Inspired by this question I attempted to pull back the 2-form $dt\wedge ds \in \bigwedge^2(\Bbb R^2)$ using the product of the stereographic projections. I'd like to have my calculations checked, for I have the faintest idea of what I'm doing here.
I'll just report the pull back by $\phi:S^1\times S^1 \setminus \{(0,1,0,1)\} \rightarrow \Bbb R^2$ with $$\phi(x,y,z,w) = \left(\frac{x}{1-y},\frac{z}{1-w}\right)$$ Now, knowing that $\phi^*$ commutes with the wedge product and $d$ I got: $$\phi^*(dt\wedge ds) = d\left(\frac{x}{1-y}\right)\wedge d\left(\frac{z}{1-w}\right) = \left(\frac{dx}{1-y} + \frac{xdy}{(1-y)^2}\right)\wedge \left(\frac{dz}{1-w} + \frac{zdw}{(1-w)^2}\right)$$ then expanding everything by linearity of $\wedge$. This form is non vanishing on $S^1\times S^1$ because $dt\wedge ds$ is non vanishing on $\Bbb R^2$ right?