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.