Showing that $F$ is not representable [closed]

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 :)

closed as off-topic by José Carlos Santos, Adrian Keister, Arnaud D., Shailesh, Alexander Gruber♦Jan 17 at 3:00

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

• Very nice answer. Thank you very much! – Greg Jan 12 at 16:00
• 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 $\mathbb{C}\to\mathbb{C}$ – Jeremy Rickard Jan 12 at 17:17