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.

As for "what does this buy us", representable functors are nice. For one thing, they are guaranteed to preserve all existing categorical limits. That's one nice property I know, and I'm sure that other people on here could tell you many more.

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.
  • Free objects do not exist (ever) in the category of fields.
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