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I'm working through the section on adjoint functors in Borceaux's Categorical Algebra. If $F: C \to D$ is a functor, and $d \in |D|$, he defines a reflection of $d$ across $F$ to be a pair $(L(d), \eta_d)$, where $L(d) \in |C|$ and $\eta_d: d \to FL(d)$ such that for any arrow $\phi: d \to F(c)$, there exists a unique $\phi^*:L(d) \to c$ such that $F(\phi^*)\circ \eta_d = \phi$. (In other words, $L(d), \eta_{d}$ is an initial object in the category $d \downarrow F$ of arrows under $d$.) The dual notion is a coreflection.

Given functors $F: C \to D$ and $G: D \to C$, he gives some equivalent characterizations for an adjunction $G \vdash F$. The following two involve the unit and counit:

  1. There exists a natural transformation $\eta: \text{Id}_D \to F \circ G$ such that for each $d \in |D|$, $G(d)$ is a reflection of $d$ along $F$.
  2. There exists a natural transformation $\epsilon: G \circ F \to \text{Id}_C$ such that for each $c \in |C|$, $F(c)$ is a coreflection of $c$ along $G$.

I'm looking for a direct proof of (1) $\implies$ (2). I'm 75% of the way there. So far I have: Let $\epsilon$ take $c \in |C|$ to ${\text{Id}_{F(c)}}^*: GF(c) \to c$ (with the asterisk notation as defined above). I've checked that it's natural using the reflection property in (1). Now I want to show that for any $c \in |C|$, we have that $F(c), \epsilon_c$ is a coreflection: i.e. given any $d \in |D|$ and $\phi:G(d) \to c$, there is a unique $\phi^\star: d \to F(c)$ such that ${\text{Id}_{F(c)}}^* \circ G(\phi^\star) = \phi$.

I'm stuck on this. I'd like to prove it as a gateway to proving all the usual adjunction stuff (i.e. I'd like to prove it on its own without taking for granted all the usual stuff about adjunctions).

Any tips?

Edit: I see that the dual question is asked and answered in this post. I will try to work it out from there.

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    $\begingroup$ Are you saying this post has a coduplicate? $\endgroup$ Sep 15, 2015 at 0:05

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You haven't got the definition of coreflection straight. $\epsilon_c$ is a coreflection if it's a final arrow in $G\downarrow c$, that is, for every $d\in|D|$ and $\phi:Gd\to c$ there should be a unique $\phi^*: d\to F(c)$ with $\epsilon_c\circ G(\phi^*)=\phi$. Is that enough to get you there?

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  • $\begingroup$ Unfortunately (?) I just wrote it wrong in the question, but I was thinking about it right, so my problem remains. $\endgroup$
    – Eric Auld
    Sep 14, 2015 at 18:49
  • $\begingroup$ Just found this post, which may be the same question upon inspection. math.stackexchange.com/questions/690971/… $\endgroup$
    – Eric Auld
    Sep 14, 2015 at 18:53
  • $\begingroup$ Yeah, that appears to be the same question, just going from left to right instead of right to left. $\endgroup$ Sep 14, 2015 at 22:13

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