I'm reading about the Choi representation from John Watrous' textbook on quantum information. On page 78, he says that for any choice of complex Euclidean spaces $\mathcal X$ and $\mathcal Y$, one may define a mapping $J: T(\mathcal X, \mathcal Y) \to L(\mathcal Y \otimes \mathcal X)$ as $$J(\Phi) = (\Phi \otimes 1_{L(\mathcal X)})(\mathrm{vec}(\mathbb 1_\mathcal X)\mathrm{vec}(1_\mathcal X)^*).$$
$\mathrm{vec}$ represents the vectorization map. $1_\mathcal X$ is the identity map on $\mathcal X$. $T(\mathcal X, \mathcal Y)$ represents linear maps from $L(\mathcal X, \mathcal X)$ to $L(\mathcal Y, \mathcal Y)$. (The $\mathrm{vec}$ mapping is a linear bijection, which implies that every vector $u ∈ X \otimes Y$ uniquely determines an operator $A \in L(Y, X )$ that satisfies $\mathrm{vec}(A) = u$.)
A paragraph later he says that:
An alternative way to prove that the mapping $J$ is a bijection is to observe that the action of the mapping $\Phi$ can be recovered from the operator $J(\Phi)$ by means of the equation $$\Phi(X) = \mathrm{Tr}_\mathcal X(J(\Phi)(1_\mathcal Y \otimes X^T)).$$
I'm not sure how to easily see that the LHS is equal to the RHS of this equation.
First of all, how to understand the expression $(\Phi \otimes 1_{L(\mathcal X)})(\mathrm{vec}(\mathbb 1_\mathcal X)\mathrm{vec}(1_\mathcal X)^*)(1_\mathcal Y \otimes X^T)$ and why should its partial trace over $\mathcal X$ be $\Phi(X)$?
It is helpful to re-express the Choi matrix in the following way: $$ J(\Phi) = \sum_{a,b \in \Sigma} \Phi(E_{a,b}) \otimes E_{a,b}. \tag{2.65} $$ Notably, this expression can be obtained by considering the fact that $\operatorname{vec}(1_{\mathcal X})\operatorname{vec}(1_{\mathcal X})^* = \sum_{a,b} E_{a,b} \otimes E_{a,b}$. Technically this only applies where $\mathcal X = \Bbb C^n$, but I claim (and am too lazy to prove) that this expression can be applied to the generalized setting with a suitable definition of $E_{a,b}$.
With that in mind, we can apply the following manipulation. \begin{align} J(\Phi)(1_\mathcal Y \otimes X^T) & = \left(\sum_{a,b \in \Sigma}\Phi(E_{a,b}) \otimes E_{a,b}\right)\left(\sum_{p,q \in \Sigma} x_{q,p} 1_{\mathcal Y} \otimes E_{p,q}\right) \\ & = \sum_{a,b \in \Sigma}\sum_{p,q \in \Sigma} x_{q,p}(\Phi(E_{a,b})\otimes E_{a,b})(1_{\mathcal Y} \otimes E_{p,q}) \\ & = \sum_{a,b,p,q \in \Sigma}x_{q,p}(\Phi(E_{a,b})1_{\mathcal Y})\otimes (E_{a,b} E_{p,q}) \\ & = \sum_{a,b,p,q \in \Sigma}\delta_{b,p} x_{q,p}\Phi(E_{a,b})\otimes (E_{a,q}) = \sum_{a,b,q \in \Sigma} x_{q,b} \cdot \Phi(E_{a,b})\otimes (E_{a,q}) \end{align} where $\delta_{b,p}$ is a Kronecker delta. Now, taking the partial trace over $\mathcal X$ yields \begin{align} \operatorname{Tr}_{\mathcal X} \Bigg(\sum_{a,b,q \in \Sigma} x_{q,b} &\cdot \Phi(E_{a,b})\otimes (E_{a,q}) \Bigg) \\ & = \sum_{a,b,q \in \Sigma} x_{q,b} \operatorname{Tr}(E_{a,q})\Phi(E_{a,b}) \\ & = \sum_{a,b,q \in \Sigma} x_{q,b}\delta_{a,q}\Phi(E_{a,b}) \\ & = \sum_{a,b \in \Sigma} x_{a,b}\Phi(E_{a,b}) = \Phi \left(\sum_{a,b \in \Sigma} x_{a,b}E_{a,b} \right) = \Phi(X). \end{align}
It might be helpful to think of the operation $A,B \mapsto \operatorname{Tr}_{\mathcal X}(AB^T)$ as a kind of "partial dot-product". In particular, consider the following alternative derivation.
\begin{align} \operatorname{Tr}_{\mathcal X}\left[J(\Phi)(1_\mathcal Y \otimes X^T)\right] & = \operatorname{Tr}_{\mathcal X}\left[ \left(\sum_{a,b \in \Sigma}\Phi(E_{a,b}) \otimes E_{a,b}\right) (1_\mathcal Y \otimes X^T)\right] \\ & = \operatorname{Tr}_{\mathcal X}\left[ \sum_{a,b \in \Sigma}(\Phi(E_{a,b})1_{\mathcal Y}) \otimes (E_{a,b}X^T) \right] \\ & = \sum_{a,b \in \Sigma}(\Phi(E_{a,b})1_{\mathcal Y})\cdot \operatorname{Tr}(E_{a,b}X^T) = \sum_{a,b \in \Sigma} x_{a,b}\Phi(E_{a,b}) = \Phi(X). \end{align} In other words, the operator $M \mapsto \operatorname{Tr}_{\mathcal X}[M(1_{\mathcal Y} \otimes X^T)]$ is applying a dot-product with $X$ to all of the $B_i$ in the sum $M = \sum_i A_i \otimes B_i$.