Uniqueness of tangent plane Let $\Sigma$ be a smooth surface defined as a surface admitting a parametrisation $\boldsymbol{r}:D\subset\mathbb{R}^2\to\mathbb{R}^3$ such that $\boldsymbol{r}$ is of class $C^1(\mathring{D})$ (and continuous on the domain: $\boldsymbol{r}\in C(D)$) and that $\forall(u,v)\in \mathring{D}$ $\|\boldsymbol{r}_u(u,v)\times\boldsymbol{r}_v(u,v)\|\ne 0$.
I wonder whether the tangent plane in a point $\boldsymbol{r}(u_0,v_0)$ of the surface is unique, as expected from graphical considerations, where tangent means orthogonal to the normal unitary vector$$\boldsymbol{N}(u_0,v_0)=\frac{\boldsymbol{r}_u(u_0,v_0)\times\boldsymbol{r}_v(u_0,v_0)}{\|\boldsymbol{r}_u(u_0,v_0)\times\boldsymbol{r}_v(u_0,v_0)\|}$$or equivalently, whether the direction of $\boldsymbol{r}_u(u_0,v_0)\times\boldsymbol{r}_v(u_0,v_0)$ is unique for any parametrisation $\boldsymbol{r}$ satisfying the above said smoothness definition. I cannot find any information in my books nor on line. Could anybody help me with a proof, if the uniqueness holds, as easy as possible or tell me that $\boldsymbol{N}$ is not unique if it is not? I $\infty$-ly thank you for any answer!
 A: You have to distinguish between the image of your map $r$ and its parametrization.
Since your definition of surface does not include the requirement of injectivity of $r$ the image set may have self intersections, and whereever that happens, the image may have several tangent planes, in particular if the self intersectios are transversal. 
Apart from that, at an interior point $(u,v)$ where $r$ is $C^1$ and regular (i.e. the condition you imposed on the norm of $n$, $||n||\neq 0$ holds, or equivalently, where the differential of $r$ has maximal rank) the tangent plane is locally unique as a two dimensional plane in $\mathbb{R}^3$ which can be thought of as a plane attached to $r(u,v)$ and can be defined as a plane normal to the normal vector. Of course there is no unique parametrization for it.
In differential geometry the definition of tangent plane is often more abstract (it is defined as a so called vector bundle), because it then allows to define the tangent space of abstract manifold (which need not be embedded or immersed into some ambient space) but the general idea is the same and the picture you have in mind is correct. 
Edit: The normal $n$ you defined is only unique up to sign. What you need to show is that if $s$ is another parametrization of the surface with $r(u_0,v_0)= s(u_1, v_1)$ then the respective normals coincide (up to sign).
It is rather difficult to prove anything here without a solid set of definitions on which you can rely, and this I can only sketch. Usually one defines the tangent space of a surface as the set of all vectors $v =\gamma^\prime(0)$ which arise as tanget vectors to smooth curves $\gamma = r\circ c$ where in turn $c$ is a smooth curve $c:(-\varepsilon, \varepsilon) \rightarrow D$ such that $c(0) =(u_0,v_0)$, so $\gamma$ is a smooth curve in $\Sigma$ with $\gamma(0) = r(u_0,v_0)$
Now the set of vectors arising that way is the same as the set of vectors arising that way when you are using $s$ in $(u_1, v_1)$ instead of $r$. If you know that then uniquness follows easily.
This is true because by definition of smooth surfaces there is a diffeomorphism $\phi$ of a neighbourhood of $U_0$ of $(u_0,v_0)$ to a neighbourhood $U_1$ of $(u_1, v_1)$ such that $s = r\circ\phi$. So each curve as described before you get when using $r$ is just reparametrized when you use $s$, hence has the same tangent vector in $\mathbb{R}^3$.
You should consult a textbook on differential geometry if you want to see this in full detail.
A: The tangent plane and normal are unique as defined, and also invariant in isometric mappings.
