Category Theory – Existence of Functor from Ring to Grp Sending Ring to Group of Units

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I am just starting with functors and all the examples about functors that I have seen don't gave explicit proofs that something is a functor. I am trying to prove that there is a functor from Ring to Grp that sends a ring $R$ to its group of units $R^*$. This exercise looks pretty tautological.

My proof:

In the present case $F$ is just a forgetful functor,right? so I have that for a ring $R$ and a ring homomorphism $f$, the functor $F:Ring \rightarrow Grp$ is defined by

$F(R)=R^* $ and $ F(f)=f|_{R^*} $

i.e. the same $f$ where I just remember the property $f(ab)=f(a)f(b)$ restricted to the subgroup $R^*$

So to prove its existence,I guess all it suffices to do it's verify that

i) $F(id_R)=id_{F(R)}$ and

ii) $F(f\circ g)=F(f)\circ F(g)$.

Then

i) $F(id_R)=id_R|_{R^*}=id_{R^*}=id_{F(R)}$

ii) Let $f: R_1 \rightarrow R_2, g: R_2 \rightarrow R_3$ be ring homomorphisms, the $g\circ f: R_1 \rightarrow R_3$ is a ring homomorphism and

$F(g\circ f)=(g \circ f)|_{R_1^*}=g|_{f(R_1^*)} \circ f|_{R_1^*}=(\square)$

Because $R_1^*$ is a subgroup and $f$ a ring homomorphism, $f(R_1^*)$ is subgroup of $R_2$ it remains to prove that $f(R_1^*)=R_2^*$, so that I can have:

$(\square)=g|_{R_2^*} \circ f|_{R_1^*}=F(g)\circ F(f)$.

1 Is this the way to go? How do I complete it (specially part ii)? It's suspicious that I don't use the homo property

2 How does one prove in general the existence of a functor? My guess is by giving the formula for the morphism and checking the two requisites, but am not sure if it is always possible to have a formula for $F(f)$

Best Answer

  1. The stuff about identities and composition in this case is trivial. The actual things you need to prove are that 1) $F(f)$ actually sends the group of units $R^{\times}$ of $R$ to the group of units $S^{\times}$ of $S$ - here you need to actually use that $f$ is a ring homomorphism, and that 2) $F(f)$ is a group homomorphism - here again you need to use that $f$ is a ring homomorphism.

  2. This is an extremely general question. Mostly you build up a set of intuitions for which sorts of constructions are "obviously" functorial. For example forgetful functors, left adjoints, right adjoints, representable functors, compositions of functors. Most examples in practice can be built like this and for the others there are more details to check. The example of the group of units is a right adjoint and also representable by the ring $\mathbb{Z}[x, x^{-1}]$.

As an exercise to keep you on your toes, show that sending a ring $R$ to its set $R \setminus R^{\times}$ of non-units is not a functor (to $\text{Set}$). (More precisely, that it's not a subfunctor of the forgetful functor.)

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