# If $f:M\to M,m\mapsto mr$ is injective, then show that $\text{Hom}_R(M,E)\stackrel{r}{\longrightarrow}\text{Hom}_R(M,E)$ is surjective

Let $$R$$ be a commutative ring, $$M$$ an $$R$$-module and $$r\in R$$. If $$f:M\to M$$ defined by $$f(m)=mr$$ is an injective $$R$$-module endomorphism, then show that the mapping $$\text{Hom}_R(M,E)\stackrel{r}{\longrightarrow}\text{Hom}_R(M,E)$$ is surjective, where $$E$$ is the injective cogenerator of $$R$$.

I have tried to search for injective generators and what I know is that

(1) an injective $$R$$-module $$E$$ is called an {\it injective cogenerator} of $$R$$ if, for every $$R$$-module $$M$$ and for every non-zero $$m\in M$$, there is a homomorphism $$\phi:M\to E$$ such that $$\phi(m)\neq0$$.

(2) since $$R$$ is commutative, $$\text{Hom}_R(M,E)$$ is also an $$R$$-module whose elements are maps $$\phi\in \text{End}_R(M)$$.

However, I cannot figure out the surjectivity of that multiplication map in $$\text{Hom}_R(M,E)\stackrel{r}{\longrightarrow}\text{Hom}_R(M,E)$$ comes about.

• Surjectivity holds simply because $E$ is an injective $R$-module. Aug 1, 2020 at 14:08
• How is the map $\text{Hom}_R(M,E)\stackrel{r}{\longrightarrow}\text{Hom}_R(M,E)$ defined? Is it like $f\mapsto r\cdot f, f\in \text{Hom}_R(M,E)$? Aug 1, 2020 at 16:28

Exercise 2.19 (Rotman): the functors $$\operatorname{Hom}_R(-, -)$$ preserve multiplication. Explicitly, if $$\mu_r : M \to M$$ is the multiplication map $$\mu_r(m) = r \cdot m$$ from a module $$M$$ over a commutative ring $$R$$ to itself, then for any $$R$$-module $$N,$$ the induced map $$\operatorname{Hom}(\mu_r) : \operatorname{Hom}_R(M, N) \to \operatorname{Hom}_R(M, N)$$ is the multiplication map $$\varphi \mapsto r \cdot \varphi.$$
Considering that $$E$$ is an injective cogenerator of $$R,$$ it is by definition an injective $$R$$-module, and this implies that $$\operatorname{Hom}_R(-, E)$$ is an exact functor, i.e., $$\operatorname{Hom}_R(-, E)$$ is right exact.
Consequently, the exact sequence $$0 \to M \xrightarrow{r \cdot} M \to \frac{M}{rM} \to 0$$ gives rise to the exact sequence $$0 \to \operatorname{Hom}_R \biggl(\frac{M}{rM}, E \biggr) \to \operatorname{Hom}_R(M, E) \xrightarrow{r \cdot} \operatorname{Hom}_R(M, E) \to 0,$$ as $$\operatorname{Hom}_R(-, E)$$ is contravariant. But this says that $$\operatorname{Hom}_R(M, E) \xrightarrow{r \cdot} \operatorname{Hom}_R(M, E)$$ is surjective.