# A submodule of a tensor product of $U_q^{\prime}(\mathfrak{g})$-modules

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.

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.

• Thank you very much! :) One more thing: the rigidity of the module category appears explicitly in your answer, but where does appear the additivity of tensor functor? – Clayton Cristiano 2 days ago