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.


1 Answer 1


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.


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