# Does Yoneda embedding reflect equivalent categories?

Let $\mathsf{Cat}$ denote the category of small categories. For categories $\mathcal A$ and $\mathcal B$ in $\mathsf{Cat}$, let $[\mathcal A,\mathcal B]$ denote the category whose objects are functors form $\mathcal A$ to $\mathcal B$ and morphisms are natural transformation between those functors. My question is

Given a functor $F:\mathcal A\to\mathcal B$. Suppose for any $\mathcal C\in\text{ob}\mathsf{Cat}$ we have $F^*:[\mathcal B,\mathcal C]\to[\mathcal A,\mathcal C]$ is an equivalence, or for any $\mathcal C\in\text{ob}\mathsf{Cat}$ we have $F_*:[\mathcal C,\mathcal A]\to[\mathcal C,\mathcal B]$ is an equivalence. Can we deduce that $F$ is an equivalence?

• Partial answer: if $F_* : [\mathcal{C},\mathcal{A}] \to [\mathcal{C},\mathcal{B}]$ is an equivalence for all categories $\mathcal{C}$, then in particular $F_* : [\mathbb{1},\mathcal{A}] \to [\mathbb{1},\mathcal{B}]$ is an equivalence, where $\mathbb{1}$ is the terminal category. But $\mathcal{A} \cong [\mathbb{1},\mathcal{A}]$ and $\mathcal{B} \cong [\mathbb{1},\mathcal{B}]$, and these isomorphisms commute with $F$ and $F_*$, so that $F$ is also an equivalence. Apr 27 '17 at 20:10
• I'm removing the "higher category theory" tag; and I question the use of the phrase "Yoneda Embedding" in the title, as this has nothing to do with Yoneda. Apr 27 '17 at 22:41
• ("Monoidal categories" is also off topic, but less egregiously.) Apr 27 '17 at 22:43
• @MaliceVidrine: It has very much to do with Yoneda! Apr 28 '17 at 7:58
• @MaliceVidrine It has something to do with Yoneda. And I think that's enough to justify my using this word in the title. period Apr 28 '17 at 19:17

Yes. This is a special case of the $2$-categorical Yoneda Lemma. Here is a direct proof.
Assume that $F^*$ is an equivalence for all categories $C$. In particular, $F^* : [B,A] \to [A,A]$ is essentially surjective. Choose some $G : B \to A$ with $GF \cong \mathrm{id}_A$. We have $FG \cong \mathrm{id}_B$ since $FGF \cong \mathrm{id}_B F$ and $F^* : [B,B] \to [A,B]$ is fully faithful.
You can use the same proof for $F_*$. Or you can give a quick argument as shown by Clive Newstead in the comments.