Let $\mathcal{C}$ be a locally small category, and $F : \mathcal{C} \rightarrow \mathcal{Set}$ functor. Then : $$F \ \text{has left adjoint} \implies F \ \text{is representable} \implies F \ \text{ preserves limits in} \ \mathcal{C}$$ Is there any counterexample for the converse implications? Explicitly :
- A non representable limits-preserving functor
- A representable functor with no left adjoint
I know that if $\mathcal{C}$ has coproducts, then having a left adjoint is equivalent to being representable, but I couldn't find a counterexample anyway.
A non representable limits-preserving functor
Take for example a poset category associated to $\mathbb Z$ with constant functor $F(x) = \{0\}$. Clearly it is limit preserving, as every limit of singletons is a singleton itself. But it is not representable - the supposed isomorphism $F(-) \simeq Hom(m, -)$ leads to an absurd $1 = F(m-1) \simeq Hom(m, m-1) \simeq \emptyset$
A representable functor with no left adjoint
Consider a category of non-empty sets. Since left adjoint functor preserve colimits, $F(\emptyset)$ is initial object in $\mathcal C$, which does not exist. So there is no functor $\mathcal C \rightarrow \mathrm{Set}$ with left adjoint.