Stably equivalent but not homotopy equivalent What are some examples of (compact, say) manifolds $X$ and $Y$ that are stably equivalent, i.e. $\Sigma^{\infty}X_+\simeq\Sigma^{\infty}Y_+$, but are not homotopy equivalent?
 A: The easiest examples are given by complements of open neighborhoods of distinct knots in $S^n$.
More generally, Spanier-Whitehead duality says that if $X$ is a compact simplicial complex and if $f,g:X \rightarrow S^n$ are embeddings, then $S^n - Im(f)$ is stabily equivalent to $S^n - Im(g)$.
See my notes here for an elementary discussion.
A: Maybe it is worth adding some simply-connected examples.
Every simply connected closed 4-manifold may be described as $X = D^4 \cup_f (S^2 \vee \cdots \vee S^2)$, where $f$ is a map $S^3 \to S^2 \vee \cdots \vee S^2$; $\pi_3$ of this wedge is known to be generated by Hopf maps and Whitehead products of two factors, so we may represent such a map by a symmetric integer matrix $A$; this integer matrix may just as well be interpreted as the cup product pairing $H^2(X;\Bbb Z) \otimes H^2(X;\Bbb Z) \to H^4(X;\Bbb Z) = \Bbb Z$. The resulting homotopy type is determined up to isomorphism by $A$ up to isomorphism of symmetric bilinear forms over $\Bbb Z$. See, for instance, Example 4.52 in Hatcher's algebraic topology book.
On the other hand, if $n > 2$, then $$\pi_{n+1}(\vee^k S^n) = (\pi_{n+1} S^n)^k = (\Bbb Z/2)^k.$$ The Whitehead product factors vanish. This follows by an inductive argument using the Hilton-Milnor theorem, as stated in this answer.
Furthermore, any map in $GL_k(\Bbb Z)$ may be realized as an automorphism of $H_n(\vee^k S^n)$ by some autoequivalence of $\vee^k S^n$, and the map $GL_k(\Bbb Z) \to GL_k(\Bbb Z/2)$ is surjective (check at the level of the generating set of elementary matrices). Because every two nonzero vectors in $(\Bbb Z/2)^k$ are related by some matrix in $GL_k(\Bbb Z/2)$, the homotopy type of $D^{n+2} \cup_f (\vee^k S^n)$ is determined entirely by whether or not $f$ is nontrivial.
Suspending the presentation given for $X$ takes the diagonals of the matrix $A$ mod 2, and so the homotopy type of the suspension is dictated by whether or not $X$ was spin (that is, whether or not its intersection form was even). If $X$ was spin, then $$\Sigma X \simeq S^5 \vee^{b_2} S^3;$$ if $X$ was not spin, then $$\Sigma X \simeq \Sigma \Bbb{CP}^2 \vee^{b_2-1} S^3.$$ In particular, the stable homotopy type is determined entirely by "even-ness" and rank of the intersection form.
