Does anyone have a proof for the following Lemma?


Let $\mathfrak{g}$ be a finite-dimensional simple Lie algebra over $\mathbb{C}$ and $U\_q^{\prime}(\mathfrak{g})$ be the quantum affine algebra over $\mathfrak{g}$. Let $M\_k$ be a finite-dimensional integrable $U\_q^{\prime}(\mathfrak{g})$-module ($k=1, 2, 3$). Let $X$ be a $U\_q^{\prime}(\mathfrak{g})$-submodule of
$M\_1\otimes M\_2$ and $Y$ a $U\_q^{\prime}(\mathfrak{g})$-submodule of $M\_2\otimes M\_3$ such that $X\otimes M\_3\subset M\_1\otimes Y$ as submodules of $M\_1\otimes M\_2\otimes M\_3$. Then there exists a $U\_q^{\prime}(\mathfrak{g})$-submodule $N$ of $M\_2$ such that $X\subset M\_1\otimes N$ and $N\otimes M\_3\subset Y$.


This is the Lemma 3.10, stated without a proof in the paper <https://www.cambridge.org/core/journals/compositio-mathematica/article/simplicity-of-heads-and-socles-of-tensor-products/0F1B26D97E9484FCCC9D023D02C509E3> The authors of this paper mention that "a similar result holds for any rigid monoidal category which is
abelian and the tensor functor is additive". However, I could not prove this result in such a general way, nor did I find this statement anywhere else.



A submodule of a tensor product of $U_q^{\prime}(\mathfrak{g})$-modules
Since $X$ lies in $M\_1\otimes M\_2$, by adjunction there is a canonical map from $M\_1^\ast\otimes X$ to $M\_2$. Let $N$ be the image of this morphism. Then the inclusion of $X$ in $M\_1\otimes M\_2$ factors through $M\_1\otimes N$, so $X\subset M\_1\otimes N$.


Now the map from $M\_1^\ast\otimes X\otimes M\_3$ to $M\_2\otimes M\_3$ has image $N\otimes M\_3$. But this map factors through the inclusion of $Y$ into $M\_2\otimes M\_3$ (as $X\otimes M\_3\to M\_1\otimes M\_2\otimes M\_3$ factors through $M\_1\otimes Y$). Therefore $N\otimes M\_3\subset Y$, as required.


