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Let $X$ and $X'$ be smooth closed manifolds. Take closed subpolyhedra $D\subset X$ and $D'\subset X'$ (with respect to some triangulations) and let $f:X\to X'$ be a homotopy equivalence such that $f(D)=D'$ and the restriction of $f$ to $D$ is also a homotopy equivalence. Is it possible for the complements $X-D$ and $X'-D'$ to have different rational homotopy types, assuming all spaces ($X,X',D,D'X-D,X'-D'$) simply-connected?

Here is some motivation behind the question: if we replace the rational homotopy type with the integral one and do not require the spaces to be simply connected, then the answer is yes, as shown in a paper by R. Longoni and P. Salvatore http://arxiv.org/abs/math/0401075; a much simpler example is in Ryan's comment below. On the other hand, additively the cohomology of $X-D$ and $X'-D'$ is obviously the same.

upd: in the first version of the question the simple connectedness condition was missing. Apologies for the mix-up.

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There is a homotopy equivalence $(S^3,K_1) \to (S^3,K_2)$ with $K_1$ the unknot, and $K_2$ the trefoil. –  Ryan Budney Apr 16 '10 at 18:20
Thanks, indeed! What if we assume everything simply connected? –  algori Apr 16 '10 at 18:57
I think Salvatore ?might? have a result analogous to his paper with Longoni using simply-connected manifolds instead of lens spaces. The argument proceeds much the same -- the configuration spaces aren't homotopy-equivalent even though the underlying manifolds are. I'll ask him about it in person next week. I'll visit Longoni as well. I don't think Salvatore has written up that paper yet. Longoni is a banker in Milano. –  Ryan Budney Apr 16 '10 at 19:47
@Ryan - +1 for the "Longoni is a banker in Milano" statement! –  Somnath Basu Apr 16 '10 at 20:19
@Paul: The map f does not necessarily induce a map between the complements. –  Oscar Randal-Williams Apr 17 '10 at 2:14

1 Answer 1

Some of the best results known along these lines are in the paper "Algebraic models of Poincare embeddings" (which used to be called "Algebraic models of complements...") by Lambrechts and Stanley. See: http://front.math.ucdavis.edu/0503.5605

Briefly, it seems that even if the complement is simply connected, there are examples of what you seek when the codimension is too small, and when the codimension is large enough the rational homotopy type of a complement is determined.

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Thanks, Dev! I am aware of that paper, but I do not know whether in the examples given in section 9 there one can construct a homotopy equivalence of couples. –  algori Apr 17 '10 at 19:28

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