, 2009 #1

    WWGD

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    Inverse functions for f:R^m-->R^m , or f:X^m-->Y^m

    Hi:
    This is , I guess a technical question:

    Given f:R^m --->R^m ; f=(f_1(x_1,..,x_m),....,f_m(x_1,...,x_m))

    Then I guess f^-1 (of course, assume f is 1-1.). Is given by a "pointwise" inverse ,

    (right?) i.e.,

    f^-1 =(f_1^-1 (x_1,..,x_m) ,....,f_m^-1(x_1,..,x_m)) ?.

    Is there some theorem on existence of inverses if we only know f to be
    continuous ( I think there is no known test for whether a function into R^m
    is onto )?

    Thanks.
     
  2. jcsd
  3. Mar 9, 2009 #2

    WWGD

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    Re: Inverse functions for f:R^m-->R^m , or f:X^m-->Y^m

    Correction:
    Conditions for f should have been (the equiv.) f_1,...,f_n are 1-1.
    Thanks.
     
  4. Mar 9, 2009 #3
    Re: Inverse functions for f:R^m-->R^m , or f:X^m-->Y^m

    Can you give an example of a continuous function R^m->R (like you f_i) that is 1-to-1? I am not sure such a function exists for m>1.
     
  5. Mar 9, 2009 #4

    WWGD

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    Re: Inverse functions for f:R^m-->R^m , or f:X^m-->Y^m

    Yes, it seems difficult, I will think about it. I also realized that the condition that the f_i
    are 1-1 is not equivalent to f:R^m-->R^m being 1-1; f :(x,y) = (Pi_1(x,y),Pi_2(x,y) )=
    (x,y), with Pi_1, Pi_2 the projections is a counterexample. Life seems to become weird
    outside of the safety of functions f:R-->R. Ontoness on each variable does not guarantee
    ontoness of f:R^m-->R^m , with the dramatic example of f:R^2-->R^2 :
    f_1(x,y) =x =f_2(x,y).

    P.S: Sorry for the wordiness of my last post on bdry. of a manifold.
     
  6. Mar 9, 2009 #5

    WWGD

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    Re: Inverse functions for f:R^m-->R^m , or f:X^m-->Y^m

    I was wrong again here : If we have f:R^n-->R^n (or f:X^n -->Y^n )

    and _just one_ of the f_i's is 1-1 , then f is 1-1 f(x_1,..,x_n)=f(y_1,..,y_n) , (x_1=(x_1i,

    etc.)then f_i(x_i)=f_i(y_i) , which cannot happen if f_i is 1-1. Rest is still tricky,

    but possible: there are continuous bijections f:X-->Y that are not homeomorphisms

    ( I need to look it up in my 'Counterexamples in Topology' book).

    These product spaces can be confusing (at least to me).
     
  7. Mar 10, 2009 #6

    WWGD

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    Re: Inverse functions for f:R^m-->R^m , or f:X^m-->Y^m

    Hi, yyat:

    I remembered the answer to your followup on existence of continuous functions
    f:R^m-->R :

    the answer is no, by k-connectivity: if h:R^m -->R is continuous, then so is

    h': R^m-{x} -->R-{h(x)} . But R^m-{x} is connected, and R-{h(x)} is not.

    I don't think that there is a nice answer for functions f:R^m-->R^k without

    using algebraic topology , i.e., fundamental groups. I will think about this, please

    let me know if you come up with some ideas.


    Another thought about ontoness of maps f:R^m-->R^m ; maybe not too deep,

    but I still think useful: given f=(f_1,..,f_n) , the functions f_i cannot be linearly

    dependent, e.g., if f_1(x,y)=x and f_2(x,y) =nx , then f will only hit points in the

    (linear) subspace (x,nx), and will miss every point (x+a,nx+b) for (a,b)=/(0,0) ; a,b in R.

    This is necessary, together with the ontoness of each f_i ( if , say, f_k misses

    y_k , then f will miss RxR..xRx{y_k}xRx..xR ) , but I don't know if these two conditions

    together are sufficient to guarantee ontoness.
     
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