The functor $\operatorname{Hom}(-,M)$, is it faithful? Let $M\neq 0$ be a left $R$-module, I want to prove that if $M$ is injective then for every left $R$-module $A$, $\operatorname{Hom}(A,M)=0$ implies $A=0$.
Indeed by contradiction if $A$ contains a non-zero element $x$, then the sequence
$$
0\to \langle x\rangle \to A \to A/\langle x\rangle \to 0
$$
is exact. Since $M$ is injective then then the sequence
$$
0\to \operatorname{Hom}(A/\langle x\rangle,M) \to \operatorname{Hom}(A,M) \to \operatorname{Hom}(\langle x\rangle,M) \to 0
$$
is exact. By hypothesis $\operatorname{Hom}(A,M)=0$, then $\operatorname{Hom}(\langle x\rangle,M)=0$ which is a contradiction. Because if $y\in M-\{0\}$ then we define $f:\langle x\rangle\to M$ by $f(x)=y$ and we extend it by linearity. It is clear that $f$ is nonzero.
Is this result true? If it is true, is the proof right?
 A: $\mathrm{Hom}(\langle x\rangle,M)\neq0$ for non-zero $x$ holds if $\mathrm{Hom}(A,M)\neq0$ for each simple module $A$, i.e. for each non-zero module whose only submodules are the zero module and itself. This is because every non-zero cyclic module (and more generaly, any finitely generated module) has a maximal proper submodule, and hence a simple quotient. If $A$ is a simple quotient of $\left<x\right>$, then $\left<x\right>\twoheadrightarrow A\to M$ is a non-zero element of $\mathrm{Hom}(\left<x\right>,A)$.
Your argument then shows that an injective module is a cogenerator, i.e.  $\mathrm{Hom}(A,M)=0$ implies $A=0$, if and only if it each simple module is isomorphic to a submodule of $M$.
For example, $\mathbb Q/\mathbb Z$ is an injective $\mathbb Z$-module (because it's divisible and $\mathbb Z$ is a PID), and is a cogenerator because it contains a copy of each simple module $\mathbb Z/p$ for $p$ a prime. Moreover, for any ring $R$, $\mathrm{Hom}_{\mathbb Z}(R,\mathbb Q/Z)$ is an injective cogenerator for left $R$-modules because $0=\mathrm{Hom}_R(M,\mathrm{Hom}_{\mathbb Z}(R,\mathbb Q/\mathbb Z))\cong\mathrm{Hom}_{\mathbb Z}(M\otimes_R R,\mathbb Q/\mathbb Z)\cong\mathrm{Hom}_{\mathbb Z}(M,\mathbb Q/\mathbb Z)$ implies $M=0$.
