TITLE: Rational homotopy type of a complement
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: Some of the best results known along these lines are in the paper

Pascal Lambrechts, Don Stanley, Algebraic models of Poincaré embeddings, Algebr. Geom. Topol. 5 (2005) 135-182, doi:10.2140/agt.2005.5.135, arXiv:math/0503605,

which used to be called "Algebraic models of complements...".
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.