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Continuous bijection that is not an embedding

  1. Apr 25, 2010 #1
    Hi:
    Just curious: a continuous function f:X-->Y ; X,Y topological spaces, can fail
    to be an embedding because it is not 1-1, or, if f is 1-1 , f can fail to be an
    embedding because, for U open in X f(U) is not open in f(X).

    Can anyone think of a "reasonable" example of the last case, by reasonable
    I mean no spaces with finitely many points, the inclusion map from a space
    into itself with a different topology (i.e.: we have (X,T) and (X,T') , and we
    use i:X-->X : i(x)=x ) , or maybe or something one could find in
    "Counterexamples" book.

    If A is a strict subspace of X , then the subspace topology on A guarantees
    that i:A-->X , the inclusion, is an embedding. Is this true for other topologies on A.?

    Thanks.
     
  2. jcsd
  3. Apr 25, 2010 #2
    X = interval [0,2 pi), Y = circle {(x,y) : x^2+y^2=1}, continuous bijection f(t) = (cos(t),sin(t)).
     
  4. Apr 25, 2010 #3
    >> If A is a strict subspace of X , then the subspace topology on A guarantees
    that i:A-->X , the inclusion, is an embedding. Is this true for other topologies on A.?

    No. In fact, that is the definition of "subspace topology" ... so for any topology on A other than the subspace topology, the inclusion i is NOT an embedding. Either because i is not continuous, or because i is not open, or both.
     
  5. Apr 25, 2010 #4
    Edgar wrote
    " No. In fact, that is the definition of "subspace topology" ... so for any topology on A other than the subspace topology, the inclusion i is NOT an embedding. Either because i is not continuous, or because i is not open, or both. "

    Thanks, Edgar. I thought the subspace topology was the initial topology with respect
    to inclusion, and, given an inclusion:

    i:A-->X

    the initial topology was (is) , by def., the smallest topology on A, which makes the
    inclusion continuous (largest topology being the discrete one, 2A), but
    I don't see how it follows that the subspace topology is the only one for which i is
    ( I am.? :) ) an embedding:


    We have A< X a strict subspace ( i.e, A is not all of X ). Given the map : i:A-->X

    We want to define a topology on A such that :

    1) Continuity of i: For U open in X , i-1(U) open in A.


    We have that i-1(U) =U/\A . So we must have U/\A open in A.

    Then any topology TA on A must contain the subspace topology.

    i.e., TA> (A, subspace) (with > meaning contains)


    2) Openness of map i: For V open in A, i(V)=V is open in X . Like you said,
    it follows that A must be an open subspace of X, i.e., A is open as a subset
    of X.

    How does it then follow that the only topology that makes i:A-->X
    into an embedding is (A, subspace) .?. I don't see how we can conclude,
    e.g.,
    (A, subspace)> TA
     
  6. May 24, 2010 #5

    Landau

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    Science Advisor

    Instead of `No. In fact, that is the definition of "subspace topology"', I think g_edgar should have written `No. In fact, that is the definition of "topological embedding"'.

    Let X,Y be topological spaces. A function f:X->Y is an embedding if
    * f is injective, and
    * f is a homeomorphism onto its image f(X), where f(X) carries the subspace topology of X.

    So the requirement of "subspace topology" is part of the definition of embedding.
     
  7. May 25, 2010 #6
    Doesn't your second *, that f is a homeomorphism onto its image f(X), automatically mean that f is injective, your first *?
     
  8. May 25, 2010 #7

    Landau

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    Science Advisor

    Yes, you are right, the first * is superfluous. So, in short, an embedding is a homemomorphism onto its image.
     
  9. May 25, 2010 #8
    That makes really good intuitive sense. But for some reason the book I have defines embedding in a slightly esoteric way. Instead of topological spaces, it talks about embedding of differentiable manifolds.

    First it defines an immersion. Basically, a smooth map f: M -> N between manifolds induces a map called the "differential map" f* between the vector spaces of the two manifolds:

    f*: T(M) -> T(N) (explicitly, this map happens to be the Jacobian).

    If f* is an injection, then f is said to be an immersion.

    In addition, if f is an injection, then f is said to be an embedding.

    This definition of an embedding seems weird to me, not as nice as your definition. Is there a reason why one would define an embedding in this way?
     
  10. May 25, 2010 #9

    Landau

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    Science Advisor

    Remember, we were talking about topological spaces, i.e. spaces with as additional structure (only) a topology. My definition was of a topological embedding, which is just called an embedding if the context (namely, topology only) should be clear.

    In your book it is about (smooth) manifolds: those are first of all topological spaces, but they also have extra structure, namely a smooth structure (atlas). An embedding between smooth manifolds, let us call it a smooth embedding to be sure, also has to deal with this smooth structure, and that is the immersion part.

    So, a smooth embedding is a topological embedding which is at the same time an immersion (= injective differential).
     
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