Maps of Wedges being Nullhomotopic 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.  
 A: We can answer this without anything fancier than cellular approximation. As user17786 points out, if the dimension of $X \vee X'$ is smaller than the dimension of the sphere $S^{n-1}$, then we win because we approximate the map $X \vee X' \rightarrow S^{n-1}$ by a cellular map, which lands in the $(n-2)$-skeleton of $S^{n-1}$, which is just the basepoint.
Moreover, we can suspend $X$ and $X'$ as many times as we like. If $X$ embeds into $S^n$ and its complement is homotopy equivalent to $X'$, then $\Sigma^l X$ embeds into $S^{n+k+l}$ and its complement is homotopy equivalent to $\Sigma^k X'$.
So, let $d$ be the dimension of the finite CW complex $X$ and let $d'$ be the dimension of $X'$. Then $\Sigma^k X \vee \Sigma^k X'$ has dimension max$(d,d')+k$, whereas $S^{n-1+k+k}$ is $(n+2k-2)$-connected. If $k$ is large then the second number is bigger than the first, and we win.
Now, can we answer the question without suspending $X$? Yes, if we allow ourselves to use Alexander duality, which tells us that $\tilde H^q(X) \cong \tilde H_{n-q-1}(X')$ and vice-versa. If $X$ is connected, then $\tilde H^0(X) = 0$ and $\tilde H_0(X) = 0$, so by Alexander duality, $H_{n-1}(X') = 0$ and $H^{n-1}(X') = 0$. (The higher homology and cohomology vanish as well.) Therefore $X'$ is homotopy equivalent to a complex of dimension at most $n-2$. (As the other user points out, this is proven by Hatcher in the section on minimal cell structures.) Anyway, if $X'$ is also connected, then the same goes for $X$, so $X \vee X'$ has dimension at most $n-2$ and we win again.
EDIT: This answer is a couple of years old but I need to correct an oversight made by past me. In this last paragraph I assumed $X$ and $X'$ are simply-connected, which is easy enough to achieve by suspending each one twice. If they are not simply-connected then there is an extra obstruction to being homotopy equivalent to a finite-dimensional complex: $H^n(X;G)$ must be zero for all twisted coefficient systems $G$. (See Wall's paper Finiteness conditions for CW-complexes.) I don't believe Alexander duality gives us anything with twisted coefficients - another good reason to just suspend and carry on with your life.
A: I guess we can suppose that all the spaces are CW-complexes, hence finite, being subsets of a sphere.
If a CW-complex $X$ (your wedge) has dimension $\leq n-2$, then all maps to a $(n-2)$-connected $CW$-complex $Y$ (the sphere $S^{n-1}$) are homotopic to a constant (because $Y$ is homotopy equivalent to a complex $Z$ whose first nontrivial cell appears in dimension $n-1$, and then we can use the $CW$-approximation theorem for maps to deduce that all maps from $X$ to $Z$ map all cells to the trivial $0$-cell).
If $H^q(X)=0$ for all $q\geq n-1$, then by the universal coefficient theorem for cohomology (Theorem 3.2 in Hatcher's), $H_q(X)=0$ for all $q\geq n-1$ and $H_{n-2}(X)$ is a free abelian group. Now, by proposition 4C.1 in Hatcher's (on minimal cell structures), $X$ is homotopy equivalent to a $CW$-complex $Z$ of dimension at most $n-1$. Moreover, if $Z$ has a cell in dimension $n-1$ it is a 'relator', as explained there, and creates a torsion subgroup in $H_{n-2}(X)$, but this group was free, so $Z$ has dimension $\leq n-2$, and we can apply the first paragraph.
EDIT: (I summarize the discussion held in the comments) The above solution fails as $\pi_1(X)$ is required to be trivial for applying Proposition 4C.1. For assuring $X$ is simply connected, we have to suspend $X$ and change the dimension of the sphere where $X$ lives. Having done that, Cary's answer is simpler than this one. If we want to avoid suspending, i.e. we want to prove that ''for a space $X$ such that $H^q(X)=0$ for all $q\geq n$, the set of homotopy classes $[X,S^{n-1}]$ is trivial", I think it is better to work with the Postnikov tower of $S^{n-1}$. In this approach, it would be a direct consequence of Corollary 4.73 in Hatcher's: If $Y$ is a connected abelian $CW$-complex and $(W,X)$ is a $CW$-pair such that $H^{q+1}(W,X;\pi_q X)$ for all $q$, then every map $X\rightarrow Y$ can be extended to a map $W\rightarrow Y$. Take $W$ to be the cone of $X$ and $Y=S^{n-1}$, and note that $H^q(X)=0$ for all $q\geq n-1$ implies that $H^q(X;G)=0$ for all $q\geq n-1$ and for any (finitely generated) abelian group $G$.
