In Spanier's "Infinite Symmetric Products, Function Spaces and Duality," he makes the following claim:
Given some $X\hookrightarrow S^n$, and X' which is an "$n$-dual" of $X$ (i.e. for some $k$, and all larger k', \Sigma^kX'\simeq\Sigma^k(S^n\setminus X)\simeq S^{n+k}\setminus X, where the first equivalence is along a deformation retract), we wish to find a map X\wedge X'\to S^{n-1}.
To do so, we remove a point from S^n\setminus(X\cup X') and so have X,X'\hookrightarrow S^n\setminus\mathrm{pt.}\cong \mathbb{R}^{n}. We define v:X\times X'\to S^{n-1},~(x,x')\mapsto\frac{x-x'}{\vert\vert x-x'\vert\vert}.
Spanier states that this map restricted to X\vee X' is nullhomotopic under the condition that $X$ and X' are connected (and the situation is in fact such that we only need to choose $X$ connected and we can get X' to be connected by suspensions, by the first three part homotopy equivalence given). Spanier claims that this is related to the fact that H^q(X\vee X')=0 for every $q\geq n-1$.
The map on the smash product comes from shrinking the wedge to a point, obviously.
Thanks for any help on this, it's sort of a complicated, classical problem.