Check my work? Equivalent functors and representability

94 Views Asked by At

This is a question about exercise 2.1.iii in Emily Riehl’s Categories in Context. The statement is the following.

Suppose $F\colon C \to \mathrm{Set}$ is equivalent to $G\colon D \to \mathrm{Set}$ in the sense that there is an equivalence of categories $H\colon C \to D$ so that $GH$ and $F$ are naturally isomorphic.

  • If $G$ is representable, then is $F$ representable?
  • If $F$ is representable, then is $G$ representable?

Recall that $G\colon D \to \mathrm{Set}$ is representable if there is an object $d$ in $D$ and a natural isomorphism $G \cong D(d,-)$. Note that for the question to make sense, $G$ representable means $D$ and therefore $C$ must be locally small.

I believe I have an argument that makes the answer to both parts “yes”, but I could use reassurance that I’m not playing too fast and loose. Check my work?


Assume $G$ is represented by $D(d,-)$. The assumption that $H\colon C \to D$ defines an equivalence of categories means that there exists $K\colon D \to C$ that witnesses this equivalence.

Claim: $F$ is represented by $C(Kd,-)$. Since $KH$ is naturally isomorphic to the identity functor $C \to C$, “whiskering” with $C(Kd,-)$ $C(Kd,-)$ yields a natural isomorphism $C(Kd,-)\cong C(Kd,KH-)$. Equivalences of categories are full and faithful, so the function sending the arrow $Kf\colon Kd\to KHx$ to $f\colon d \to Hx$ is a natural isomorphism $C(Kd,KH-)\cong D(d,H-)$. (This is the part I’m nervous about.) By assumption, this latter is naturally isomorphic to $GH$, which is naturally isomorphic to $F$.

1

There are 1 best solutions below

0
On BEST ANSWER

Yes, the argument works.

The content of $K\colon D \to C$ being full and faithful between locally small categories is that the map $f\mapsto Kf$ is an isomorphism $D(x,y)\to C(Kx,Ky)$. This isomorphism is natural in $y$ because $K$ is a functor.