# Understanding representable functors

I'm trying to wrap my head around the concept of representable functors - even though I know the definitions. I'm referencing the second page here for the example I want to understand about the forgetful functor from the category of Group to Set. Why doesn't the trivial group $\{1\}$ satisfy the representation following a similar diagram to that of the identity functor on Set? Using the same notation as in that example, I can take a group homomorphism $f : S \rightarrow T$ (where $S$ and $T$ are groups now), and for any group homomorphism $g \in \textbf{Hom}(\{1\}, S)$ and $h \in \textbf{Hom}(\{1\}, T)$ it and seemingly replicate the argument. I feel the place where it may break down is that the components can only ever map to the identity elements this way but how exactly does using $\mathbb{Z}$ fix this? And what does this representability buy us?

In the case of a map $\mathbb{Z} \to S$, we are still required to map the identity (i.e. zero) to the identity. However, the element $1$ can map to any element in $S$ we like, and any such choice uniquely determines a group homomorphism. Therefore the maps $\mathbb{Z} \to S$ are in bijection with the elements of $S$, just as in the case of sets, where the maps $\{1\} \to A$ are in bijection with the elements of the set $A$. See below for more exposition on this.
This is an instance of a more general pattern: $\mathbb{Z}$ is the free object on a single element in the category of groups. Any time such a thing exists in a category with a forgetful functor to $\mathsf{Set}$, the forgetful functor is represented by that object. Other examples are
• $\mathbb{Z}$ is the free object on one element in the category of abelian groups (just like in groups).
• $\mathbb{Z}[x]$ is the free object on the element $x$ in the category of rings with unity, and in the category of commutative rings with unity.
• If $R$ is a commutative ring, $R[x]$ is the free object on the element $x$ in the category of $R$ algebras.