Question: > 
> Given a group homomorphism $R \otimes\_{\Bbb{Z}} M \rightarrow M$, I need to show that this makes $M$ into a left $R$-module. 
> 
> 
> 

We can see that

$\bullet$ $r(m+m') = r \otimes (m+m') = r \otimes m + r \otimes m' = rm + rm'$

$\bullet$ $(r+r')m = (r+r') \otimes m = r \otimes m + r' \otimes m = rm + r'm$

$\bullet$ $(rr')m = rr' \otimes m = r \otimes r'm = r(r'm)$

$\bullet$ $1m = 1 \otimes m = m$

I have a question about the third one, where I used $rr' \otimes m = r \otimes r'm$. Is this valid? Because we are tensoring over $\Bbb{Z}$ and not over $R$, right? 

Thanks in advance

Answer: 
I don't think this is true, take for instance $R=\Bbb Z[X]$, then you can define a morphism $\Bbb Z[X]\otimes\_{\Bbb Z}M$ by sending $P\otimes m$ to $$\left(\sum\_{i\geq 0} a\_ic\_i\right)m$$ where $(c\_n)\_{n\in\Bbb N}$ is some sequence of integers, and $P=\sum\_{i\geq 0}a\_iX^i$. This, in general, does not define a $\Bbb Z[X]$-module structure on $M$.

---

Actually, there is no reason to assume a general map of groups should map $1\otimes m$ to $m$ either. For instance, this fails when $R=\Bbb Z/2\Bbb Z$ and $M=\Bbb Z/3\Bbb Z$, since $\Bbb Z/2\Bbb Z\otimes\_{\Bbb Z}\Bbb Z/3\Bbb Z=0$.
One could try with $rm=\varphi(r\otimes m)$, where $\varphi\colon R\otimes\_{\mathbb{Z}}M\to M$ is the given group homomorphism.

Then $(r+s)m=\varphi((r+s)\otimes m))$, $r(m+n)=\varphi(r\otimes(m+n))$ don't pose problems.

The problem is with
$$
r(sm)=\varphi(r\otimes(sm))=\varphi(r\otimes(\varphi(s\otimes m)))
$$
while
$$
(rs)m=\varphi((rs)\otimes m))
$$

So, without some more hypotheses on $\varphi$, I'd say this is not true.