Let $X\subseteq B(H)$ be an operator system and let $M\subseteq B(K)$ be a von Neumann algebra. We form the Fubini-tensor product $$X \otimes_\mathcal{F} M := \{z \in B(H\otimes K): (\sigma\otimes \iota)(z) \in M \text{ and } (\iota \otimes \tau)(z)\in X \text{ for all }\sigma\in B(H)_*, \tau \in B(K)_*\}.$$ We have a natural injective linear map $$X \otimes_\mathcal{F}M \to B(M_*,X): z \mapsto (\omega \mapsto (\iota \otimes \omega)(z)).$$

Is it true that the image of this map consists of all completely bounded maps $M_*\to X$?

Note that this is at least true if $X$ itself is a von Neumann algebra (However, this result is not very easy to find in the literature. See e.g. the article "Completely positive maps in the tensor product of von Neumann algebras" by Nagisa and Tomiyama - theorem 2).

Let $X\subseteq B(H)$ be an operator system and let $M\subseteq B(K)$ be a von Neumann algebra. We form the Fubini-tensor product $$X \otimes_\mathcal{F} M := \{z \in B(H\otimes K): (\sigma\otimes \iota)(z) \in M \text{ and } (\iota \otimes \tau)(z)\in X \text{ for all }\sigma\in B(H)_*, \tau \in B(K)_*\}.$$ We have a natural injective linear map $$X \otimes_\mathcal{F}M \to B(M_*,X): z \mapsto (\omega \mapsto (\iota \otimes \omega)(z)).$$

Is it true that the image of this map consists of all completely bounded maps $M_*\to X$?

Note that this is at least true if $X$ itself is a von Neumann algebra (However, this result is not very easy to find in the literature).