Showing that $F$ is not representable

Question

As I'm trying to find (counter)examples of representable functors, I tried looking up some instructive examples.

One of the counterexamples I'm having trouble with, is the following: Show that the functor $$ F:CRings \rightarrow Sets: R \mapsto \left\{r^2 \rvert r \in R\right\}$$ is not representable. Any help is appreciated :)

Answer 1

A functor $F$ is representable if and only if its category of elements has an initial object.

So let $\mathcal{C} = \mathbf{CRing}$ and let $F : \mathcal{C} \to \mathbf{Set}$ be as in your question. The objects of its category of elements $\int^{\mathcal{C}} F$ are pairs $(R, r^2)$ where $r \in R$, and a morphism $f : (R, r^2) \to (S, s^2)$ is a ring homomorphism $f : R \to S$ such that $f(r^2) = s^2$.

We prove $\int^{\mathcal{C}} F$ has no initial object.

Let $f : (R, r^2) \to (\mathbb{C}, -1)$ be an arbitrary morphism in $\int^{\mathcal{C}} F$.

Then $f : R \to \mathbb{C}$ is a ring homomorphism such that $f(r^2) = -1$, and so $f(r) = \pm i$.

Define $g : R \to \mathbb{C}$ by $g(x) = \overline{f(x)}$ for each $x \in R$. Then:

• $g$ is a ring homomorphism since it is a composite of two ring homomorphisms;
• $g \ne f$, since $g(r) = \overline{f(r)} = \overline{\pm i} = \mp i \ne f(r)$;
• $g(r^2) = \overline{f(r^2)} = \overline{-1} = -1$;

So $g$ is a morphism $(R, r^2) \to (\mathbb{C}, -1)$ in $\int^{\mathcal{C}} F$ distinct from $f$.

But this means that there is no initial object in $\int^{\mathcal{C}} F$, since if there were, there would be a unique morphism from that object to $(\mathbb{C}, -1)$ in $\int^{\mathcal{C}} F$, contrary to what we just showed.