# Rational homotopy type of a complement

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.

• 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