Anderson’s exercise 8.3 on “generating and cogenerating” I am reading “Anderson and Fuller - Rings and Categories of Modules (2nd edition, 1992)” 
and I am trying to find a solution of exercise 8.3 on page 112, which says:
Let $M$ be a left $R$-module. Show that if $M$ cogenerates a generator, then $M$ is faithful.
So, $M$ cogenerates a generator, say $M$ cogenerates $X$ and $X$ is a generator, where $X$ is a left R-module.
(1) $M$ cogenerates $X$, by Corollary 8.13.(1) (page 109) the reject of $M$ in $X$ is zero, that is:
$$ Rej_X (M) = \bigcap \{ ker(h) \text{  |  } h \in Hom_R (X,M) \} = 0 $$
(2) $X$ is a generator, this means that $X$ generates $_R R$, that is, $R$ as a left $R$-module.
Then, by Corollary 8.13.(2) (page 109) the trace of $X$ in $R$ equals $R$, that is:
$$ Tr_R (X) = \sum \{ im(h) \text{  |  } h \in Hom_R (X,R) \} = R $$
(3) It follows that there are finitely many $f_i \in Hom_R (X,R)$ and $x_i \in X$, $i=1, 2, \cdots ,n$ such that 
$$1 = f_1 (x_1) + f_2 (x_2) + \cdots + f_n(x_n)$$
( I do not know if we need this in the proof. )
(4) Proposition 8.22 on page 111 says that the annihilator of $M$ in $R$, that is:
$$ l_R (M) = \{ r \in R \text{  |  } rM = 0 \} $$
equals the reject of $M$ in $R$, that is:
$$ Rej_R (M) = \bigcap \{ ker(h) \text{  |  } h \in Hom_R (R,M) \} $$
Thus 
$$ l_R (M) = Rej_R (M) $$
(5) Now we have to show that M is faithful, that is, the annihilator is zero, that is,
$$ l_R (M) = Rej_R (M) = 0 $$
So, we have to prove that if $s \in l_R (M) = Rej_R (M)$ then $s = 0$
Take $s \in Rej_R (M) = \cap \{ ker(h) \text{  |  } h \in Hom_R (R,M) \}$
Then, for all $ h \in Hom_R (R,M)$ we have $h(s) = 0$
Also, because $ l_R (M) = Rej_R (M) = 0 $, we have $s \in l_R (M)$
So, for all $m \in M$ we have $sm = 0$
Take $x \in X$
Then, for all $ h \in Hom_R (X,M)$ we have $h(sx) = sh(x) = 0$, because $h(x) \in M$
Thus, for all $ h \in Hom_R (X,M)$, $sx \in ker(h)$
Thus, by (1) above, 
$sx \in Rej_X (M) = \cap \{ ker(h) \text{  |  } h \in Hom_R (X,M) \} = 0$
Therefore, $sx = 0$, for all $x \in X$
So, $s \in l_R (X) = Rej_R (X) = \cap \{ ker(h) \text{  |  } h \in Hom_R (R,X) \} = 0$
Then, for all $ h \in Hom_R (R,X)$ we have $h(s) = 0$
Now I am stuck and I do not know where I am heading to.
Can someone, please, help me to finish this proof ?
 A: Saying that $M$ cogenerates a generator $G$ means that $G$ embeds in a direct product of copies of $M$. Indeed, since $\operatorname{Rej}_G(M)=0$, if we take all homomorphisms $f\colon G\to M$, the natural map
$$
G\to M^{\operatorname{Hom}_R(G,M)}
$$
defined by $x\mapsto (f(x))_{f\in\operatorname{Hom}_R(G,M)}$ has zero kernel.
If $r$ annihilates $M$, then it also annihilates $G$. Since a generator is necessarily faithful, then $r=0$. Therefore $M$ is faithful as well.
Why is a generator $G$ faithful? Because $R$ (as a left $R$-module) is an epimorphic image of a direct sum of copies of $G$, thus any $r$ that annihilates $G$ has to annihilate $R$.
A: I found the answer in line of my argument in my original post, 
and I am embarrassed because I was so close and I did not see it.
We proved $sx = 0$, for all $x \in X$, and we have to show that $s=0$
In (3) we found
$$1 = f_1 (x_1) + f_2 (x_2) + \cdots + f_n(x_n)$$
So
$$s = f_1 (s x_1) + f_2 (s x_2) + \cdots + f_n(s x_n)$$
Then, for each $i=1, 2, \cdots ,n$: $ s x_i=0 $, because $x_i \in X$ thus
$$s = f_1 (0) + f_2 (0) + \cdots + f_n(0) = 0$$
Now, we are ready, because thie proves that $M$ is faithful
