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Consider the functor $F: B \rightarrow Set$ where $B$ is a locally small category.

Is it true that if $F$ has a left adjoint then it is representable?

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  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. $\endgroup$ – José Carlos Santos Jan 26 '18 at 8:38
  • $\begingroup$ @JoséCarlosSantos I looked for a counterexample among some common categories but couldnt find one, so I'm suspecting it's true. I understand we'd like to construct the natural isomorphism between $F$ and $Hom(A,-)$ using the adjoint somehow, but I honestly have no ideas where to start. Hints would be as appreciated as an explanation. $\endgroup$ – user525096 Jan 26 '18 at 8:58
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Yes, it is. See for instance "Algebraic Theories: A Categorical Introduction to General Algebra" by J. Adámek, J. Rosický, E. M. Vitale, page 7 (chapter 0, section 0.10 about representable functors).

Proof: Let $L \dashv F$ and $1 = \{*\}$. Then, \begin{align} Fb \cong \mathbf{Set}(1,Fb) \cong B(L1,b) \end{align} naturally in $b \in B$, hence $F \cong H^{L1}$.

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  • $\begingroup$ Thanks for the reference. Does this remark have a number? All I see on page 7 is a lemma about Yoneda embedding preserving finite coproducts, am I looking at the wrong page? $\endgroup$ – user525096 Jan 26 '18 at 9:27
  • $\begingroup$ Could you explain the notation in the last symbol, $H^{F1}$? I see that you showed $F \cong B(L1, -)$ which is what we wanted, but I'm not sure how $H^{F1}$ expresses that. Did you mean $H^{L1}$ or am I missing something? $\endgroup$ – user525096 Jan 26 '18 at 9:46
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    $\begingroup$ Maybe we have different editions. I refer to chapter 0 (Preliminaries), section 0.10 (Representable functors). $\endgroup$ – Taroccoesbrocco Jan 26 '18 at 9:47
  • $\begingroup$ Yeah I think this might have changed across editions, the one under this link iti.cs.tu-bs.de/~adamek/algebraic_theories.pdf seems to be missing what you posted. $\endgroup$ – user525096 Jan 26 '18 at 9:49
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    $\begingroup$ Of course, thank you! It is $H^{L1}$. I fixed the typo. $\endgroup$ – Taroccoesbrocco Jan 26 '18 at 9:49

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