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 ^^^^^^^^^<-(Remove) |--------BEGIN GEEK CODE BLOCK--------| Ros: I don't believe in it anyway. |Version: 3.1 | |GM d-() s+:+++ a--- C++ UIA P+>++++ | Guil: What? |L E--- W+ N++ o? K w>--- !O M-- V-- | |PS? PE? Y PGP- t+(*) 5 X+ R- tv+(-) | Ros: England. |b+@ DI++++ D--- G e(*)>++++ h->++ !r | |!y->+++ | Guil: Just a conspiracy of |---------END GEEK CODE BLOCK---------| cartographers, you mean?