$\newcommand{\Z}{\mathbb{Z}}$ I want to make sure I'm understanding the introduction to category theory I'm reading.
I know that the association of a ring to its group of units is functorial, but I need to show that it is not faithful. From what I understand, it suffices to find a pair of distinct morphisms whose images under the functor agree. Does the following approach make sense?
Atempt: Let $f: \Z/2\Z \rightarrow \Z/3\Z$ and $g:\Z/2\Z \rightarrow \Z/6\Z$ be two distinct ring homomorphisms. Then $(\Z/2\Z)^\times = \{1\}$, and since $(\Z/3\Z)^\times = \{1,2\}$ and $(\Z/6\Z)^\times = \{1,5\}$, we know that as two element groups $(\Z/3\Z)^\times \cong (\Z/6\Z)^\times \cong C_2$. Therefore $Ff:\{1\} \rightarrow C_2$ and $Fg:\{1\} \rightarrow C_2$ and since there is only one map from the trivial group into $C_2$ we have $Ff=Fg$ while $f \neq g$, demonstrating that $F$ is not faithful.
Thanks in advance for any clarifications.
Edit: I think the misconception came from the fact that I had understood the definition of being faithful as being injective on morphisms, which isn't technically correct, as evidenced by Daniel Schepler's answer. A faithful functor is injective on parallel pairs of morphisms $f,g: c \rightrightarrows c'$ in a category $\mathbf{C}$.
No, the definition of a faithful functor involves two morphisms $f, g \in \operatorname{Hom}_{\mathcal{C}}(X, Y)$ with the same domain and the same codomain in the "source category" of the functor. It is not sufficient just to have the codomains map to isomorphic (or even equal) objects under the functor in the "target category" of the functor. (Though a somewhat related question here is whether a functor reflects isomorphisms: this is defined to mean that whenever $F(f)$ is an isomorphism, then $f$ is also an isomorphism. It turns out that the group of units functor also does not reflect isomorphisms.)
Also, in your example, it turns out that there are no morphisms of unital rings $f : \mathbb{Z} / 2 \mathbb{Z} \to \mathbb{Z} / 3 \mathbb{Z}$.
As a hint for how to proceed, maybe try looking at: