Equivalent characterizations of the essential image of reflective subcategories

Question

I'm currently trying to prove the third item of the following exercise from Category Theory in Context,

Exercise 4.5.vii. Consider a reflective subcategory inclusion $D \hookrightarrow C$ with reflector $L: C \to D$.

1. Show that $\eta L = L\eta$, and that these natural transformations are isomorphisms.
2. Show that an object $c \in C$ is in the essential image of the inclusion $D\hookrightarrow C$, meaning that it is isomorphic to an object in the subcategory $D$, if and only if $\eta_c$ is an isomorphism.
3. Show that the essential image of $D$ consists of those objects $c$ that are local for the class of morphisms that are inverted by $L$. That is, $c$ is in the essential image if and only if the pre-composition functions $$ C(b, c) \xrightarrow{f^*} C(a, c) $$ are isomorphisms for all maps $f : a \to b$ in $C$ for which $Lf$ is an isomorphism in $D$. This explains why the reflector is also referred to as “localization.”

Here $\eta$ is the unit of the adjunction. I have managed to prove the first two items. However, reflective subcategories were just briefly introduced and so I am not sure how to relate item $(3)$ with $(2)$, if this is the right path to begin with. Any hints? I would also appreciate if someone can shed some light on the term localization and why $(3)$ 'explains' this nomenclature.

Answer 1

Hint: First prove that $iLc$ satisfies the condition in 3, where $i:D\to C$ is the inclusion. Then use apply 2 to conclude.

Regarding "localization" the point is that $L$ is characterized by the arrows it inverts, so the language is being imported from commutative algebra. Specifically, the motivating situation is that of $R[S^{-1}]$-modules, where $S$ is a multiplicative set in a commutative ring $R$. Every $R[S^{-1}]$ is functorially an $R$-module by restricting the scalars, and this functor is fully faithful, with the left adjoint $R[S^{-1}]\otimes_R (-)$. So this is a reflective subcategory. Furthermore an $R$-module admits an $R[S^{-1}]$-action if and only if it is local for those $R$-module maps inverted by tensoring with $R[S^{-1}]$, as in Riehl's point 3. The classical algebraic framework would reduce those maps to the maps $s:R\to R$ determined by elements of $S$, while the general categorical framework instead asks for locality with respect to all maps $\eta_A:A\to A[S^{-1}]$, for $A$ an $R$-module. We can bridge these frameworks by observing that locality with respect to each $s$ is equivalent to locality with respect to the single unit map $R\to R[S^{-1}]$, which implies locality with respect to every $\eta_A$ by considering the action of $L$ on a presentation of $A$.

Answer 2

Here's a proof of part 3. Denote $I:\mathsf{D}\to \mathsf{C}$ to the inclusion.

$(\Rightarrow)$. Let $f:a\to b\in\mathsf{C}$ be such that $Lf$ is an iso. It suffices to show that $\mathsf{C}(b,Id)\xrightarrow{f^*}\mathsf{C}(a,Id)$ is an iso. We have a naturality commutative square $$ \require{AMScd} \begin{CD} \mathsf{C}(b,Id)@>{f^*}>>\mathsf{C}(a,Id)\\ @V{\cong}VV@VV{\cong}V\\ \mathsf{D}(Lb,d)@>{\smash{Lf^*}}>>\mathsf{D}(La,d) \end{CD} $$ Since three out of four arrows is an iso, so is the fourth one.

$(\Leftarrow)$. By part 1, $L\eta_{c'}$ is an iso for all $c'\in\mathsf{C}$. Hence, by hypothesis, $\mathsf{C}(ILc,c)\xrightarrow{\eta_c^*}\mathsf{C}(c,c)$ is bijective, so there is $f:ILc\to c$ with $f\cdot\eta_c=1_c$. By $(\Rightarrow)$, the map $\mathsf{C}(ILc,ILc)\xrightarrow{\eta_c^*}\mathsf{C}(c,ILc)$ is a bijection. But this last map sends both $\eta_c\cdot f$ and $1_{ILc}$ to the same morphism, whence $\eta_c\cdot f=1_{ILc}$.

Answer 3

Judging from the comments, I feel like that this addendum to @Kevin Arlin's answer may be useful.

Here's a proof for one direction of the third item.

Consider $\eta_c : c \to iLc$; by the first item $iL\eta_c : iLc \to iLc$ is iso so that $\eta^{*}_c: \mathsf{C}(iLc,c) \to \mathsf{C}(c,c)$ is a bijection. In particular there is a unique $s: iLc \to c$ such that $s \eta_c = 1_c$. This implies that $iLs \cdot iL\eta_c = 1_{iLc}$; now, $iL\eta_c=\eta_{iLc}$ is an isomorphism, and by the triangular identities, its inverse is $i\epsilon_{Lc}$. Therefore $iLs=i\epsilon_{Lc}$ and \begin{equation*} \eta_c s= iLs \cdot \eta_{iLc} = i\epsilon_{Lc} \cdot \eta_{iLc} = 1_{iLc} \end{equation*} where the first equality follows from naturality.