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Mapping Torus of a Manifold is a Manifold.

  1. Nov 21, 2012 #1


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    Hi, All:

    I'm trying to show that the Mapping torus of a manifold X is a manifold, and I'm trying to see what happens when X has a non-empty boundary B.

    Remember that the mapping torus M(h) of a space X by the map h is constructed like this:

    We start with a homeomorphism h:X-->X (we can add conditions to h like, say,
    respect a possible PL structure, or for h to be a diffeomorphism, etc.) , then we
    do the quotient on X x I ; I=[0,1]:

    X/~ : (x,0)~ (h(x),1 )

    i.e., we glue the top- and bottom levels about the homeomorphism h.

    In the boundaryless case, the possible trouble points are those that are identified,
    since XxI is itself a manifold. Some trivial examples-- h is the identity -- are manifolds,
    but I cannot see clearly the case for general h. Say (x,0)~(x',1) . Then there are
    charts for x, x' respectively in X:

    For x: (Ux, Phi_x) , and (Ux', Phi_x' ) for x'

    How do we get a chart for the identified point (x,0)~(x',1) in X/~ ?

    The case where X has boundary I suspect, the identification process has the effect
    of capping the boundary. If X has boundary B, then h: X-->X takes B to B.

    But I can't think of how to make this more rigorous. Any ideas?
  2. jcsd
  3. Nov 22, 2012 #2


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    i suggest you go back an work the exercise i recommended for your question on quotient spaces, as you seem to need practice working with homeomorphisms. you might even benefit from working through a more elementary book on topology like munkres or mendelson or gemignani.
  4. Nov 22, 2012 #3


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    Is this such a trivial result? I've asked around in my school and some people knew
    the answer was yes, but could not give me an answer why. I think you just decided
    at some point that I'm not too sharp instead. Moreover, no one had an actual answer
    for my previous. I'm not questioning their/your knowledge nor talent, I just don't
    think the result is as trivial as you believe it is. It is not extremely hard either; but
    I don't think it is trivial either.
  5. Nov 22, 2012 #4


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    I'm not saying you are not sharp. I am suggesting you have not played around enough with these ideas. But ask yourself, did you try the exercise I just gave, or are you just in the habit of asking your friends, or looking in your book? get your pencil working.
    Here are some suggestions for practice:

    In all mathematics involving mappings, especially therefore geometry and algebra, it is important to practice mappings along with spaces.

    In particular every construction of a new space from old ones, also involves constructing new mappings from old ones. Whenever one learns a new construction for spaces, and hence also mappings, one should learn the relationship between the new mappings and the old.

    E.g. a product of two spaces X,Y is a new space XxY, but also it comes with a pair of maps XxY-->X and XxY-->Y such that a function Z-->XxY is admissible (e.g. continuous) as a map if and only if both composed functions Z-->XxY-->X and Z-->XxY-->Y are admissible mappings. Moreover every pair of continuous mappings Z-->X and Z-->Y induces a unique continuous map Z-->XxY, with compositions Z-->XxY-->X and Z-->XxY-->Y equal to the original pair of maps.

    This is the characteristic property of the product topology.

    As for a quotient space of a space X by a subspace A, the topology on the space X/A is defined as all sets whose preimages under the natural projection X-->X/A are open in X. This is the largest topology on X/A making that projection continuous.

    Then the key to understanding maps out of the quotient space X/A is this: a function X/A-->Z is continuous if and only if the composition X-->X/A-->Z is continuous.

    As a consequence we learn to form continuous maps out of X/A as follows:

    Any function from X-->Z that is continuous and which sends all points of A to a single point of Z, induces a continuous map X/A-->Z, (and conversely). check this, and all other claims made here. (i am not working out the proofs as i go except in my head, so i could make a mistake here, but there is a pretty fair probability i am getting it mostly correct. but the only way to be sure is to do the proof yourself.)

    As a consequence any continuous map X-->Y such that the subspace A of X maps into the subspace B of Y, induces a continuous map X/A-->Y/B.

    So what is a homeomorphism? It is a continuous map with a continuous inverse. Suppose X-->Y is a homeomorphism mapping the subspace A homeomorphically to the subspace B. Then also the inverse g of f maps Y-->X homeomorphically to X and maps the subspace B to A. hence we get two induced continuous mappings X/A-->Y/B which you should check are also inverse to each other.

    In particular this answers lavinia’s conjecture in the affirmative, i.e. under these conditions, X/A is homeomorphic to Y/B. Since the result follows immediately from the basic definitions and properties of quotients, it might be called elementary, but i will not call it trivial, because to become so first we have to master the concepts, which takes some work. also one has to be taught properly. the word trivial is used mostly as an insult in my experience.

    Here are some more exercises on quotient spaces. Check that in R/Q, the quotient of the reals by the rationals, that the point [0] is dense. I.e. the only closed subset of R/Q containing [0] is the whole space. or prove otherwise.

    in the same spirit, if ≈ is an equivalence relation on X, how are maps out of X/≈ related to maps out of X? Is a function X/≈-->Z continuous whenever the composition X-->X/≈-->Z continuous?

    Does a map out of X-->Z induce a continuous map out of X/≈ whenever it send equivalent points of X to the same point of Z?

    Can you use this to describe a map out of a nbhd of a point on the problematic set of a mapping torus, that defines a chart?
    Last edited: Nov 22, 2012
  6. Nov 22, 2012 #5


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    Thanks, mathwonk, I'll work out your suggestions.
  7. Nov 22, 2012 #6


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    enjoy! the point of view i am giving here is called the "functorial" point of view, and is the modern approach, not always emphasized in classical books. it makes many things rather easy, but does not solve every problem. if we want to use the tricky word trivial, the functorial point of view does render some types of questions "trivial", or at least plausibly clear.

    e.g. if a construction is functorial (i.e. if it takes spaces to spaces and maps to maps, and preserves identities and compositions), then it always takes isomorphisms to isomorphisms. so here is quick argument (with huge gaps) affirming lavinia's conjecture:

    The quotient construction is a functor from pairs of spaces (X,A) and maps of pairs (X,A)-->(Y,B) to single spaces X/A and maps of those spaces X/A-->Y/B.

    Therefore, since a homeomorphism of pairs (X,A)-->(Y,B) is a homeomorphism of X that carries A homeomorphically onto B, such a map also induces a homeomorphism X/A-->Y/B QED.

    So to a person who thinks functorially, the conjecture is in some sense quite clear.

    the torus case is harder for me to see, but maybe post thanksgiving it will become clearer. but i think it is just a matter of showing that if f:U-->V is a homeomorphism, then (-1/2,1/2) x U is homeomorphic to (-1/2,0]xU + [0,1/2)xV /{ (0,x) ≈ (0,f(x)) }.

    what do you think??? well i haven't made it clear enough to answer.
    Last edited: Nov 22, 2012
  8. Nov 30, 2012 #7


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    The counterexample I was thinking about is this:

    A= two homologically-trivial loops in T^2 , e.g., loops that contain neither meridians nor


    B=two homologically non-trivial loops in T^2. e.g., two meridians.

    Then T^2/A ~ T^2 , but

    T^2/B ~ / T^2 , since we can disconnect T^2/B by removing two points, but

    we cannot do the same for T^2~ T^2/A .
  9. Dec 1, 2012 #8


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    Right, the issue of maps descending is helpful, in that a continuous map in the
    "top space"X , may descend into a continuous map on the quotient space X/~.

    Still, if we want to show that the quotient is a differentiable manifold, (at least C^1)
    then we want a differentiable map X-->R^k to descend into a differentiable map
    X/~ -->R^k . But I don't think differentiable maps descend into differentiable maps,
    since quotients sometimes don't even preserve Hausdorff. So, how do we define a
    differentiable map in a space that is not even Hausdorff? Let me check what we
    need as conditions to guarantee that the quotient of Hausdorff is itself Hausdorff.
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