Is this function from $M_g \to \mathbb{Z} \otimes M$ well defined? Given a $G$-module $M$ (a $\mathbb{Z}G$-module), let $M_g$ denote the group of coinvariants, that is $M/\langle m -gm \rangle$. 
If we regard $\mathbb{Z}$ as a right $\mathbb{Z}G$-module with trivial $G$-action, is the function $f:M_g \to \mathbb{Z} \otimes_{ZG} M,$ where $f(\bar{m}) = 1 \otimes m$ well defined?
 A: Actually you want to prove $M_G \cong \mathbb{Z} \otimes_{\mathbb{Z}G} M$, right?
Low-level answer:
We have a homomorphism of $\mathbb{Z}G$-modules $M \to \mathbb{Z} \otimes_{\mathbb{Z}G} M$ given by $m \mapsto 1 \otimes m$. It maps $gm$ to $1 \otimes gm = g1 \otimes m = 1 \otimes m$ (since $G$ acts trivially on $\mathbb{Z}$). Hence, it maps $m-gm$ to zero. The fundamental theorem for homomorphisms gives us an extension to a $\mathbb{Z}G$-linear map $M_G \to \mathbb{Z} \otimes_{\mathbb{Z} G} M$. Actually it is an isomorphism, one can construct an inverse homomorphism by means of the universal property of the tensor product. I won't explain this, because it is a good exercise, and on the other hand the other answers below show that, in fact, no computation is needed.
Middle-level answer:
We have an isomorphism of $\mathbb{Z}G$-modules $\mathbb{Z} \cong \mathbb{Z}G / \langle 1-g : g \in G\rangle$. Now, use the general isomorphism $R/I \otimes_R M \cong M/IM$ to get the claim.
High-level answer:
Obviously $G$ acts trivially on $M_G$, and if $N$ is a $\mathbb{Z}$-module, regarded as a $\mathbb{Z}G$-module on which $G$ acts trivially, we have
$$\hom_{\mathbb{Z}}(\mathbb{Z} \otimes_{\mathbb{Z}G} M,N) \cong \hom_{\mathbb{Z}G}(M,N) =\{f \in \hom_{\mathbb{Z}}(M,N) : f(gm)=g f(m)\} =\{f \in \hom_{\mathbb{Z}}(M,N) : f(gm-m)=0\} = \hom_{\mathbb{Z}}(M_G,N).$$
The Yoneda Lemma implies $\mathbb{Z} \otimes_{\mathbb{Z}G} M \cong M_G$.
In other words, both functors $M \mapsto M_G$ and $M \mapsto \mathbb{Z} \otimes_{\mathbb{Z}G} M$ are left adjoint to the functor from $\mathbb{Z}$-modules to $\mathbb{Z}G$-modules given by the trivial $G$-action, thus they are isomorphic.
A: Suppose $\bar{m} = \bar{n}$.  Then $1 \otimes m - 1 \otimes n = 1 \otimes (m - n)$.  By definition, $m -n \in M_g$.  Let's assume without loss of generality that $m - n = a - ga$ where $a \in M$, $g \in M$.  So $1 \otimes (m - n) = 1 \otimes (a - ga) = 1 \otimes a - 1 \otimes ga =  1 \otimes a - 1 \otimes a = 0$, and the function is well defined.
