Help to find all different cases need for proof about homomorphism from Z to R I am a bit confused about why my professor approached the following a certain way, and also why it cannot be done differently.
The question is to prove that for any ring R we there is a unique homomorphism $$\phi : \mathbb{Z} \to R$$
Now here is my approach:
Define $\phi(0)=0_{R}$ and $\phi(1)=1_{R}$
and define $\phi(x)=x\phi(1)$ $\forall x \in \mathbb{Z}$
Then it is just a matter of verification that
$\phi(x+y)=(x+y)\phi(1)=\phi(x)+\phi(y)$
and $\phi(xy)=xy\phi(1)=\phi(x)\phi(y)$
as integers all obey these rules and we are just applying ring axioms.
However, the proof given in class took into account seemingly endless many cases
for example $x,y \ge 0$ and $\phi(x+y)$
$x,y \ge 0$ and $\phi(-x-y)$
$x,y \ge 0$ and $x \ge y$ etc
I am having a lot of trouble understanding why this is needed at all, and why we cannot just do the general case. Furthermore I am not even sure if I were to do it by cases what all the different cases would be.
Does anyone have any insight or opinions for me?
Thanks
Update: 
I am just going to learn the proof using cases as it seems not doing so only makes things even more complicated .
I am now looking for someone to help me to understand all the diffirent cases I must consider
 A: You say that "this is just a matter of verification", but you need to define $x\phi(1)$ recursively, using the homomorphism property. Therefore we have $\phi(1)=1_R$, and then we conclude that $\phi(n)=\sum_{1=1}^n\phi(1)$ for all positive $n$. Indeed $\phi(2)=\phi(1+1)=\phi(1)+\phi(1)$, etc. Also $\phi(0)=0$ and then $\phi(n)=\sum_{i=1}^{|n|}-\phi(1)$ for all negative $n$.
A: How do you define $x\phi(1)$, for $x\in\mathbb{Z}$?  Maybe you are taking for granted that this is already defined, but if you think about how you would define it rigorously, you would probably do something like the following.  Define $x\phi(1)$ for $x\geq 0$ by induction via the formula $$(x+1)\phi(1)=x\phi(1)+\phi(1),$$ and then define $x\phi(1)$ to be the additive inverse of $(-x)\phi(1)$ if $x<0$.  If you use this definition, then to prove a statement like $(x+y)\phi(1)=x\phi(1)+y\phi(1)$ for all $x,y\in\mathbb{Z}$, you will necessarily have to split into cases according to the signs of $x$, $y$, and $x+y$, since the very definition of the expressions involved are given by splitting into cases.
If you can give a definition of $x\phi(1)$ that does not treat the cases $x\geq 0$ and $x<0$ separately, you can avoid having to consider cases here, but it is not obvious how you would make such a definition.
