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Inverse of continuous bijection

  1. May 16, 2009 #1
    Under which conditions is an inverse of a continuous bijection continuous?

    I'm not seeking for "the" answer. There probably are many. But anyway, I'm interested to hear about conditions that can be used to guarantee the continuity of the inverse.

    So far I don't know anything else than the open mapping theorem, but I'm not interested in getting restricted to linear mappings now.
     
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  3. May 16, 2009 #2

    CompuChip

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    I think it's always true? For example, you can use the characterization
    f(x) is continuous at x = a whenever limx -> a f(x) = f(a)
    and consider
    f-1(f( limy -> a f-1(y) ))
    using that f is continuous to pull the limit outside (and that it is bijective for f-1 to be defined, of course).
     
  4. May 16, 2009 #3
    If [itex]f^{-1}[/itex] is not continuous, it can be that [tex]\lim_{y\to a}f^{-1}(y)[/tex] does not converge.

    There are some very trivial counter examples. For example equip [itex]X=\mathbb{R}[/itex] with the discrete topology where [itex]\{x\}[/itex] are open, and [itex]Y=\mathbb{R}[/itex] with the Euclidean topology with basis consisting of [itex]]a,b[[/itex]. Now [itex]\textrm{id}:X\to Y[/itex] is continuous, but [itex]\textrm{id}:Y\to X[/itex] is not.

    The open mapping theorem states that if [itex]X,Y[/itex] are Banach spaces, and [itex]f:X\to Y[/itex] is a bounded surjective linear mapping, then [itex]f[/itex] is also open. So in particular if [itex]f[/itex] is also bijective, then [itex]f^{-1}[/itex] is bounded.

    I believe that there exists bounded linear bijections [itex]f:X\to Y[/itex] between non-complete norm spaces [itex]X,Y[/itex] so that [itex]f^{-1}[/itex] are not bounded. I don't know examples of these, but it seems reasonable to assume that the assumption about the completeness of [itex]X,Y[/itex] in the open mapping theorem is not redundant.
     
  5. May 16, 2009 #4
    an idea

    I'll now assume that [itex]X,Y[/itex] are metric spaces, and that [itex]X\times Y[/itex] is equipped with some natural metric like [itex]\sqrt{d_X^2 + d_Y^2}[/itex].

    If [itex]f:X\to Y[/itex] is continuous, then the graph [itex]G=\{(x,f(x))\in X\times Y\;|\; x\in X\}[/itex] is closed.

    One way to approach the original problem would be to try to prove that closedness of [itex]G[/itex] could be used to prove the continuity of [itex]f[/itex] back.

    In general [itex]G[/itex] being closed will not imply [itex]f[/itex] being continuous, because there are examples like [itex]f:\mathbb{R}\to\mathbb{R}[/itex], [itex]f(0)=0[/itex], [itex]f(x)=\frac{1}{x}[/itex], [itex]x\neq 0[/itex].

    In this example the graph escapes to infinity. So this raises a following question: If [itex]G[/itex] is closed, and [itex]Y[/itex] is compact, will [itex]f[/itex] then be continuous? I could not prove anything else than that for all converging sequences [itex]x_n\to x[/itex] there exists a subsequence so that [itex]f(x_{n_k})\to f(x)[/itex].
     
  6. May 16, 2009 #5

    quasar987

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    tHE MOST USEFUL THEOREM is that if f:X-->Y is a continuous bijection with X a compact topological space and Y a Hausdorff topological space, then f^-1 is continuous.

    A counter-example to a bijective continuous map whose inverse is not continuous is given by f:[0,1)-->S^1, f(t)=e^{2pi*i*t}. Then z_n-->1 from "above", f^-1(z_n) approaches 0, but when z_n-->1 from "below", then f^-1(z_n)-->1
     
  7. May 16, 2009 #6
  8. May 16, 2009 #7
    From the study of topological spaces or metric spaces this function must be an open map and this will imply that the function must be a homeomorphism.
     
  9. May 16, 2009 #8
    These were very helpful comments.

    My knowledge on topology relies on the courses and lecture notes. I have not gone through any books. Should I find the quasar987's theorem if searched for it in literature? Does the theorem have any name?
     
  10. May 16, 2009 #9

    quasar987

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    I haven't really read any book on general topology either so I can't give you a reference, but I did saw it stated and proved in the first chapter of Bredon's "Topology and Geometry".

    But actually, the proof is trivial: Let F be some closed set in X. Then F is compact, so f(F) is compact. But a compact set in a Hausdorff space is closed. QED

    Of course the proof relies on the facts that
    (1) a closed set in a compact set is compact.
    (2) continuity preserves compactness.
    (3) a compact set in a Hausdorff space is closed.
    but all three follow quite easily from the relevant definitions.
     
  11. May 16, 2009 #10
    I see. In fact that result was mentioned on the fourth lowest row of this wikipedia article,

    http://en.wikipedia.org/wiki/Open_map

    being called the "closed map lemma".

    A final fact that is needed is that open and closed maps are the same, when we restrict the attention to bijections?
     
  12. May 17, 2009 #11

    quasar987

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    Yes, but this follows from the following relatively more elementary fact: By definition, a map g is continuous if g^{-1}(U) is open as soon as U is open. But by "duality" open/closed, this condition is easily seen to be equivalent to requiring that g^{-1}(F) is closed as soon as F i closed.

    If f:X-->Y is a bijection, then the above implies that f is closed iff f is open because both statement are equivalent to "f^-1 is continuous" (since (f^{-1})^{-1}=f).
     
    Last edited: May 17, 2009
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