Linking number and cup product Let $S^p$ and $S^q$ be disjoint spheres in $\mathbb{R}^n$ with $n=p+q+1$ and let $X= \mathbb{R}^n- (S^p\cup S^q)$. By Alexander duality, their fundamental classes yield cohomology classes in $\tilde{H}^p(X)$ and $\tilde{H}^q(X)$, respectively, say $\alpha^p$ and $\alpha^q$.
The cup product $\alpha^p \cup \alpha^q$ lies in $\tilde{H}^{p+q}(X)\cong \tilde{H}_0(S^p \cup S^q) \cong \mathbb{Z}$, so there's an integer assigned to this cup product.
Now, consider the case $p=q=1, n=3$.
Can I and if yes, how can I show that this integer is the linking number of the spheres defined by the "usual" counting of signs at the crossings?
Can this be generalized to a linking number in higher dimensions?
Edit:
Thanks for your answer,  Aloizio Macedo.
I think saying that Alexander duality is given by capping with an orientation class is not totally correct in this case since I need compact manifolds with boundary for Lefschetz duality. But I still tried to use your ideas for the case $p=q=1, n=3$ using $S^3$ instead of $\mathbb{R}^3$:
Call the spheres $S_1, S_2$.
Let $\beta_1 = \alpha^1 \cap [S^3]$, $\beta_2 = \alpha^2 \cap [S^3] \in H_2(S^3,S_1\cup S_2)$. 
(Then $[S_i] = \partial(\beta_i) \in H_1(S_1\cup S_1)$ for $i=1,2$.)
Let $D_1$ be the inverse of $\cdot \cap [S^3]$.
We have 
$\alpha^1 \cup \alpha^2 = D_1(\beta_1) \cup D_1(\beta_2) = D_1(\beta_2 \circ \beta_1)$, where $\circ$ is the intersection product from Bredon.
Now, if $\beta_i=[N_i]$ for submanifolds $N_i$, $i=1,2$, I get
$\alpha^1\cup \alpha^2 = D_1(\beta_2 \circ \beta_1) = D_1([N_1]\circ [N_2])=D_1([N_1 \cap N_2])$.
But $N_i$ would have to be two-dimensional such that $N_1 \cap N_2$ is $1$-dimensional and I'm not quite sure how to get a linking number from here... (An intersection of a Seifert surface and a knot usually is $0$-dimensional, a set of points, right?)
 A: The OP has already worked out an answer, but I wanted to provide how I think about this. Let $L$ be the link, made up of circles $\alpha$ and $\beta$.
We have an Alexander duality map $D: H^1(S^3\setminus L)\rightarrow H_1(L)$, given by $\gamma\mapsto\partial(\omega\frown\gamma)$, where $\omega$ is a fundamental class. This has an inverse, $D^{-1}: H_1(L)\rightarrow H^1(S^3\setminus L)$, which can be described as follows: for any 1-cycle $\sigma\subset S^3\setminus L$, it is the boundary of a 2-cycle in $S^3$, which we denote by $\partial^{-1}\sigma$. Then
$$ D^{-1}(\tau)(\sigma) = [\tau\cap\partial^{-1}\sigma]\in\mathbb{Z} $$
where $[\tau\cap\partial^{-1}\sigma]$ is just the oriented intersection. I don't know a reference for this.  @CheerfulParsnip mentions this in his answer here; maybe he knows a source?
Anyway, from here it is easy. We have
\begin{align*}
D\left(D^{-1}(\alpha)\smile D^{-1}(\beta)\right) &= \partial(\omega \frown(D^{-1}(\alpha)\smile D^{-1}(\beta)))\\
&= \partial(\omega\frown D^{-1}(\alpha))\frown D^{-1}(\beta)\\
&= \alpha\frown D^{-1}(\beta)\\
&= D^{-1}(\beta)(\alpha)\\
&= [\beta\cap\partial^{-1}\alpha]
\end{align*}
and this is just the intersection of $\beta$ with the Seifert surface of $\alpha$.  If we choose orientations right, this is the linking number of $\alpha$ and $\beta$.
