Dimension of tangent spaces Consider the set 
$$ S^{2} = \{ (x,y,z) \in \mathbb{R} \mid x^{2} + y^{2} + z^{2}=1 \}$$
as a manifold with two charts given by the stereographic projections from the points N $= (0,0,1)$ and S $= (0,0,-1)$.
The chart obtained from the projection from N is a homeomorphism
$$ \phi : S^{2} \setminus \{N\} \longrightarrow \mathbb{R^{2}} $$
and, if $p\in S^{2} \setminus \{N\}$, the tangent spaces $T_{p}S^{2}$ and $T_{\phi(p)}\mathbb{R^{2}}$ are both vector spaces of dimension 2.
Consider now the differential of $\phi $ in $p$, the linear application
$$ d \phi_{p}:T_{p}S^{2}\longrightarrow T_{\phi(p)}\mathbb{R^{2}}.$$
In my opinion, this application should be represented by a $2 \times 2$ matrix. Nevertheless, I am working out a solved exercise in which that application is represented by a $3 \times 2$ matrix, whose coefficients are functions of $x,y$ and $z$.
Why is this? Is it because $\phi$ is viewed as a function from a subset of $\mathbb{R^{3}}$ to $\mathbb{R^{2}}$? Why then this apparent inconsistency? 
 A: The two-dimensional tangent space $T_pS^2$ is naturally a subspace of the three-dimensional space $T_p\mathbb R^3$.  Since stereographic projection extends to (a neighborhood of $p$ in) $\mathbb R^3$, I suspect that you have computed the derivative of the extension.  This restricts to the derivative you are after along the embedding $T_pS^2 \hookrightarrow T_p\mathbb R^3$.
Note that $T_pS^2$ doesn't have a canonical basis.  Without choosing a basis, linear maps $T_pS^2 \to \dots$ cannot be written as matrices.  For any choice of basis they can be.  For example, if $p = (x,y,z) \neq S,N$, you could decide to use latitude and longitude $-z(\partial_x + \partial_y) + r \partial z$ and $y \partial_x - x \partial_y$, where $r = \sqrt{x^2 + y^2}$ is the distance to the pole (and $\partial_x,\partial_y,\partial z$ are the standard basis in $T_p\mathbb R^3 \cong \mathbb R^3$).  Or you could decide that you prefer unit vectors, and rescale the vector pointing along latitude lines $y \partial_x - x \partial_y \leadsto r^{-1}(y \partial_x - x \partial_y)$.
