Suppose to have an adjunction $F \dashv G$, where $G: \mathcal{B} \to \mathcal{A}$ and $F: \mathcal{A} \to \mathcal{B}$. Suppose moreover that $G$ is a faithful functor injective on objects (i.e. an embedding in Borceux' terminology). Hence, I can consider the essential image induced by $G$, i.e. $G(\mathcal{B})$, whose objects are $Obj(G(\mathcal{B}))=\{G(B): \, B \in Obj(\mathcal{B})\}$ and whose hom-sets are $Hom(G(B),G(B'))=\{G(f): \, f \in Hom_{\mathcal{B}}(B,B')\}$.

In general, the essential image of a functor is not a category. But it turns out to be a category if $G$ is full (and this is not the case!) or if $G$ is injective on objects (and this is the case). Thus, under my assumptions, $G(\mathcal{B})$ is a category.

I have the following some questions:

  1. In our case, $G(\mathcal{B})$ is a subcategory of $\mathcal{A}$, that is neither wide, nor full. Now, in Adamek's Joy of Cats terminology, we may ask for the possible reflectivity of $G(\mathcal{B})$.

With the definitions at hand (and taking into account the comment by @Daniël Apol), I think that $G(\mathcal{B})$ is a reflective subcategory of $\mathcal{A}$. In fact, by our choice of the essential image, I can identify $\mathcal{B}$ with $G(\mathcal{B})$. Such an identification is an isomorphism of categories, that composed with $F$ gives the reflection of $\mathcal{A}$ in $\mathcal{B}$. Is my argument correct?

  1. Following @Daniël Apol 's comment, let me change the notion of essential image induced by functor. For instance, assume that the essential image induced by $G$ is the full subcategory of $\mathcal{A}$ generated by the objects $\{G(B): \, B \in Obj(\mathcal{B})\}$. I call it the strong essential image induced by $G$. In such a case, I cannot identify $\mathcal{B}$ with its essential image. Then, what about the reflectivity of the strong essential image induced by $G$, taking into account the existence of the adjunction $F \dashv G$?

  2. Assume the argument of part $(i)$ to be correct. Suppose moreover that $G \circ F=Id_\mathcal{A}$. Is it possible to derive some further consequences on $G$ and $F$? I have no clue.

  3. Assume that in part $(ii)$ I can show the reflectivity of the strong essential image induced by $G$. Suppose moreover that $G \circ F=Id_\mathcal{A}$. Is it possible to derive some further consequences on $G$ and $F$?

Please, answer point-by-point.

  • $\begingroup$ 1) Which subcategory of $\mathcal{A}$ do you exactly mean when you talk about the essential image of $G$, if you say that you are not requiring it to be full? There are some ways to define a non-full essential image for which $\mathcal{B}$ is not identified with this version of the essential image of $G$. You have make sure that $G\colon\mathcal{B}\to\mathrm{im}\,G$ is full. 2) The essential image will generally not be a wide subcategory: those are subcategories that contain all objects of $\mathcal{A}$. $\endgroup$ Nov 11, 2023 at 20:11
  • $\begingroup$ @DaniëlApol Thank you for your comment, I'll edit my question $\endgroup$ Nov 12, 2023 at 10:02

1 Answer 1

  1. Your argument works. A (not necessarily full) reflective subcategory of $\mathcal{A}$ is a category $\mathcal{C}$ equipped with a faithful functor $\mathcal{C}\to\mathcal{A}$ that admits a left adjoint, and indeed that holds for $G(\mathcal{B})$ since you can identify it with $\mathcal{B}$ in this case.
  2. It is generally false that the strong essential image of $G$ will be a reflective subcategory. For example, if $\mathcal{A}=\mathrm{sGrp}$ is the category of semigroups and $\mathcal{B}=\mathrm{Mon}$ is the non-full subcategory of monoids, and if $G\colon\mathrm{Mon}\to\mathrm{sGrp}$ is the inclusion, then $G$ is faithful and injective on objects (namely, given a semigroup, then if it admits a monoid structure, this monoid structure is unique). We have a left adjoint $F$ to $G$, sending a semigroup $S$ to $S\sqcup\{e\}$ with a monoid structure which has $e$ as the unit and acts inside $S$ like the semigroup did. Write $\mathcal{C}$ for the ''strong essential image'', which is the full subcategory of $\mathrm{sGrp}$ on all semigroups that admit a monoid structure. We will show that the inclusion $i\colon\mathcal{C}\to\mathrm{sGrp}$ is not a right adjoint (and this shows that $\mathcal{C}$ is not a reflective subcategory of $\mathrm{sGrp}$). Indeed, if it was a right adjoint, then $\mathcal{C}$ would be closed under limits in $\mathrm{sGrp}$. Consider the two maps $f,g\colon\mathbb{N}\to F\mathbb{N}$ in $\mathcal{C}$ (where $F$ was defined above as the left adjoint to $G$) that are uniquely determined by $f(0)=e$, $f(1)=1$, $g(0)=0$ and $g(1)=1$. Recall that 0 is no longer the unit of $F\mathbb{N}$, so $f$ and $g$ are genuinely different maps. Also note that $f$ is a map of monoids, but $g$ only exists in $\mathcal{C}$ and not in $\mathrm{Mon}$. The equalizer $\mathrm{eq}(f,g\colon\mathbb{N}\to F\mathbb{N})$ in $\mathrm{sGrp}$ is the semigroup $\mathbb{N}^+$ of positive integers equipped with addition, which does not admit a monoid structure. Therefore $\mathcal{C}$ is not closed under limits in $\mathrm{sGrp}$, and this then shows that $\mathcal{C}$ is not a reflective subcategory of $\mathrm{sGrp}$. Again, this argument breaks down for $\mathrm{Mon}$ instead of $\mathcal{C}$ because the map $g$ does not exist in $\mathrm{Mon}$.
  3. If $G\circ F=\mathrm{id}_\mathcal{A}$, then $G$ is surjective on objects. Since $G$ is bijective on objects now, we just assume $G$ acts like the identity on objects (identifying $\mathcal{B}$ with $G(\mathcal{B})$), and consequently we can assume $F$ acts like the identity on objects. Then $G\circ F=\mathrm{id}_\mathcal{A}$ tells us that $G$ is full as well. Therefore, $G$ is an essentially surjective fully faithful functor, so an equivalence of categories (and $F$ is its inverse equivalence). In particular, the unit of the adjunction defines an isomorphism $\eta\colon\mathrm{id}_\mathcal{A}\cong G\circ F$. (Note that just having an equality $G\circ F=\mathrm{id}_\mathcal{A}$ does not a priori mean that this equality is witnessed by the adjunction unit).
  4. Since point 2. generally fails, we don't really need to answer this, but in all cases where somehow point 2. doesn't fail, you can repeat the argument in 3.

P.S. It doesn't matter because you made clear which definitions you had in mind, but I think it is more usual to have something called the ''essential image'' be closed under isomorphisms, so that every object $A\in\mathcal{A}$ for which $A\cong GB$ for some $B\in\mathcal{B}$ is also contained in the essential image. (And if you would do something like this, it is also clear why it is unclear what the exact definition should be if $G$ is not full.) I would have called the essential image categories you defined something like the ''strict image of $G$''.

  • $\begingroup$ Thank you very much for you answer. On the other hand, if in the last two parts, I have $F \circ G=Id_\mathcal{B}$, what can I say in both cases on the functors $F$ and $G$? $\endgroup$ Nov 14, 2023 at 15:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .