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Let $\mathfrak{g}$ be a complex reductive Lie algebra. Let $M,N,L$ be finite dimensional $\mathfrak{g}$-modules.

Suppose $M$ is a quotient of $N$ and $N$ is a quotient of $L$.

My question: Is $M$ a quotient of $L$?

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  • $\begingroup$ Where presumably "being a quotient" means "being isomorphic to a quotient"? (Otherwise it's almost never true.) $\endgroup$ – Torsten Schoeneberg Mar 23 at 10:43
  • $\begingroup$ Yes. That is what I meant. Thank you for pointing out this issue. $\endgroup$ – James Cheung Mar 26 at 8:37
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If $q: L \twoheadrightarrow M$ is the canonical quotient map for $M$, and $p: M \twoheadrightarrow N$ is the canonical quotient map, then $pq: L \to N$ is surjective, and thus $N \cong L / \mathrm{ker}(pq)$.

Note that this did not use any structure of $\mathfrak{g}$ or the finite-dimensionality of the modules $L$, $M$, and $N$; it follows only from the fact that if $\varphi: A \twoheadrightarrow B$ is a surjective morphism of "modules," then $B \cong A / \mathrm{ker}(\varphi)$. This is true for abelian groups, and a "module" of any kind should be an abelian group with extra structure. In fact, this fact is true* in any abelian category, which are the most general categories of modules.

*: modulowink wink replacing surjectivity with the categorical notion of epimorphism.

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