Non-free projective pearls (general and Abelian)

Question

A pearl is an ordered pair $\ \mathbf P:=(G\,\ S),\ $ where $\ G\ $ is a group, and $\ S\ $ is a non-empty subset of G which does not contain the neutral element of $\ G\ $ (i.e. not 1 in the multiplicative notation, nor 0 in the additive notation).

If $\ \mathbf Q:=(H\,\ T)\ $ is another pearl then a morphism $\ f:\mathbf P\rightarrow \mathbf Q\ $ is defined as a group homomorphism $\ f:G\rightarrow H\ $ such that $\ f(S) \subseteq T$.

This defines the (general) category of pearls. Pearls for which their group is Abelian, form a full subcategory (with all their morphisms between them).

Let's define a free pearl $\ \mathbf P\ $ (as above) as one for which a subset $\ F\subseteq G\ $ is a set of the free generators of the group $\ G,\ $ and $\ S\subseteq F.\ $ Obviously Possibly, every free pearl is projective (in the category of pearls).

The definition of an abelian-free pearl (and the issue of projectivity) is similar as for the category of abelian pearls.

QUESTION   Can you provide examples of projective pearls which are not free? The same for the case of the category of abelian pearls.

REMARK This requires paying close attention to pearl epimorphisms.

                M O T I V A T I O N

1. Categories of enhanced groups are among the most important. After groups, pearls are among the simplest among them. Thus it provides a view onto the next categories which already appear;

2. There are problems on the general categories (or a general class of them) when we would like to obtain a positive result or else a counter-example. Then pearls may provide a relatively simple testing ground and they may project onto other categories.

3. Truly, the notion of pearls appeals to me, I find them attractive.

EDIT   After reading @ArturoMagidin comments, I replaced my careless obviously by careful possibly. (I'd like to see Arturo's comments posted by him as one or more answers to make them more readable, please).

Answer 1

Włodzimierz Holsztyński suggests, in comments to my other answer, a modification/subcategory of pearls: call a morphism of pearls $f\colon (G,S)\to (H,T)$ sharp if and only if $f(S)=T$. Now consider the sharp subcategory to be the category of pearls with sharp morphisms between them (and similarly for abelian pearls). Now asking if in this subcategory free pearls might be projective, and if there are projectives that are not free. I am writing a separate answer because the previous one was already a bit long, and this deals with different concept of morphisms.

As far as epimorphisms in the sharp categories go, my arguments given for the general category do not work. I do not yet know if all epimorphisms in the sharp category have underlying group homomorphism that is surjective. However, this still holds for the abelian pearls. In addition, it is still the case that neither sharp subcategory has projectives.

Theorem. Let $f\colon (G,S)\to(H,T)$ be a sharp morphism of abelian pearls. Then $f$ is an epimorphism in the sharp subcategory of abelian pearls if and only if $f\colon G\to H$ is a surjective morphism of abelian groups.

Proof. Assume $f$ is surjective. Let $(K,U)$ be another abelian pearl, and $g,h\colon (H,T)\to(K,U)$ be sharp morphisms such that $gf = hf$. Since $f$ is surjective, we can conclude that $g=h$ as group homomorphisms, hence as pearl (sharp) homomorphisms. Conversely, suppose that $f$ is not surjective. Let $K = H\times (H/f(G))$, and let $U=\{(f(s),0+f(G))\mid s\in S\}$. Note that since $f$ is a (sharp) pearl morphism, $f(s)\in T$ for all $s$, so $(0,0)\notin U$. Now let $g,h\colon H\to K$ be defined as follows: \begin{align*} g(x) &= (x,x+f(G)),\\ h(x) &= (x,0+f(G)). \end{align*} Note that $g,h\colon (H,T)\to (K,U)$ are sharp morphisms: for all $t\in T$ there exists $s\in S$ such that $f(s)=t$ (because $f$ is sharp), hence $t\in f(G)$; thus, $(t,t+f(G)) = (t,0+f(G)) = (f(s),0+f(G))\in U$; i.e., $g(T)\subseteq U$ and $h(T)\subseteq U$. Moreover, given $(f(s),0+f(G))\in U$, we have $f(s)\in T$ so $(f(s),0+f(G)) = g(f(s))=h(f(s))$, so $g(T)=U$ and $h(T)=U$. Thus, both are sharp.

Since $f$ is not surjective, there exists $x\in H$ such that $x\notin f(G)$. Therefore, $g(x) = (x,x+f(G))\neq (x,0+f(G)) = h(x)$. Hence, $g\neq h$. However, for all $a\in G$, $g(f(a)) = (f(a),f(a)+f(G)) = (f(a),0+f(G)) = h(f(a))$, so $g\circ f = h\circ f$. Thus, $f$ is not an epimorphism. $\Box$

However, free and free-abelian pearls are still not projective in the corresponding sharp subcategories. In fact, once again there are no projectives at all. It is clear that if the underlying group homomorphism is onto then the pearl homomorphism is an epimorphism (even in the sharp category), though for nonabelian pearls there may be other epimorphisms.

Let $(H,T)$ be any pearl with $T$ finite, and take the identity map $i\colon (H,T)\to (H,T)$. Now let $x$ be an element not in $H$ of infinite order, and let $G=H*\langle x\rangle$ be the free product of $H$ with the infinite cyclic group; in the abelian case, take $G=H\times \langle x\rangle$. Let $S=T\cup\{x\}$, and let $t\in T$ be arbitrary. Let $f\colon (G,S)\to (H,T)$ be the map induced by the group homomorphism that maps $H$ to itself via the identity, and sends $x$ to $t$. The universal property of the free product (resp. the direct product/direct sum) guarantees such a homomorphism exists. This is a pearl homomorphism, and is sharp, since $T\subseteq f(T)\subseteq f(S)\subseteq T$. The underlying group homomorphism is onto, so the pearl map is an epimorphism. However, there are no sharp pearl homomorphisms at all from $(H,T)$ to $(G,S)$ (let alone one that factors $i$ through $f$), because the cardinality of $S$ is strictly larger than that of $T$. Thus, $(H,T)$ is not projective.

If $T$ is infinite, then a similar construction will do by replacing $\langle x\rangle$ with a free group $F$ (resp. free abelian group) of rank $\kappa$, where $\kappa\gt |T|$, replacing $S=T\cup\{x\}$ with the (disjoint) union of $T$ and the free generating set for $F$, and mapping all elements of the free generating set to elements of $T$ arbitrarily.

Answer 2

Actuallly, I'm not sure that free pearls as defined are in fact projective; for that matter. I'm pretty sure that there are no projectives at all. I have a characterization of epimorphisms, then a specific example of a free pearl that is not projective, and finally an argument that no pearl is projective.

Theorem. A homomorphism $f\colon (G,S)\to(H,T)$ of pearls is an epimorphism if and only if $f\colon G\to H$ is surjective. The same is true for the category of abelian pearls. However, if $f$ is an epimorphism then it need not be the case that $f(S)=T$.

Proof. Let $f\colon (G,S)\to (H,T)$ be a pearl homomorphism such that $f\colon G\to H$ is onto. Let $(K,U)$ be a pearl, and let $g,h\colon (H,T)\to (K,U)$ be pearl homomorphisms such that $g\circ f = h\circ f$. If $x\in H$, then there exists $a\in G$ such that $f(a)=x$. Therefore, $g(x) = g(f(a)) = h(f(a)) = h(x)$. So $g=h$ as morphisms $H\to K$, hence $g=h$ as morphisms of pearls. Thus, $f$ is an epimorphism.

Conversely, suppose that $f\colon (G,S)\to(H,T)$ is a homomorphism and $f\colon G\to H$ is not onto. In the abelian case, we can construct the amalgamated direct product $K= H\times H/\{ (a,-a)\mid a\in f(G)\}$, and in the general case we can construct the amalgamated coproduct $K=H\amalg_{f(G)} H$; both these groups have embeddings $\lambda,\rho\colon H\to K$ into the "left" and "right" cofactors, such that the equalizer of $\lambda$ and $\rho$ is $f(G)$; (in fact, $\lambda(H)\cap\rho(H) = f(G)$). In particular, $\lambda\circ f = \rho\circ f$, but since $f(G)\neq H$, $\lambda\neq\rho$.

Moreover, since $\lambda$ and $\rho$ are embeddings, $\lambda(T)$ and $\rho(T)$ cannot contain the identity of $K$. Thus, $(K,\lambda(T)\cup \rho(T))$ is a pearl, and we have two induced maps $\lambda,\rho\colon (H,T)\to (K,\lambda(T)\cup \rho(T))$. These maps satisfy $\lambda\circ f = \rho \circ f$ but $\lambda\neq \rho$. Therfore, $f$ is not an epimorphism of pearls. This shows that homomorphisms of pearls (resp. abelian pearls) are epimorphisms (in the respective category) if and only if the underlying group homomorphism is surjective.

On the other hand, an epimorphism $f\colon (G,S)\to (H,T)$ need not induce a surjective map $f|_S\colon S\to T$. Indeed, let $G$ be any group with more than two elements, and let $S$ and $T$ be subsets of $G-\{e\}$ such that $S\subseteq T$ but $S\neq T$ (that's where we need $|G|>2$). Then $i\colon (G,S)\to (G,T)$ induced by the identity map is an epimorphism, since the identity map is onto, but the induced map $S\hookrightarrow T$ is not surjective. Same argument holds in the category of abelian pearls $\Box$

Now, a pearl $(P,S)$ is projective if and only if for every pearl homomorphism $h\colon (P,S)\to (H,U)$, and every pearl epimorphism $f\colon (G,T)\to (H,U)$, there exists a pearl homomorphism $g\colon (P,S)\to(G,T)$ such that $f\circ g = h$. I claim that there are free pearls that are not projective, in both the general and abelian case.

Proposition. Let $F_2$ be the free group of rank $2$, with free generators $x$ and $y$. Then $(F_2,\{x\})$ is a free pearl, but is not projective in the category of all pearls. Replacing the free group with its abelianization, we obtain a free-abelian pearl that is not projective in the category of abelian pearls.

Proof. I will use $\mathbb{Z}^2$ for the abelianization of $F_2$, and identify $x$ and $y$ with their images in the abelianization.

That $(F_2,\{x\})$ is a free pearl and $(\mathbb{Z}^2,\{x\})$ is an abelian-free pearl follows from the definition: the underlying group is free (resp. free abelian), and the underlying subset is a subset of a free generating set.

By way of contradiction, assume that $(F_2,\{x\})$ is projective.

Consider the pearl homomorphism $i\colon (F_2,\{x\}) \to (F_2,\{x,y\})$ induced by the identity map of $F_2$. Now let $f\colon (F_2,\{y\})\to (F_2,\{x,y\})$ be the pearl epimorphism induced by the identity map $F_2\to F_2$. Then there must exist a pearl homomorphism $g\colon (F_2,\{x\})\to (F_2,\{y\})$ such that $f\circ g = i$. Since the underlying group homomorphisms of $f$ and $i$ are the identity, it follows that the underlying group homomorphism of $g$ is the identity. But since $g$ is a pearl homomorphism, we must have $g(x) \in \{y\})$, which is impossible. Thus, $(F_2,\{x\})$ is not projective in the category of pearls.

By taking the abelianizations, we likewise show that $(\mathbb{Z}^2,\{x\})$ is not projective in the category of abelian pearls. $\Box$

In fact, I don't think there are any projectives in these categories. Let $(G,S)$ be any pearl. Now let $H=G\times \langle x\rangle$, with $x$ a nontrivial element of whatever order you want, the group written multiplicatively. Let $T=\{(s,1)\in H\mid s\in S\}\cup \{(e_G,x)\}$. Then $(H,T)$ is a pearl. Let $h\colon (G,S)\to (H,T)$ be the map induced by the embedding $G\hookrightarrow H$. Now let $(K,U)$ be the pearl with $K=H$ and $U=\{(e_G,x)\}$, and $f\colon (K,U)\to (H,T)$ be the homomorphism induced by the identity $K\to H$. This is an epimorphism. If $g\colon G\to K$ is such that $f\circ g = h$, then we must have $g=h$; but then $f(S)$ is not contained in $U$, so $g$ cannot induce a pearl homomorphism. Thus, $(G,S)$ cannot be a projective object. If $G$ is abelian, the example is a diagram of abelian pearls, and so $(G,S)$ is also not projective in the category of abelian pearls.

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I tried to think through the free objects and why the usual proof that free implies projective would fail.

The underlying objects of pearls seem to be pairs, $(X,S)$, where $X$ is a pointed set and $S$ is a nonempty subset of $X$ that does not contain the disinguished object. Morphisms $(X,S)\to(Y,T)$ are set theoretic maps $f\colon X\to Y$ of pointed sets with $f(S)\subseteq T$. A free object would be a pair $(G,\mathcal{S})$ and a morphism of underlying pairs $\iota\colon (X,S)\to (G,\mathcal{S})$ such that for every underlying pair morphism $f\colon (X,S)\to (H,T)$ with $(H,T)$ a pearl, there exists a unique pearl homomorphism $\mathfrak{f}\colon (G,\mathcal{S})\to (H,T)$ such that $\mathfrak{f}\circ \iota = f$. Indeed, the free object will be the free group on $X-\{*\}$ $\iota X\to G$ is the embedding of $X$ into the free generators of $G$ and $*$ to $e_G$: for then the induced group homomorphism from the set-theoretic map $f\colon X\to H$ will satisfy the requirements.

The usual proof that "free implies projective", though, breaks down. Given a free pearl $(F,S)$, with $X$ a free generating set for $F$ that contains $S$, and a homomorphism $h\colon (F,S)\to (H,U)$, and an epimorphism/surjective map $g\colon (G,T)\to (H,U)$, we can use $h$ and $g$ to construct a set map $X\to G$ that will satisfy $f\circ g = h$. However, we may not be able to realize such a map to one in which $g(S)\subseteq T$, so that we cannot actually construct a map of pairs $(X,S)\to (G,T)$ to which we can apply the universal property of the free object. That is what happens with the examples above.

Answer 3

Pearls may be defined as first-order structures $\mathbf P = \langle P; *, {}^{-1},e,S\rangle$ in the language of one binary operation, one unary operation, one 0-ary operation, and one unary relation, axiomatized by

1. identities saying that $\langle P; *, {}^{-1},e\rangle$ is a group,
2. the universal Horn sentence $\neg S(e)$, and
3. the existential sentence $\exists x S(x)$.

Pearl morphisms have been defined so that they are exactly the structure-preserving maps.

Comment 1: Once you define the objects and morphisms of your category, and some underlying set functor is understood, you no longer have the right to define your free objects. If they exist, they are already determined.

Comment 2: The category of pearls, as defined in the problem, has no free objects.

Comment 3: If you delete Axiom 3, the one that asserts that $S$ interprets as a nonempty relation, then the category is a universal Horn class, so free objects exist. The free object over the set $X$ is the free group over $X$ with $S$ interpreted as the empty relation. These are exactly the projective objects in this category, too.