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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.

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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 $A$.

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  • $\begingroup$ Thanks for your reply! I think I've got your hint, but still can't manage to finish this off. If $c$ is on the essential image via $w : c \simeq Ld$, then $Ld$ verifies the condition in $(3)$ and thus so does $c$ via factoring $f^*$ through $f^* : C(b,Ld) \to C(a,Ld), w_* : C(a,c) \to C(a,Ld), w_* : C(b,c) \to C(b,Ld)$. $\endgroup$ – qualcuno Feb 18 '19 at 4:19
  • $\begingroup$ (cont.) However the converse is still unclear to me: if $c$ verifies the condition on $(3)$, we can apply this to $\eta_c$ (as $(1)$ guarantees that $L\eta_c$ is an iso). I see how this shows that $\eta_c$ is a split epi: if $a$ is the inverse of $\eta_c^*$, then $a(1_c)\eta_c =( \eta_c^* a)(1_c) = 1_c$. However, can we assert that this arrow is actually an iso? Having $Lc \simeq c$ for any $c$ in the essential image seems too strong, but I may have bad intuition on this. $\endgroup$ – qualcuno Feb 18 '19 at 4:19
  • $\begingroup$ @GuidoA. Actually, $Lc$ is indeed isomorphic to $c$ in that case. Given an iso $a:c\to Lc'$, we have a naturality square involving $a,La, $ and $\eta_{Lc'}$, all of which are iso, as well as $\eta_c$. So the latter is an iso as well. $\endgroup$ – Kevin Arlin Feb 18 '19 at 21:08

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