Proof of unique coordinatization of tensor space once bases are chosen I am trying to prove the statement:

If $K^{n_1} \times K^{n_2} \times \dots \times K^{n_d} \xrightarrow{G} K^{n_1 \times n_2 \times \dots n_d}$ is isomorphic as a tensor space to $K^{n_1} \times K^{n_2} \times \dots \times K^{n_d} \xrightarrow{\otimes} K^{n_1} \otimes K^{n_2} \otimes \dots \otimes K^{n_d}$, then f = G up to some permutation of the indices ${i_1}{i_2}\ldots{i_d}$. 

Let me define everything precisely. 

Let $V^1$, $V^2$, $\ldots$ , $V^d$, $W$ be vector spaces over $K$ and let $\otimes$ be the multilinear mapping $\otimes :V^1 \times V^2 \times \dots \times V^d \rightarrow W$ such that $W$ equals the space spanned by the image of $\otimes$, and for any multilinear mapping $f: V^1 \times V^2 \times \dots \times V^d \rightarrow H$ for any vector space $H$, there exists a unique linear mapping $F: T \rightarrow H$ s.t. $F(\otimes({v^1 \times v^2 \times \dots \times v^d})) = f(v^1 \times v^2 \times \dots \times v^d)$.

Let $V^1 \times V^2 \times \dots \times V^d \xrightarrow{\otimes} W$ and $V^1 \times V^2 \times \dots \times V^d \xrightarrow{\otimes '} W'$ be two tensor products of ${V^1 \times V^2 \times \dots \times V^d}$. $\eta: W \rightarrow W'$ is a tensor space morphism if $\eta$ is a vector space morphism and ${\eta (v^1 \otimes v^2 \otimes \ldots \otimes v^d) = \eta (v^1 \otimes ' v^2 \otimes ' \ldots \otimes ' v^d)}$ for all $v^1 \times v^2 \times \dots \times v^d \ \in \ V^1 \times V^2 \times \dots \times V^d$, where  $\otimes(v^1 \times v^2 \times \dots \times v^d)$ is denoted $v^1 \otimes v^2 \otimes \dots \otimes v^d$. 

The $d$-fold tensor space is isomorphic as a tensor space to the space of $d$-fold hypermatrices. A d-fold hypermatrix over K is a function from $(1, 2, \ldots, n_1) \times (1, 2, \ldots, n_2) \times \ldots \times (1, 2, \ldots, n_d)$ to K, and the space of all such hypermatrices is denoted $K^{n_1 \times n_2 \times \ldots \times n_d}$. Hypermatrices in $K^{n_1 \times n_2 \times \ldots \times n_d}$ are written as $A = (a_{{i_1}{i_2}\ldots{i_d}})$ where each $a_{{i_1}{i_2}\ldots{i_d}} \in K$. 

To see that these spaces are isomorphic as tensor spaces, choose ordered bases $(e^k_i)_{i=1}^{n_k}$ of $V^{k}$ for ${k = 1, \ldots, d}$, and consider the mapping
\begin{align*} 
V^1 \times V^2 \times \ldots \times V^d \ &amp;\xrightarrow{f} \quad  K^{n_1 \times n_2 \times \ldots n_d} \\
v^1 \times v^2 \times \ldots \times v^d \ &amp;\longmapsto \quad  (a_{{i_1}{i_2}\ldots{i_d}} = a^1_{i_1}a^2_{i_3}\ldots a^d_{i_d} )
\end{align*} 
where $v^k = \sum_{i = 1}^{n_k} a^k_i \ e^k_i  $ for k = 1, 2, $\dots$ d. Note that f satisfies the universal property of the tensor product, and thus the map F below must be an isomorphism of tensor spaces. 
\begin{align*} 
V^1 \otimes V^2 \otimes \ldots \otimes V^d \ &amp;\xrightarrow{F} \quad  K^{n_1 \times n_2 \times \ldots n_d} \\
v^1 \otimes v^2 \otimes \ldots \otimes v^d \ &amp;\longmapsto \quad  (a_{{i_1}{i_2}\ldots{i_d}})
\end{align*} 
where $v^1 \otimes v^2 \otimes \ldots \otimes v^d = \sum_{i_1, i_2, \ldots, i_d = 1}^{n_1, n_2, \ldots, n_d} \  a_{{i_1}{i_2}\ldots{i_d}} \ e_{i_1}^1 \otimes e_{i_2}^2 \otimes \ldots \otimes e_{i_d}^d  $.
I did not read your post too closely, but at the end you are listing products of components. That is exactly what the coefficients are for a basis of a tensor product space built up with simple tensors from a basis of the original two spaces. Do you know how to make a basis of a tensor product space?
I did not read your post too closely, but at the end you are listing products of components. That is exactly what the coefficients are for a basis of a tensor product space built up with simple tensors from a basis of the original two spaces. Do you know how to make a basis of a tensor product space?