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 :)

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    $\begingroup$ do you assume that the rings have a unit element ? $\endgroup$ – Tsemo Aristide Jan 12 '19 at 14:02
  • $\begingroup$ Yes, I assume that they have a unit element $\endgroup$ – Jonas Jan 12 '19 at 14:19
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    $\begingroup$ Yes, I think they are assumed to be commutative. I will edit the question. $\endgroup$ – Jonas Jan 12 '19 at 14:55
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    $\begingroup$ I found these short notes on Google: pi.math.cornell.edu/~zbnorwood/ucla/files/repfunctors.pdf . The given example is exactly yours. $\endgroup$ – Crostul Jan 12 '19 at 16:36

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.

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  • $\begingroup$ Very nice answer. Thank you very much! $\endgroup$ – Jonas Jan 12 '19 at 16:00
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    $\begingroup$ Or (and I suspect this is essentially the same argument in a different form) if $F$ were representable then it would preserve equalizers. But the “squares” functor doesn’t preserve the equalizer of the identity and complex conjugation as ring homomorphisms $\mathbb{C}\to\mathbb{C}$ $\endgroup$ – Jeremy Rickard Jan 12 '19 at 17:17

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