Why is $\text{Hom}(V,W)$ the same thing as $V^* \otimes W$? I have a couple of questions about tensor products:
Why is $\text{Hom}(V,W)$ the same thing as $V^* \otimes W$? 
Why is an element of $V^{*\otimes m}\otimes V^{\otimes n}$ the same thing as a multilinear map $V^m \to V^{\otimes n}$? 
What is the general formulation of this principle?
 A: The result is generally wrong for infinite-dimensional spaces: see this question.
For finite dimensional space $V$, let's build an isomorphism $f : V^* \otimes W \to \hom(V,W)$ by defining
$$f(\phi \otimes w)(v) = \phi(v) w$$
This clearly defines a linear map $V^* \otimes W \to \hom(V,W)$ (it's bilinear in $V^* \times W$). Reciprocally, take a basis $(e_i)$ of $V$, then define $g : \hom(V,W) \to V^* \otimes W$ by:
$$g(u) = \sum_i e_i^* \otimes u(e_i)$$
Where $(e_i^*)$ is the dual basis to $(e_i)$ (I will use a few of its properties in $\color{red}{red}$ below). This is well-defined because $V$ is finite-dimensional (the sum is finite). Let's check that $f$ and $g$ are inverse to each other:


*

*For $u : V \to W$, $$f(g(u))(v) = \sum e_i^*(v) u(e_i) = u \left( \sum e_i^*(v) e_i \right) \color{red}{=} u(v)$$ and so $f(g(u)) = u$.

*For $\phi \otimes w \in V^* \otimes W$, $$g(f(\phi \otimes w)) = \sum e_i^* \otimes f(\phi \otimes w)(e_i) = \sum e_i^* \otimes \phi(e_i) w = \sum \phi(e_i) e_i^* \otimes w \color{red}{=} \phi \otimes w$$
And so $f$ and $g$ are isomorphisms, inverse to each other.

It is known that for finite dimensional $V$, then $(V^*)^{\otimes m} = (V^{\otimes m})^*$. Then an element of $V^{* \otimes m} \otimes V^{\otimes n}$ is an element of $(V^{\otimes m})^* \otimes V^{\otimes n} = \hom(V^{\otimes m}, V^{\otimes n})$. So by definition / universal property of the tensor product, it's a multilinear map $V^m \to V^{\otimes n}$.
A: One general form of that assertion (noting that it cannot be quite as simple as one might imagine) is the Cartan-Eilenberg adjunction
$$
\mathrm{Hom}(X\otimes Y,Z)\;\approx\;\mathrm{Hom}(X,\mathrm{Hom}(Y,Z))
$$
in some reasonable additive (or whatever) category, where, significantly, the tensor product must be a genuine categorical tensor product, as opposed to a "projective" or "injective" tensor product, which have only half the properties of a genuine tensor product. So, for example, there is no genuine tensor product of (infinite-dimensional) Hilbert spaces in any reasonable category of topological vector spaces. (The thing often called the "Hilbert-space tensor product" has only half the requisite properties.
A: Well, the actual result is $V^*\otimes W\cong \{\varphi\in \mathcal{L}(V,W):\dim \text{range }\varphi<\infty\}$, via the canonical map $\Phi: \varphi\otimes w\mapsto (v\mapsto \varphi(v)w)$ as given in the answer above (note that this is always an embedding of $V^*\otimes W$ into $\mathcal{L}(V,W)$).
On one hand, every element of $V^*\otimes W$ can be written as $\sum^{n}_{i=1} \varphi_i\otimes w_i$, and its image under $\Phi$ is included in $\mathrm{span}(w_1,\cdots,w_n)$; on the other hand, if $\dim \text{range }\varphi<\infty$, pick a basis $w_1,\cdots,w_n$ of $\text{range }\varphi$, then $\varphi = \sum^{n}_{i=1} \varphi_iw_i$ for $\varphi_i\in V^*$, and $\sum^{n}_{i=1} \varphi_i\otimes w_i$ is a preimage of $\varphi$.
