From: Dave Rusin <rusin@math.niu.edu>
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" <dgiaimo@ix.netcom.com>
Subject: Re: Embeddings
Date: Tue, 15 Jun 1999 19:08:50 -0700
Newsgroups: [missing]
To: "Dave Rusin" <rusin@math.niu.edu>

----- Original Message -----
>From: Dave Rusin <rusin@math.niu.edu>
>To: <dgiaimo@ix.netcom.com>
>Cc: <rusin@math.niu.edu>
>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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