A generalization of Cauchy's mean value theorem. The following simple theorem is known as Cauchy's mean value theorem. Let $\gamma$ be an immersion of the segment $[0,1]$ into the plane such that $\gamma(0) \ne \gamma(1)$. Then there exists a point such that the tangent line at that point is parallel to the line passing through $\gamma(0)$ and $\gamma(1)$. So the boundary values of an immersion determine a direction such that for any immersion of a segment with given boundary values there exists a tangent line parallel to the direction.
I would like to propose the following conjecture, generalizing this statement. Instead of immersions of a segment we consider immersions of a compact manifold $M^n$ with non-empty boundary $\partial M$ into ${\mathbf R}^{n+1}$. For a map $f\colon \partial M \to {\mathbf R}^{n+1}$ we consider the space $L(f)$ of all immersions $g\colon M \to {\mathbf R}^{n+1}$ such that $g|_{\partial M}=f$.
The conjectural claim is the following: If $f$ is sufficiently generic, then for every connected component $L_0$ of the space $L(f)$ there exists a hyperplane direction $l$ such that for any immersion $g$ from $L_0$ $l$ is parallel to the tangent plane to $g(M)$ at some point.
If the conjecture is true then it is very interesting how $l$ depends on $g$.
 A: The answer is no for $n=2$. It is sufficient to construct 3 surfaces with common boundary (say $\Sigma_i$, $i\in\{1,2,3\}$)
such that there is no choice of points $p_i\in\Sigma_i$ with pairwise parallel tangent planes.
Let us take a smooth function $f:S^1\to \mathbb R$, $f(t)\approx\sin(2\cdot t)$ with one little bump near zero
so it has 3 local minima and maxima.
We want to construct three functions $h_1,h_2,h_3$ from unit disc $D$ to $\mathbb R$ such that
each has $f$ as boundary values and


*

*if $\nabla h_1(x)=\nabla h_2(y)$ then  $\nabla h_1(x)=0$

*$\nabla h_3\not=0$ anywhere in the disc.
Then graphs of functions give the needed surfaces.
The graphs of $h_1$ and $h_2$ are parts of boundary of 
convex hull of graph of $f:\partial D\to\mathbb R$; it is easy to check (1).
The graph of $h_3$ is a ruled surface which formed by lines passing 
through points $(u,f(u))$, $(\phi(u),f(\phi(u))\in\mathbb R^3$, $u\in S^1$ for some involution diffeomorphism $\phi: S^1\to S^1$.
To have the property one has to choose  $\phi$ with two fixed points (say at global minima of $f$)
so that if $f(\phi(x))=f(x)$ for some $x$ then $f'(\phi(x)\cdot f'(x)<0$.
The later is easy to arrange, that is the place we need the bump of $f$.
P.S. Hopefully it is correct now :)
A: I will consider another counterexample, based on Anton's ideas. I think it looks simpler.
We will construct a closed domain $D$ in a plane and three functions on it with same restrictions to the boundary of the domain such that there is no choice of points on graphs of functions with pairwise parallel tangent planes.
First observation is the following: Let function $f$ on the coordinate plane depends on the coordinate $x$ only and function $g$ depends on the coordinate $y$ only. Then $df(a)=dg(b)$ if and only if $df(a)=dg(b)=0$. (It seems to me that the crucial Anton's idea is to consider functions with very degenerate gradient map, having value on a curve).
Now we take the functions $f(x,y)=-x^2$ and $g(x,y)=P_4(y)$ where $P_4$ is a polynomial of degree $4$ with two zeroes, three pairwise different critical values and leading term $y^4$.
We set the domain $D$ to be a set of solutions to an inequality $f \ge h$. It is a closed subset in a plane diffeomorphic to a unit disk. We define functions now: the first function is $f$, the second function is $g$. Third function is a function without critical points on $D$ coinciding with a restriction of $f$ (or $g$) to the boundary. Such a function exists! (it is an exercise from Morse theory. I mention here that we need a polynomial of degree 4 (degree 2 is unsufficient) to satisfy that extension without critical points property).
