Why are the two definitions of reflective subcategory equivalent?

Question

From Wikipedia: Let $A \subseteq B$, where $B$ is a category and $A$ is a full subcategory. On Wikipedia there are two equivalent definitions of reflective subcategory:

1. the inclusion functor $i$ has a left adjoint $F : B \rightarrow A$
2. For any $b \in B$, there is an $a_b \in A$ and a morphism $t : b \rightarrow i(a_b)$, such that for any $a \in A$ and morphism $f : b \rightarrow i(a)$, there is a unique morphism $\tilde{f} : a_b \rightarrow a$ such that $i(\tilde{f}) \circ t = f$

I'm trying to prove that the first definition implies the second. Suppose $F$ is left adjoint to $i$. Let $\tau_{a,b}$ be the bijections $Mor_A(F(b), a) \rightarrow Mor_B(b, i(a))$. I think that $a_b = F(b)$, and the morphism $t : b \rightarrow i(F(b))$ is the unit $\eta_b$.

I'm also guessing that $\tilde{f} = \tau_{a,b}^{-1}(f)$. Therefore, it is required to prove $\tau_{a,b}^{-1}(f) \circ \eta_b = f$.

I'm stuck trying to prove this. Both $\tau$ and $\eta$ appear in this. I'm also not sure how to use the fact that $i$ is an inclusion from a subcategory, because I don't think this theorem is true for a general adjoint pair.

Answer 1

$\newcommand{\A}{\mathsf{A}}\newcommand{\B}{\mathsf{B}}\require{AMScd}$You say:

I don't think this theorem [$\tau_{a,b}^{-1}(f)\circ\eta_b=f$] is true for a general adjoint pair

This can't even be stated for an adjoint pair since it relies on the fact that $i(x)=x$ and $i(F(x))=F(i(x))$ making $\epsilon$ composable with $\eta$, which isn't usually possible.

However, the general theorem at play here is true: if $L:\B\rightleftarrows\A:R$ are a left-right pair of adjoint functors then for any $b\in\B$ there is an $a_b\in\A$ with an arrow $t:b\to R(a_b)$ such that for all $a\in\A$, $f:b\to R(a)$, there is a unique $\tilde{f}:L(b)\to a$ with $R(\tilde{f})\circ t=f$.

In your case, $R(f)=f$ and $R(a)=a$ etc. so the equation $\tau_{a,b}^{-1}(f)\circ\eta_b=f$ agrees with this but has a minor abuse of notation (strictly, we should write $i(\tau_{a,b}^{-1}(f))\circ\eta_b=f$).

You pick $a_b:=L(b)$ and $t=\eta_b$. You pick $\tilde{f}:=\epsilon_a\circ L(f)$ and you find out that: $$R(\tilde{f})\circ t=R(\epsilon_a)\circ RL(f)\circ\eta_b=R(\epsilon_a)\circ\eta_{R(a)}\circ f=f$$And conversely if $\tilde{f}$ has that property then $\epsilon_a\circ L(f)=\epsilon_a\circ LR(\tilde{f})\circ L(\eta_b)=\tilde{f}\circ\epsilon_b\circ L(\eta_b)=\tilde{f}$ so this arrow $\tilde{f}$ is unique. At least, if you define adjunctions by a pair of natural transformations satisfying the triangle identities.

If you use the definition, "there exists a family of isomorphisms $\tau_{a,b}:\A(L(b),a)\cong\B(b,R(a))$ which are natural in both variables" then we can obviously say there is a unique $\tilde{f}:=\tau_{a,b}^{-1}(f)$ which makes $\tau_{a,b}(\tilde{f})=f$, so it suffices to prove that $\tau_{a,b}(g)=R(g)\circ\eta_b$ for all $g:L(b)\to a$, where $\eta_b=\tau_{L(b),b}(1_{L(b)})$.

But, we can just recognise this as the Yoneda lemma applied to the natural transformation $\tau_{\bullet,b}:\A(L(b),-)\implies\B(b,-)\circ R$.

$$\begin{CD}1@>\mathrm{id}_{L(b)}>>\A(L(b),L(b))@>\A(1,g)>>\A(L(b),a))\\@V=VV@V\tau_{L(b),b}VV@VV\tau_{a,b}V\\1@>>\eta_b>\B(b,RL(b))@>>\B(1,R(g))>\B(b,R(a))\end{CD}$$By naturality and by definition this diagram commutes, but it is stating precisely the equation $\tau_{a,b}(g)=R(g)\circ\eta_b$.