# Why are the two definitions of reflective subcategory equivalent?

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.

• Small comment. The notations $Mor_A(x,y)$ or $Hom_A(x,y)$ are quite archaic. Usually $A(x,y)$ is used instead. Cheers! Aug 12 at 13:50

$$\newcommand{\A}{\mathsf{A}}\newcommand{\B}{\mathsf{B}}\require{AMScd}$$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 $$\tau_{a,b}(g)=R(g)\circ\eta_b$$.

• Thank you, there is one thing that I don't understand: Does $RL(f) \circ \eta_b = \eta_b \circ f$? It's used in an equation in the middle. Why is this true? Also, did you mean $\eta_{R(a)}$ on the right hand side? Otherwise I can't get composition to work. The codomain of $f$ is $R(a)$ and the domain of $\eta_b$ is $b$. Aug 12 at 9:25
• @NaomiZhang Yes, $\eta$ is a natural transformation $\mathrm{Id}\implies RL$. To explain why that's true is tricky without more context. There are at least three ways to define "adjunction". In one of them, $\eta$ is natural by definition: in the others, you need to prove it, but the method of proof is different depending on your starting definition. So I should ask you to say what the definition of adjunction is that you're using. Btw I showed two ways to calculate $\tau_{a,b}(g)=R(g)\circ\eta_b$ using two different starting definitions. Aug 12 at 10:09
• And yes I did mean $\eta_{R(a)}$ that was a careless typo Aug 12 at 10:10
• That solves it, that $\eta$ (defined as $\tau_{a, F(a)}(id_{F(a)}$) is a natural transformation was what I was missing. I checked that. Aug 13 at 3:39