I have just read that over a noetherian local ring $(A,\mathfrak{m})$, there is always a nontrivial map between two finitely generated modules such that their support is exatly $\mathfrak{m}$. The brief explanation I found is that any such a module always surjects onto $A/\mathfrak{m}$ and $A/\mathfrak{m}$ injects into $M$.
I am fine wiht the explanation and I see how I can surject $M$ onto $A/\mathfrak{m}$ (indeed $M/\mathfrak{m}M$ is an $A/\mathfrak{m}$ vector space, and I can choose one basis vector to project onto).
What I am missing is the injection $A/\mathfrak{m} \rightarrow M$. How could I prove it?
Following the hint I received as comment, we can give a complete answer. If a module is finitely generated over $(A,\mathfrak{m})$, then $M/\mathfrak{m}M$ is a finite dimensional vector space over $A/\mathfrak{m}$ (which is not zero by Nakayama's lemma), so we can surject it onto any one dimensional subspace, which is isomorphic to $A/\mathfrak{m}$.
Now, to define an injection $A/\mathfrak{m} \rightarrow M$ we recall that $Ass_A(M) \subset Supp_A(M)$. But $Supp_A(M)=\mathfrak{m}$ by hypothesis and since $M$ is noetherian $Ass_A(M)$ is not empty. Hence $\mathfrak{m} \in Ass_A(M)$, which means that there is a submodule of $M$ isomorphic to $A/\mathfrak{m}$. This gives the injection we were looking for.
Now, if we fix two modules $M$ and $N$ with such properties, we first surject $M$ onto $A/\mathfrak{m}$ with the first map and then we inject $A/\mathfrak{m}$ into $N$ with the second one.