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 $\mathbb R^{d_1}\otimes\mathbb R^{d_3}$.

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$^1$).

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 $E_1\otimes E_3$?

*Remark*: Tensors $u$ of order 3 like this are so-called *cores* in the *tensor-train decomposition* which I'm trying to understand.

$^1$ 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 $E^I$.

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.

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.