Question about $\text{Hom}_{\mathfrak{g}}(M,L)$ Let $\mathfrak{g}$ be a complex semisimple Lie algebra. 
Suppose $M,N,L$ are $\mathfrak{g}$-modules, $N$ is a $\mathfrak{g}$-submodule of $M$.
Does this implies $\text{Hom}_{\mathfrak{g}}(N,L)\le \text{Hom}_{\mathfrak{g}}(M,L)$ as a vector subspace?
If yes, how to prove it? 
 A: Firstly, since the functor Hom is contravariant in the first variable, what you expect is a map going the other way. For a general abelian category, the only situation in which you expect to have an inclusion $$\mathrm{Hom}(A,B) \subseteq \mathrm{Hom}(C,B)$$ is when $A$ is a quotient of $C$. For a semi-simple Lie algebra $\mathfrak{g}$ over $\mathbf{C}$, the category of finite-dimensional modules is semi-simple, so there are (in general, many) ways to represent a given submodule $N \subseteq M$ as a quotient. Choosing one of them gives you the desired inclusion (which is not canonical in general!).
However, the category of all $\mathfrak{g}$-modules is far from semi-simple, so there is no such inclusion in the generality you are asking for. For instance, if $\mathfrak{g}=\mathfrak{sl}_2(\mathbf{C})$ and you take $N=L=\mathbf{C}$ to be the trivial representation and $M$ to be the co-Verma module containing it as a submodule, then 
$$\mathrm{Hom}_\mathfrak{g}(M,\mathbf{C})=0,$$ so there can be no inclusion as in your question.
