# Module of homomorphisms as injective module

We know that for any $R$-module exist injective $R$-module $\overline{M}$ such that there is inclusion $i:M\rightarrow \overline{M}$, where we treat $M,\overline{M}$ as $\mathbb{Z}$-modules.

Show that $\widetilde{M}:=\mathrm{Hom}_{\mathbb{Z}}(R,\overline{M})$ is injective $R$-module.

$\widetilde{M}$ is injective module iff functor $\mathrm{Hom_{R}(\cdot,\widetilde{M}})$ is exact.

Let $0\rightarrow N'\rightarrow N\stackrel{\alpha}{\rightarrow} N''\rightarrow 0$ be exact sequence of $R$-modules. Then we obtain sequence $0\rightarrow \mathrm{Hom}_R(N',\widetilde{M})\rightarrow \mathrm{Hom}_R(N,\widetilde{M})\stackrel{\alpha^{*}}{\rightarrow}\mathrm{Hom}_R(N'',\widetilde{M})\rightarrow 0$

We know that this functor left-exact, so we should prove that $\mathrm{im}\alpha^*=\mathrm{Hom}_R(N'',\widetilde{M})$, but we know that $\mathrm{Hom}_R(N'',\widetilde{M})=\mathrm{Hom}_R\left(N'',\mathrm{Hom}_{\mathbb{Z}}(R,\overline{M})\right)\cong \mathrm{Hom}_{\mathbb{Z}}\left(N''\otimes_RR,\overline{M}\right)\cong \mathrm{Hom}_{\mathbb{Z}}(N'',\overline{M})$

So, I should prove that $\mathrm{im}\alpha^*=\mathrm{Hom}_{\mathbb{Z}}(N'',\overline{M})$.

How to do it?

• you may do the same thing for the middle term and lift the surjection via the isomorphisms. – Math137 Jul 5 '15 at 22:09

## 1 Answer

You forgot the functor $\operatorname{Hom}_R(\text{--},M)$ is contravariant.

Let $0\longrightarrow N'\xrightarrow{\ u\ }N$. What has to be proved is $$\mathrm{Hom}_R(N,\widetilde{M})\rightarrow \mathrm{Hom}_R(N',\widetilde{M})$$ is surjective. For this, note first $$\operatorname{Hom}_R(N,\widetilde{M})=\operatorname{Hom}_R(N,\operatorname{Hom}_\mathbf Z(R,\overline{\!M\mkern-0.5mu})\simeq\operatorname{Hom}_\mathbf Z(N\otimes_R R,\overline{\!M\mkern-0.5mu})\simeq \operatorname{Hom}_\mathbf Z(N,\overline{\!M\mkern-0.5mu}).$$ Now consider the commutative diagram: $$\begin{matrix} \operatorname{Hom}_R(N,\widetilde{M})&\xrightarrow{\operatorname{Hom}_R(u,1)}&\operatorname{Hom}_R(N',\widetilde{M})\\ \downarrow &&\downarrow\\ \operatorname{Hom}_\mathbf Z(N,\overline{\!M\mkern-0.5mu})&\xrightarrow{\operatorname{Hom}_\mathbf Z(u,1)}& \operatorname{Hom}_\mathbf Z(N',\overline{\!M\mkern-0.5mu}) \end{matrix}$$ The two vertical maps are isomorphisms, and the bottom horizontal map is surjective since $\overline{\!M\mkern-0.5mu}$ is an injective $\mathbf Z$-module, hence the top horizontal map is surjective, as was to be proved.