I have to prove that there is a functor Ring $\rightarrow$ Grp that sends a ring $R$ to its group of units $R^×$. In order to to that I would start like this:
Let $R,S,T$ be rings, $h$ a homo of rings from $R$ to $S$ and define
$F: \text {Obj}(\text {Ring)}\rightarrow \text {Obj}(\text {Grp})$ as $F(R)=R^×$ ?
and
$F: \text {Hom}_{\text { Ring} }(R,S)\rightarrow \text{Hom}_{\text { Grp}}(F(R),F(S))$ as $F(h)=h|_{R^×}$, meaning the ring hom $h$ restricted to the groups of units and seen as a group hom
Is this beginning of the proof, written rigorously and correctly ? I know I have to check that
i)- F(h) is a group hom
ii)- F maps the identity to the identity
iii)- there is compatibility with composition
1 Can I really start off by defining $F(R)=R^×$ ? I mean can I regard it as true "by construction" or is it something to prove?. That looks like assuming part of the thesis is true. But then how can I proof that part if I can't even write a formula for $F(R)=\cdot$ ? What should I have written instead of "define $F(R)=R^×$"? I already know how to prove the other parts but I am struggling with this part. Actually I assumed this part to prove them
2 Is there something else I am missing?
Best Answer
The thing you've been given to prove might suffer from poor wording. I would rephrase it as:
And now it is clear that the burden here is to define $F$ on morphisms, since it is already defined on objects.
Still the sentence "show that there is functor such that $F(R)=R^\times$" is fine. You can still simply define $F(R)=R^\times$ on objects and it will satisfy the requirement. Of course this is kind of trivial, its like saying "show that there is function $f:\mathbb{R}\to\mathbb{R}$ such that $f(x)=x^2$". Yeah, it is simply $f(x)=x^2$, it is a valid definition.
That's why we can safely assume that the intention of the author was to focus on morphisms, not on objects.
Now, the devil is always in details. Given a ring homomorphism $h:R\to S$, your idea to consider $h$ restricted to $R^\times$ is correct. However that's not enough. Formally $h_{|R^\times}$ is a function $R^\times\to S$. But that is not correct, our $F(h)$ is supposed to be a function $F(R)\to F(S)$, i.e. a function $R^\times \to S^\times$. So the correct definition of $F(h):R^\times\to S^\times$ is simply $F(h)(x)=h(x)$, i.e. $F(h)$ is a restriction of $h$ on both domain and range.
However, for that definition to be valid we need to know that $h(R^\times)\subseteq S^\times$. Whether this is true depends on what $Ring$ actually is. Sometimes homomorphisms between rings are defined as functions $h:R\to S$ such that $h(x+y)=h(x)+h(y)$ and $h(xy)=h(x)h(y)$. Also known as non-unital homomorphisms. Such homomorphisms in general won't satisfy $h(R^\times)\subseteq S^\times$, for example when $h(x)=0$ is constant. And such homomorphism does not induce a homomorphism $R^\times\to S^\times$. So you need to be careful.
This is true however if we consider $Ring$ as category with unital homomorphisms as morphism. A unital homomorphism additionally satisfies $h(1)=1$ condition. Now, if $x\in R^\times$ then
$$1=h(1)=h(xx^{-1})=h(x)h(x^{-1})$$ $$1=h(1)=h(x^{-1}x)=h(x^{-1})h(x)$$ which implies that $h(x)$ is invertible (with $h(x^{-1})$ as its inverse), i.e. $h(x)\in S^\times$.
Note that some authors simply define ring homomorphism as unital ring homomorphism and omit the "unital" word.