Let $d\_i\in\mathbb N$, $I\_i:=\{1,\ldots,d\_i\}$ and $u\in\mathbb R^{d\_1}\otimes\mathbb R^{d\_2}\otimes\mathbb R^{d\_3}$. It's somehow clear to me that we may regard $u$ as a three-dimensional array (see Corollary 3 below) and hence consider its entry $u\_{i\_1i\_2i\_3}$ at the index $(i\_1,i\_2,i\_3)\in I\_1\times I\_2\times I\_3$. By fixing the middle index $i\_2\in I\_2$, we may define $$v^{(i\_2)}\_{i\_1i\_3}:=u\_{i\_1i\_2i\_3}\;\;\;\text{for }(i\_1,i\_3)\in I\_1\times I\_3$$ and I guess we can again consider $v^{(i\_2)}$ as being an element of $\mathbb R^{d\_1}\otimes\mathbb R^{d\_3}$.



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> However, I'm really struggling to see that all the identifications involved are really legitimate. I'm used to think of tensor products in a more abstract fashion (see my definition of the tensor product below$^1$).
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> Is the process of "fixing the middle index", which is effectively a (surjective?) transformation from $\mathbb R^{d\_1}\otimes\mathbb R^{d\_2}\otimes\mathbb R^{d\_3}$ to $\mathbb R^{d\_1}\otimes\mathbb R^{d\_3}$, a special case of a more general result of somehow "folding" a tensor product space $E\_1\otimes E\_2\otimes E\_3$ to $E\_1\otimes E\_3$?
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*Remark*: Tensors $u$ of order 3 like this are so-called *cores* in the *tensor-train decomposition* which I'm trying to understand.




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$^1$ If $E\_i$ is a $\mathbb R$-vector space, I'm defining $$(x\_1\otimes x\_2)(B):=B(x\_1,x\_2)\;\;\;\text{for }B\in\mathcal B(E\_1\times E\_2)\text{ and }x\_i\in E\_i,$$ where $\mathcal B(E\_1\times E\_2)$ is the space of bilinear forms on $E\_1\times E\_2$, and $$E\_1\otimes E\_2:=\operatorname{span}\{x\_1\otimes x\_2:E\_i\in E\_i\}\subseteq{\mathcal B(E\_1\times E\_2)}^\ast.$$



> 
> **Lemma 1**: If $I$ is a finite nonempty set, $(e\_i)\_{i\in I}$ is the standard basis of $\mathbb R^I$ and $E$ is a $\mathbb R$-vector space, then the linear extension $\iota\_1$ of $$a\otimes x\mapsto(a\_ix)\_{i\in I}\tag1$$ is an isomorphism between $\mathbb R^I\otimes E$ and $E^I$.
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> **Lemma 2**: If $I\_i$ is a finite nonempty set, then $$\iota\_2:\left(\mathbb R^{I\_2}\right)^{I\_1}\to\mathbb R^{I\_1\times I\_2}\;,\;\;\;a\mapsto\left(a(i\_1)(i\_2)\right)\_{(i\_1,\:i\_2)\in I\_1\times I\_2}$$ is an isomorphism.
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> **Corollary 3**: If $k\in\mathbb N$ and $d\_1,\ldots,d\_k\in\mathbb N$, then $$\bigotimes\_{i=1}^k\mathbb R^{d\_i}\cong\mathbb R^{d\_1\times\cdots\times d\_k}\tag2.$$
> *Proof*: This follows by induction from Lemma 1 and Lemma 2.
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Cores in the tensor-train decomposition
Yes, mapping $A \otimes B \otimes C$ to $A \otimes C$ by choosing a coordinate of $B$ (equivalently, basis element of dual space $B^\*$) is indeed a special case of choosing any element of $B^\*$, considering it as a map $B \to k$ (where $k$ is the field), and then extending that map to $A \otimes B \otimes C \to A \otimes k \otimes C \cong A \otimes C$.


Your $v^{(i\_2)}$ is sometimes called a *slice*, or more specifically a $2$-slice, since the index in the $2$nd position was fixed. This map, that sends a tensor to its $i\_2$'th $2$-slice, is surjective: any matrix arises as the $2$-slice of a tensor given by sticking that matrix into the appropriate entries, and filling the rest with any old thing (zeros, whatever).


