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 :
Let $R,S,T$ be rings, $h$ a homo of rings from $R$ to $S$ and I 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
I have already proven that
i)- F(h) is a group hom
ii) $F$ maps $R^\times$ to $S^\times$
iii)- F maps the identity to the identity
I am missing:
iv)- there is compatibility with composition
I am struggling with iv). Keeping track of the domains of composition is making it complicated. This is my try:
I have to prove that $F(g \circ f)=F(g)\circ F(f)$
where $f: R \rightarrow S$ and $g: S \rightarrow T$ For that I have to check that the domains are the same and that they map to the same elements:
domains:
$D=\text{dom} F(g)\circ F(f)$
$=\{x \in \text{dom}F(f) s.t. F(f)(x)\in \text{dom} F(g)\}$
$=\{x \in R^\times s.t. F(f)(x)\in S^\times\}$.....
But how do I conclude that it is equal to $=\text {dom}(F(g \circ f)) $? To start with what is $\text {dom}(F(g \circ f))=?$ It is for sure a subset of $F(R)=R^\times$, but I am not sure wether it is actually equal to it.
mapping: Provided the domains are equal,
$\forall x \in D$
$F(g \circ f)(x)=(g \circ f)|_{R^\times}(x)= (g \circ f )(x)=g(f(x))=g(f|_{R^\times}(x))=g|_{S^\times}(f|_{R^\times}(x))=F(g)(F(f)(x))=(F(g)\circ F(f))(x)$
How do I complete this part? Is the rest OK?
Lets start with $$R \xrightarrow[]{f} S \xrightarrow[]{g} T$$
morphisms. Now $F(R)=R^\times$. And so $F(f):R^\times \to S^\times$ and $F(g):S^\times\to T^\times$. Lets assume these are restrictions (which are well defined if we consider unital homomorphisms btw).
So, the situation is that we have a following sequence:
$$R^\times \xrightarrow[]{F(f)} S^\times \xrightarrow[]{F(g)} T^\times$$
which clearly shows that $F(g)\circ F(f)$ has $R^\times$ as its domain.
On the other hand, apply $F$ to $g\circ f$ morphism, which is a morphism $R\to T$. This gives us $F(g\circ f)$ which (by definition) is a morphism $R^\times\to T^\times$.
This shows that $F(g\circ f)$ and $F(g)\circ F(f)$ at least share domain and range.
Concrete formulas are also the same. Easy to check, because $F(f)(x)$ is literally defined as $f(x)$, it only has domain and range restricted. Which you've explicitely checked in your last sequence of equalities.