If $\nabla$ is the covariant derivative, why $\nabla: \mathcal{X}(S) \rightarrow \mathcal{X}(S) \otimes \Omega^{1}(S)$? I'm studying differential geometry, and my professor defined covariante derivative in this way:
Let $S \subset \mathbb{R}^3$ be a surface, $$\mathcal{X}(S) = \left\{ \xi:S\rightarrow \bigcup\limits_{p\in S} T_p S; \mbox{ }\xi(p) \in T_pS\hspace{0.1cm} \forall\hspace{0,1cm} p \in S \right\}$$ (the set of the fields over the surface $S$), and $\Omega^{1}(S)$ the set of 1-forms over $S$. 
Consider $p$ $\in$ $S$ and $y \in T_pS,$ choosing a curve $\alpha: I \rightarrow S$, satisfying $\alpha(0) = p$ and $\alpha'(0) = y$, we can define the covariant derivative at $p$ of the vector field $\xi$ relative to the vector $y$ as :
$$\nabla_y \xi (p) := \pi_{T_pS} \circ \left(\left.\frac{d\xi(\alpha(t)) }{dt} \right|_{t=0}\right).  $$
Where $\pi_{T_pS}$ is the projection of $\mathbb{R}^3$ over $T_p S$. It's easy to verify that this definition doesn't depend of the parametrization of the curve $\alpha$.
Now, we can expand this notion of covariant derivative by changing the vector $y$ for a field $X$ $\in$ $\mathcal{X}(s)$, so
$$\nabla_{X} \xi (p) := \nabla_{X(p)} \xi(p). $$  
My problem is here. My teacher said that this caracterion show us that the operator $\nabla$ acts in $\mathcal{X}(S)$ and return tensor of $\mathcal{X(S)}\otimes \Omega^{1}(S),$ i. e;
$$\nabla: \mathcal{X}(S) \rightarrow \mathcal{X}(S)\otimes\Omega^{1}(S)$$
$$\xi \mapsto \nabla\xi $$
I really want to know why $\nabla \xi$ $\in$ $\mathcal{X}(S)\otimes\Omega^{1}(S)$. From my knowledge and understanding of the circumstances, I see $\nabla \xi$ as
$$\nabla \xi : \mathcal{X}(S) \rightarrow \mathcal{X}(S)  $$
which implies  $\nabla \xi$ $\in$ $\mathcal{F}(\mathcal{X}(S),\mathcal{X}(S))= \{f:\mathcal{X}(S)\rightarrow \mathcal{X}(S) \}$. Is $\mathcal{F}(\mathcal{X}(S),\mathcal{X}(S)) = \mathcal{X}(S)\otimes\Omega^{1}(S)$ (or isomorphic as linear space) ? If yes, how I see this?
 A: This is true if you consider the objects involved as $C^{\infty}(S)$-modules. 
Namely, given a vector field $\xi \in \mathcal{X}(S)$ and a smooth function $f \in C^{\infty}(S)$, you can form the product $f\xi$ which is defined pointwise by $(f\xi)(p) = f(p)\xi(p)$ (the product of the tangent vector $\xi(p) \in T_pM$ by the number $f(p)$). This also works for differential forms and more generally for sections of a vector bundle over $S$ and gives $\mathcal{X}(S)$ and $\Omega^1(S)$ the structure of a $C^{\infty}(S)$-module. Then we have a natural isomorphism of $C^{\infty}(S)$-modules between
$$ \mathcal{F}(\mathcal{X}(S), \mathcal{X}(S)) = \operatorname{Hom}_{C^{\infty}(S)}(\mathcal{X}(S), \mathcal{X}(S)) := \{ T \colon \mathcal{X}(S) \rightarrow \mathcal{X}(S) \, | \, T(fX) = fX \,\,\, \forall f \in C^{\infty}(S) \} $$
and
$$ \mathcal{X}(S) \otimes_{C^{\infty}(S)} \Omega^1(S). $$
To define the isomorphism, note that we have a natural $C^{\infty}(S)$ bilinear map 
$$\Phi \colon \mathcal{X}(S) \times \Omega^1(S) \rightarrow \mathcal{F}(\mathcal{X}(S), \mathcal{X}(S))$$
given by
$$ \Phi(\xi, \omega)(\nu) = \omega(\nu) \xi.$$
By the universal property of the tensor product, this gives a map (denoted by the same symbol) 
$$\Phi \colon \mathcal{X}(S) \otimes_{C^{\infty}(S)} \Omega^1(S) \rightarrow \mathcal{F}(\mathcal{X}(S), \mathcal{X}(S)) $$
and it turns out that this map is an isomorphism.

Note that the map defined above is the same map you would define to show that $V \otimes_{\mathbb{R}} V^{*}$ is isomorphic to $\operatorname{Hom}_{\mathbb{R}}(V,V)$ as real vector spaces. However, the argument as to why this map is an isomorphism is more delicate because you are working with (not necessarily free) modules over a ring and not with finite dimensional vector spaces. For full details, see Chapter 7 in the book "Differentiable Manifolds" by Lawrence Conlon.
A: A linear transformation between finite-dimensional vector spaces $V$ and $W$ can be seen as an element of $W \otimes V^*$. The canonical map
$$
    \phi\colon W \times V^* \to \operatorname{Hom}(V,W),\
    \phi(w,\alpha)(v) = \alpha(v)w
$$
is bilinear, so factors through a homomorphism $W \otimes V^*\to \operatorname{Hom}(V,W)$.  By checking on a basis you can see this is an isomorphism.
For any $\xi\in\mathcal{X}(S)$ and $p\in S$, the operator $\nabla \xi$ is a linear transformation from $T_pS$ to $T_pS$.  So it is an element of $T_pS \otimes T_p^* S$ for each $p$.  Taken globally, $\nabla \xi$ is a section of $TS \otimes T^*S$, which is isomorphic to $\mathcal{X}(S) \otimes \Omega^1 (S)$.
