# Quotient of a quotient.

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 $$L$$?

• Where presumably "being a quotient" means "being isomorphic to a quotient"? (Otherwise it's almost never true.) – Torsten Schoeneberg Mar 23 at 10:43
• Yes. That is what I meant. Thank you for pointing out this issue. – James Cheung Mar 26 at 8:37

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.