Trivial module functor is $\text{Hom}_\Bbb Z(\Bbb Z,-)$

Question

Weibel - Homological Algebra - Page 161

Considering $\Bbb Z$ as a $\Bbb Z-\Bbb ZG$-bimodule, the "trivial module functor" from $\Bbb Z$-mod to $\Bbb ZG$-mod is the functor $\text{Hom}_\Bbb Z(\Bbb Z,-)$.

Can someone tell me why this is so?

• I understand that if I am given a $\Bbb Z$-module $M$, then $\text{Hom}_\Bbb Z(\Bbb Z,M)\cong M$, since I can take $\varphi_m(1)=m$ for any $m\in M$ (and these are all the homs).
• I understand that we want to have trivial $G$-module structure on $M$, i.e. $g\cdot m= m$ for every $g\in G$ and $m\in M$.
• I understand taking $\Bbb Z$ as a $\Bbb Z-\Bbb ZG$-bimodule makes $\Bbb Z$ a trivial right $G$-module.

Answer 1

For an Abelian group $\textrm{Hom}(\Bbb Z,A)\cong A$ as an $A$-module. Then $a\in A$ corresponds to $\varphi_a:n\mapsto na$. Then as a right $\Bbb ZG$-module $n\cdot g=g$ for all $n\in\Bbb Z$. Then $\textrm{Hom}(\Bbb Z,A)$ becomes a left $\Bbb ZG$-module by $g\cdot \psi n\mapsto\psi(n\cdot g)$. Under this action it's easy to check that $g\cdot\varphi_a=\varphi_a$: the $G$-action is trivial.

Note that the side on which the ring acts on the module is reversed for the first argument of the hom-functor.