Can the tensor product of two function spaces be regarded as a function space? Let $K,T$ be fields and $V:=\{g:K\to T\}$ be a vector space over T. Then take $W:=V\otimes V$, is this $W$ isomorphic to some function space?
Little background: In quantum mechanics the the state of a one-electron (half spin fermion) system at a given time is a function $\mathbb{R}^{3}\to \mathbb{C}^4$ (with possible constraints I can't recall). Then the state of an $N$-electron system is an element of $\{\mathbb{R}^{3}\to \mathbb{C}^4\}^{\otimes N}$, then mysteriously the state is regarded as a function $\mathbb{R}^{3N}\to (\mathbb{C}^4)^{\otimes N}$.
Full disclosure: I posted this question on Mathematics as well.
 A: The set $V_N$ of functions $\mathbb{R}^{3N}\to (\mathbb{C}^4)^{\otimes N}$ has a natural structure of complex vectors space induced by  $$(a\psi + b\phi)(\vec{x}) := a \psi(\vec{x})+ b\phi(\vec{x}) \quad \mbox{for every $\vec{x} \in \mathbb{R}^{3N}$}$$ where $a,b \in \mathbb  C$ and $\psi,\phi$ are maps $\mathbb{R}^{3N}\to (\mathbb{C}^4)^{\otimes N}$. In particular this applies to $V_1$. Therefore $V_1\otimes \cdots \mbox{($N$-times)}\cdots\otimes V_1$ is a well defined complex vector space. It is easy to prove that the map
$$f: V_1\times  \cdots \mbox{($N$-times)}\cdots \times  V_1 \ni (\psi_1, \ldots, \psi_N) \to \Psi \in V_{N}$$
such that
$$\Psi(\vec{x}, \ldots, \vec{x}_n) := \psi_1(\vec{x}_1) \otimes \cdots \otimes \psi_1(\vec{x}_N) \quad \mbox{for every $\vec{x_i}\in \mathbb R^3$}$$
(the tensor product is here the one between the spaces $\mathbb C^4$) is multilinear. The universal property of the tensor product implies that there is a unique linear map
$$f_\otimes: V_1\otimes  \cdots \mbox{($N$-times)}\cdots \otimes  V_1  \to V_{N}$$
such that 
$$f_\otimes: V_1 \otimes  \cdots \mbox{($N$-times)}\cdots \otimes  V_1 \ni (\psi_1, \ldots, \psi_N) \to \Psi \in V_{N}\:.$$
(The tensor product above is that of spaces of functions $V_1$ not of fibers $\mathbb C^4$.)
The map $f_\otimes$ is the linear homomorphism you are looking for. Indeed,  $f_\otimes$ is  injective and thus you have the identification you are looking for.
Unfortunately I do not have much time to fix details but the way to prove injectivity should be like this.
If $\{\psi_k\}_{k \in K}$ is a Hamel basis of $V_1$ (Zorn's lemma assures it exists), it is easy to prove that
$\{\psi_{k_1} \otimes \cdots \otimes \psi_{k_N}\}_{k_1,\ldots, k_N \in K}$ is a Hamel basis of $V_1 \otimes \cdots \otimes V_1$ $N$ times.
By construction, it turns out that the kernel of $f_\otimes$ is made of the finite linear combinations
$$\sum_{k_1,\ldots, k_N \in K} C^{k_1\ldots k_N} \psi_{k_1} \otimes \cdots \otimes \psi_{k_N}=0$$
Since $\{\psi_{k_1} \otimes \cdots \otimes \psi_{k_N}\}_{k_1,\ldots, k_N \in K}$ is a Hamel basis, we conclude that every $C^{k_1\ldots k_N}=0$ and thus $Ker(f_\otimes)= \{0\}$.
ADDENDUM. The construction survives the introduction of natural topologies and referring to associated tensor-product topological structures. E.g., $V_N$ can be equipped with a natural Hilbert space structure and $f_\otimes$ can be defined, with elementary adaptations, as a Hilbert space homomorphism.
A: Conforming to Valter's notation $V_N=\{\mathbb R^{3N}\to (\mathbb C^4)^{\otimes N}\}$ I think there is a one-to-one correspondence between $V_N$ and $V_1^{\otimes N}$ because $$\mathrm{dim } \,V_1=(\mathrm{dim }\, \mathbb C^4)^{\left|{\mathbb R^3}\right|}$$ moreover $$\mathrm{dim }\,V_1^{\otimes N}=(\mathrm{dim }\,V_1)^N=(\mathrm{dim } \,\mathbb C^4)^{N\left|{\mathbb R^3}\right|}$$ whereas $$\mathrm{dim }\,V_N=(\mathrm{dim }\,((\mathbb C^4)^{\otimes N}))^{\left| \mathbb R^{3N}\right|}=(\mathrm{dim }\, \mathbb C^4)^{N\left|{\mathbb R^{3N}}\right|}$$ but since $\left|\mathbb R^3\right| =\left|\mathbb R^{3N} \right|$ they have the same dimension ($2^\mathfrak c$) thus are isomorphic.
