dg.differential geometry - Two discs with no parallel tangent planes

If I understood your question correctly, I think that the answer is no. In fact, even more is true: if you choose any identification $varphi colon Sigma_1 to Sigma_2$ in such a way that the boundary are compatibly identified, then for any embedding of $Sigma_1$ and $Sigma_2$ there is a point $p$ of $Sigma_1$ such that the tangent plane to $Sigma_1$ is the same as the tangent plane to $Sigma_2$ at $varphi(p)$. To see this, let $mathbb{R}P^2$ denote the real projective plane, the space parameterizing linear subspaces of dimension one in $mathbb{R}^3$. Suppose by contradiction that you found embeddings with the mentioned property. Let $gamma colon Sigma_1 cup Sigma_2 to mathbb{R}P^2$ be the function defined by sending $p$ to the linear subspace spanned by $N_p wedge N_{varphi(p)}$, where $N_p$ is the normal direction to the image of $p$ and similarly for $N_{varphi(p)}$, and the wedge product is the usual wedge product in $mathbb{R}^3$. The function $varphi$ is indeed a function, since never the normal directions at $p$ and $varphi(p)$ are parallel. But observe that $gamma(p)$ is always a vector contained in the tangent plane at $p$. Thus, $gamma$ determines a global vector field on the image of $Sigma_1 cup Sigma_2$ that is never zero. This is obviously impossible!

Note that the above argument is essentially the one used in the Borsuk-Ulam Theorem (http://en.wikipedia.org/wiki/Borsuk–Ulam_theorem).

• Next amateur observing - Can you see city lights on the Moon from Earth?
• Previous soft question - Can you prove equivalence without being able to calculate it?