A subcategory $D$ of a category $C$ is called reflective, if the embedding $D \hookrightarrow C$ has a left adjoint $L:C \to D$. The left adjoint $L$ is called the reflector. If the category $C$ is cartesian closed and $D$ is an exponential ideal, then $L$ preserves finite products.

The category of groups is not cartesian closed. But all examples of reflectors that I know preserve finite products:

  • quotient by $n$th term of the lower central series;
  • quotient by $n$th term of the derived series;
  • $HR$-localisation of Bousfied for any commutative ring $R$;
  • ...

Question: Is there a reflective subcategory of the category of groups whose reflector does not preserve finite products?

UPD: Here I do not assume that the subcategory $D$ is full. Probably, it is better to call such subcategories "weak reflective".

UPD2: I am stupid, I am sorry. I need a full subcategory $D$. I need this functor $L$ to be idempotent functor.

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    $\begingroup$ Just to understand your first example, do you view the category of profinite groups as a subcategory of the category of groups? There's an obvious forgetful functor from (profinite groups, continuous group homomorphisms) to (groups, group homomorphisms) but I wouldn't view it as an embedding of a subcategory. $\endgroup$ – YCor Jul 2 '16 at 13:05
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    $\begingroup$ I agree that it is a bad example because the subcategory of profinite groups is not full. Usually people assume that a reflective subcategory is full. However, it is a (non-full) subcategory and the embedding functor has a left adjoint. I changed the place of this example. But an example of such a week reflective subcategory is interesting as well. $\endgroup$ – Sergei Ivanov Jul 2 '16 at 15:55
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    $\begingroup$ I wasnt referring to fullness. I understand a subcategory $D$ of $C$ as picking a subcollection of objects, and a subset of arrows including identities of the chosen objects, and in a way that's stable under composition. But to pass from groups to profinite groups, one also needs to endow them with some extra-structure. OK, there's a stupid trick to fix this, namely one considers the category whose objects are topological groups and arrows are abstract group homomorphisms. Then it's equivalent to the category of groups, and profinite groups with continuous homomorphisms form a subcategory. $\endgroup$ – YCor Jul 2 '16 at 15:56
  • $\begingroup$ Ok, you convinced me. It is not a subcategory. I forgot that the topology is defined by the structure of discrete group only in the case of topologically finitely generated profinite group. $\endgroup$ – Sergei Ivanov Jul 2 '16 at 16:16
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    $\begingroup$ So, probably you want $D$ to be a category endowed with a faithful functor to $C$? $\endgroup$ – YCor Jul 2 '16 at 16:20

The answer is no: every such full subcategory is stable under direct products, or equivalently $L$ preserves direct products.

Here all Hom are in the category of groups. The definition of $L$ means that $L$ is a covariant functor and for every group there is a group homomorphism $G\to LG$, so that the obvious squares commute, and such that the natural map $\mathrm{Hom}(LG,U)\to \mathrm{Hom}(G,U)$ induced by $G\to LG$ is a bijection for all $U\in D$.

Idea of the proof. Categorically usually describes $G\times H$ by showing that it represents $K\mapsto\mathrm{Hom}(K,G)\times \mathrm{Hom}(K,H)$. But one needs here to describe it on the left, i.e. to describe $\mathrm{Hom}(G\times H,K)$, which can be described as the set of pairs $(u,v)$ in $\mathrm{Hom}(H,K)\times\mathrm{Hom}(G,K)$ whose images centralize each other.

Next, one observe that for $U\in D$, both $\mathrm{Hom}(L(G\times H),U)$ and $\mathrm{Hom}(LG\times LH,U)$ can be described as the set of pairs in $\mathrm{Hom}(H,K)\times\mathrm{Hom}(G,K)$ whose images centralize each other; more precisely, we have a canonical homomorphism $\pi:L(G\times H)\to LG\times LH$ inducing a bijection $\mathrm{Hom}(LG\times LH,U)$ and $\mathrm{Hom}(L(G\times H),U)$ for all $U\in D$ (given by $f\mapsto f\circ \pi$). One would like to conclude from Yoneda's lemma; however we cannot because $LG\times LH$ might not be in $D$. Nevertheless, we can argue as in the proof of Yoneda's lemma, picking $U=L(G\times H)$ and consider a preimage of the identity map of $L(G\times H)$. This yields $f$ such that $f\circ\pi$ is the identity of $L(G\times H)$ and thus $\pi$ is injective.

So to conclude we need to show that $\pi$ is surjective and this is actually the beginning of the proof below, since it uses very little.

Here's now the detailed proof.

First observe that $L$ maps the trivial group 1 to itself (edit: alternatively, see the comment by Todd Trimble). Indeed we have a bijection (the previous one for $G=1$, $U=L1$) $\mathrm{Hom}(L1,L1)\to \mathrm{Hom}(1,L1)$; the latter is a singleton and the former contains the identity of $L1$ and the trivial endomorphism and hence is a singleton iff $L1=1$. It follows that $L$ maps the trivial (=constant) homomorphisms to themselves (because these are the one factoring through the trivial group).

For any groups $G,H$ use the notation for the canonical injections and projections: $$ G\stackrel{i_G}\to G\times H\stackrel{i_H}\leftarrow H,\quad G\stackrel{p_G}\leftarrow G\times H\stackrel{p_H}\to H,$$ inducing homomorphisms $$ LG\stackrel{Li_G}\to L(G\times H)\stackrel{Li_H}\leftarrow LH,\quad LG\stackrel{Lp_G}\leftarrow L(G\times H)\stackrel{Lp_H}\to LH.$$ We have $$p_Gi_G=\mathrm{id}_G,\quad p_Gi_G=\mathrm{id}_G, \quad p_Gi_H=1, \quad p_Hi_G=1,$$ thus $$Lp_GLi_G=\mathrm{id}_{LG},\quad Lp_HLi_H=\mathrm{id}_{LH}, \quad Lp_GLi_H=1_{LH,LG}, \quad Lp_HLi_G=1_{LG,LH},$$ where $1_{A,B}$ is the trivial (constant) homomorphism from $A$ to $B$. From the first two equalities, we deduce that $Li_G$ and $Li_H$ are injective and we have semidirect decompositions $$L(G\times H)=Li_G(LG)\ltimes \mathrm{ker}(Lp_G)=Li_H(LH)\ltimes \mathrm{ker}(Lp_H)$$ (where the projection to the left-hand factor is given by $Li_GLp_G$, respectively $Li_HLp_H$); from the last two equalities we deduce that $Li_G(LG)\subset \mathrm{ker}(Lp_H)$ and $Li_H(L_H)\subset \mathrm{ker}(Lp_G)$. It follows that the map $$L(G\times H)\stackrel{Lp_G\times Lp_H}\to LG\times LH$$ is surjective: indeed the left-hand map being surjective in restriction to $Li_G(LG)$, it amounts to see that the right-hand map is surjective in restriction to the kernel of the left-hand map, and this kernel is $\mathrm{ker}(Lp_G)$ (by the first semidirect decomposition), and this kernel contains $Li_H(L_H)$, in restriction to which the right-hand map is surjective. This proves that $Lp_G\times Lp_H$ is surjective (I used very little of the hypotheses about $L$: only that it's an endofunctor mapping the trivial groups to themselves).

Now we proceed to show that $Lp_G\times Lp_H$ is also injective.

If $f:A\to B$ is a group homomorphism and $K$ another group, denote by $q_K[f]$ the resulting map $\mathrm{Hom}(B,K)\to\mathrm{Hom}(A,K)$ given by $g\mapsto g\circ f$.

Consider the following maps: $$[*]\quad \mathrm{Hom}(LG\times LH,K)\stackrel{q_K[Lp_G\times Lp_H]}\to \mathrm{Hom}(L(G\times H),K)\stackrel{q_K[Li_G]\times q_K[Li_H]}\to \mathrm{Hom}(LG,K)\times\mathrm{Hom}(LH,K).$$

Let $s$ be the composite map. Let us show that $s$ is injective and describe its image. Since $q_K$ is a contravariant functor, the composite map $s$ is given by $$s=q_K[(Lp_G\times Lp_H)\circ Li_G]\times q_K[(Lp_G\times Lp_H)\circ Li_H]$$ $$=q_K[Lp_GLi_G\times Lp_HLi_G]\times q_K[Lp_GLi_H\times Lp_HLi_H]$$ $$=q_K[\mathrm{id}_{LG}\times 1_{LG,LH}]\times q_K[1_{LH,LG}\times \mathrm{id}_{LH}].$$ Explicitly, $\mathrm{Hom}(LG\times LH,K)$ is given (via the embedding $s$) by the set of pairs of homomorphisms $(u_G,u_H)\in\mathrm{Hom}(LG,K)\times\mathrm{Hom}(LH,K)$ whose images in $K$ centralize each other; here if $u\in \mathrm{Hom}(LG\times LH,K)$, $u_G$ is its restriction to $G$ and $u_H$ its restriction to $H$, and $u(x,y)=u_G(x)u_H(y)$ for all $(x,y)\in LG\times LH$. We see (from the above cumbersome but trivial computation) that $s(u)=(u_G,u_H)$.

Now assume that $K=U$ belongs to $D$. Then the map $$\mathrm{Hom}(L(G\times H),U)\stackrel{q_K[Li_G]\times q_K[Li_H]}\to \mathrm{Hom}(LG,U)\times\mathrm{Hom}(LH,U)$$ is isomorphic (by the functorial property of $L$) to the map $$\mathrm{Hom}(G\times H,U)\stackrel{q_K[i_G]\times q_K[i_H]}\to \mathrm{Hom}(G,U)\times\mathrm{Hom}(H,U),$$ which itself is injective, for the same reason as the previous one (a homomorphism on a direct product is determined by its restriction to factors) and also, for the same reason its image consists of pairs of homomorphisms whose images centralize each other.

So, writing symbolically [$*$] as $X\stackrel{v}\to Y\stackrel{w}\to Z$ we have shown that $w\circ v$ and $w$ are both injective with the same image (**edit: [$**$]). This implies that $v$ is bijective. Namely, for every $U$ in $D$, $$\mathrm{Hom}(LG\times LH,U)\stackrel{q_K[Lp_G\times Lp_H]}\to \mathrm{Hom}(L(G\times H),U)$$ is bijective.

Apply this (the surjectivity) to $U=L(G\times H)$ and the identity of $U$, we obtain that there exists a homomorphism $g:LG\times LH\to L(G\times H)$ such that $g\circ (Lp_G\times Lp_H)=\mathrm{id}_{L(G\times H)}$. This shows that $Lp_G\times Lp_H$ is injective and concludes the proof.

Edit: in [$**$] I have to justify that they have the same images. I was initially not careful about this: if $G\to LG$ is surjective for all $G$, then this is fine because the above description really provides the same description for pairs of homomorphisms extending to a homomorphism on the direct product. But if $G\to LG$ is not surjective, it's a priori not clear that if we have $G\times H\to U$, then the images of the factored maps $LG\to U$ and $LH\to U$ centralize each other. However this is true by the lemma below, and thus it's fine.

Lemma If $G,U$ are groups with $U\in D$ and $f:G\to U$ induces $f':LG\to U$, then $C_U(\mathrm{Im}(f))=C_U(\mathrm{Im}(f'))$, where $C_U$ denotes the centralizer in $U$. In particular, if $g:H\to U$ is another such homomorphism and induces $g'$ and if the images of $f$ and $g$ centralize each other, then the images of $f'$ and $g'$ centralize each other.

Proof: first note that $\mathrm{Im}(f)\subset\mathrm{Im}(f')$, so the inclusion $C_U(\mathrm{Im}(f))\supset C_U(\mathrm{Im}(f'))$ is clear. Now consider $t\in C_U(\mathrm{Im}(f))$. Then $h:y\mapsto tf'(y)t^{-1}$ is a homomorphism from $LG$ to $U$ and the composite map $G\to U$ equals $f$. By the universal property of $L$ (namely, the injectivity of $\mathrm{Hom}(LG,U)\to\mathrm{Hom}(G,U)$), it follows that $h=f'$. So $t$ centralizes the image $\mathrm{Im}(f')$. This proves the first statement.

The second statement follows by applying the first statement twice: if $u$ belongs to the image of $f$, then it centralizes the image of $g$, hence centralizes the image of $g'$. Now if $u'$ belongs to the image of $g'$, then it centralizes the image of $f$, hence centralizes the image of $f'$.

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    $\begingroup$ I haven't read through all the details, but one minor point is that of course $L$ preserves the terminal because the terminal in $Grp$ is initial and left adjoints $L$ preserve initials. $\endgroup$ – Todd Trimble Jul 3 '16 at 15:05
  • $\begingroup$ Sure. Since I don't know these facts by heart, I gave the trivial 2-line argument. $\endgroup$ – YCor Jul 3 '16 at 15:16
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    $\begingroup$ It's fine; it's not a criticism actually; I'm just interested in determining the generality of the argument (does it extend to semi-abelian categories, say: ncatlab.org/nlab/show/semi-abelian+category). $\endgroup$ – Todd Trimble Jul 3 '16 at 15:23
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    $\begingroup$ I've just edited to fill a missing point in the argument. Indeed although I see no examples, I see no reason why $G\to LG$ should be surjective for all $G$. As a consequence, the condition that the image of two homomorphisms centralize each other could fall apart when replacing homomorphisms $G\to U\leftarrow H$ by the factored homomorphism $LG\to U\leftarrow LH$, which might have a larger images. So I added an additional argument to show that these images still centralize each other. $\endgroup$ – YCor Jul 3 '16 at 16:55
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    $\begingroup$ It is very nice! The only thing that I didn't understand (and didn't believe) is the lemma about centralizers. The fact that a reflective subcategory is closed under products is standard. It's not a criticism, I think, I had to add it to my question. $\endgroup$ – Sergei Ivanov Jul 3 '16 at 21:02

If you don't want a full subcategory, then an easy example would be to take $\mathsf{Grp}$ as a subcategory of itself via the squaring functor $S: \mathsf{Grp} \to \mathsf{Grp}$, $G \mapsto G \times G$. A left adjoint is given by the co-squaring functor $L: \mathsf{Grp} \to \mathsf{Grp}$, $G \mapsto G \amalg G$ (here $\amalg$ is the free product, i.e. the coproduct). For we have

$\mathsf{Grp}(LG,H) = \mathsf{Grp}(G \amalg G, H) = \mathsf{Grp}(G,H)^2 = \mathsf{Grp}(G,H \times H) = \mathsf{Grp}(G,SH)$.

This reflector doesn't preserve finite products. For example, the canonical map $L(\mathbb{Z} \times \mathbb{Z}) \to L(\mathbb{Z}) \times L(\mathbb{Z})$ is not injective.

It would be nice to have a more natural example, admittedly.

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    $\begingroup$ Trivial comment: another way of seeing the adjunction is to write the squaring as $\prod \circ \Delta$, and use $\amalg \dashv \Delta \dashv \prod$, so $\amalg \circ \Delta \dashv \prod \circ \Delta$ where the left adjoint is co-squaring. $\endgroup$ – Todd Trimble Jul 3 '16 at 2:05
  • $\begingroup$ I don't even see how you view it as a subcategory (it's not just choosing objects and arrows, you have to specify how you view a certain group as the square of some other group). $\endgroup$ – YCor Jul 3 '16 at 8:37
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    $\begingroup$ @YCor What is wrong with objects $\{G\times G\mid G\in\mathsf{Grp}\}$ and morphisms $\{f\times f\mid f:G\to G'\}$? $\endgroup$ – მამუკა ჯიბლაძე Jul 3 '16 at 9:22
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    $\begingroup$ @მამუკაჯიბლაძე oh, it seems you're right. So the objects are those groups that are equal and not only isomorphic to a direct square. So it's sensible to the bare set-theoretic definition of the direct product. But it seems to work! $\endgroup$ – YCor Jul 3 '16 at 9:28
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    $\begingroup$ I don't feel this answer should have been downvoted, since it did give a correct answer to the prior version of the question at which it was aimed. $\endgroup$ – Todd Trimble Jul 4 '16 at 14:46

This has been withdrawn, as the idea that groupoids can be useful in this case is still conjectural!

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    $\begingroup$ I don't see how this is relevant to the question. The question is about reflective subcategories of $\text{Grp}$, not categories in which $\text{Grp}$ is a reflective subcategory. $\endgroup$ – Qiaochu Yuan Jul 4 '16 at 19:51
  • $\begingroup$ Have I made an elementary mistake? I was assuming that if $A$ is a reflective subcategory of $B$ and $B$ is a reflective subcategory of $C$ then$A$ is a reflective subcategory of $C$. $\endgroup$ – Ronnie Brown Jul 5 '16 at 14:05
  • $\begingroup$ A reflective subcategory $D \subseteq \mathsf{Grp}$ sits in a diagram $D \overset{j}{\to} \mathsf{Grp} \overset{i}{\to} \mathsf{Grpd}$ with reflectors $D \overset{L}{\leftarrow} \mathsf{Grp} \overset{U}{\leftarrow} \mathsf{Grpd}$. Since $\mathsf{Grpd}$ is ccc, if $D \overset{ij}{\to} \mathsf{Grpd}$ is an exponential ideal, then $LU$ preserves finite products. If we knew that $U$ reflects finite products (I'm not sure if this is true), we could conclude that $L$ preserves finite products. But the hypothesis that $D \overset{ij}{\to} \mathsf{Grpd}$ is an exponential ideal is very restrictive. $\endgroup$ – Tim Campion Jul 6 '16 at 10:48
  • $\begingroup$ ... It requires that for $G \in D$, and any groupoid $X$, there is exactly one conjugacy class of groupoid morphisms $X \to G$. Actually, I think the only $D \subseteq \mathsf{Grp}$'s that will satisfy this are the empty $D$ (which is not reflective) and the singleton $D$ consisting of the trivial group. This exponential ideal / cartesian closed category fact was mentioned by the OP, but I think it's a red herring. $\endgroup$ – Tim Campion Jul 6 '16 at 10:48
  • $\begingroup$ My response was too hasty, and not looking up facts in the "Elephant", so I have withdrawn it. The category $\mathsf{Grpd}$ does have over that of $\mathsf{Grp}$ some features which many have found useful in topology, but it is pure conjecture that they are useful in this case. $\endgroup$ – Ronnie Brown Jul 7 '16 at 14:15

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