From: Dave Rusin
Subject: Re: Embeddings
Date: Tue, 15 Jun 1999 01:45:52 -0500 (CDT)
Newsgroups: [missing]
To: dgiaimo@ix.netcom.com
Keywords: knots and pairs of spaces (category TOP^2 )
>> There is an old question in knot theory about whether two embeddings f,g
>of
>> a manifold M into R^n having homeomorphic complements are necessarily
>> the same knot (i.e. are the pairs (R^n,f(M)) and (R^n,g(M)) homeomorphic).
>
> Can you please explain what you mean by the pairs being homeomorphic?
That means "isomorphic" in the category Top^2 of pairs of spaces:
there should exist morphisms F, G with FoG=Id, GoF=Id. Here you just
need to know that the objects in this category are the pairs (X,Y) with X
a top. space and Y a subspace of X; the morphisms are the maps
F:X->X' for which F(Y) is contained in Y'.
If (X,Y) and (X',Y') are homeomorphic, it follows in particular that
X-Y and X'-Y' are homeomorphic spaces, as are X and X', and Y and Y'.
You're essentially asking about the converse: you have pairs with X=X'
and Y=Y' (up to homeomorphism), and you want to know whether assuming
X-Y=X'-Y' is enough to give homeomorphisms of pairs (x,Y) and (X',Y').
That's true in some "knot-like" situations, but certainly not "trivial"
since it's not true for every pair of objects in Top^2. For example,
take X=X'=R^2,
Y=x-axis union {-1}x[0,1] union {1}x[0,+1],
Y'=x-axis union {-1}x[0,1] union {1}x[-1,0]
Then Y and Y' are homeomorphic, as are X-Y and X'-Y' (slick proof: they're
simply-connected proper open subsets of the complex plane, hence _complex-
analytic_ equivalent to the unit disc and thus to each other !)
But given any homeomorphism F : R^2 -> R^2 taking Y to Y' it's easy
to show F({ (-1,1), (1,1) }) = { (-1,1), (1,-1) } from which we see
F( [-1,1]x{1} ) has to cross the horizontal axis, violating the
injectivity of F.
dave
==============================================================================
From: "Daniel Giaimo"
Subject: Re: Embeddings
Date: Tue, 15 Jun 1999 19:08:50 -0700
Newsgroups: [missing]
To: "Dave Rusin"
----- Original Message -----
>From: Dave Rusin
>To:
>Cc:
>Sent: Monday, June 14, 1999 11:45 PM
>Subject: Re: Embeddings
>
>For example,
> take X=X'=R^2,
> Y=x-axis union {-1}x[0,1] union {1}x[0,+1],
> Y'=x-axis union {-1}x[0,1] union {1}x[-1,0]
> Then Y and Y' are homeomorphic, as are X-Y and X'-Y' (slick proof: they're
> simply-connected proper open subsets of the complex plane,
Aren't X-Y and X'-Y' not connected? Then how can they be simply
connected?
> hence _complex-
> analytic_ equivalent to the unit disc and thus to each other !)
> But given any homeomorphism F : R^2 -> R^2 taking Y to Y' it's easy
> to show F({ (-1,1), (1,1) }) = { (-1,1), (1,-1) } from which we see
> F( [-1,1]x{1} ) has to cross the horizontal axis, violating the
> injectivity of F.
>
> dave
>
--Daniel Giaimo
Remove nospam. from my address to e-mail me. | dgiaimo@(nospam.)ix.netcom.com
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