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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.

Actuallly, I'm not sure that free pearls as defined are in fact projective; for that matter. I do have some observations on epimorphisms of pearls first.

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.

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.

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.

Actually, I'm not sure that free pearls are projective.

Consider the free pearl $(F_2,\{x\})$, where $F_2$ is the free group on $x$ and $y$. Now consider the epimorphism $f\colon (F_2,\{y\})\to(F_2,\{x,y\})$ induced by the identity map on $F_2$, and likewise $g\colon (F_2,\{x\})\to (F_2,\{x,y\})$. If $(F_2,\{x\})$ were projective, then there would be a homomorphism $h\colon (F_2,\{x\}) \to (F_2,\{y\})$ such that $fh = g$. But then the map on underlying sets satisfies $\mathrm{id}\circ h = \mathrm{id}$, so $h=\mathrm{id}$; yet we also need $h(x) = y$, which is clearly impossible.

Actuallly, I'm not sure that free pearls as defined are in fact projective; for that matter. I do have some observations on epimorphisms of pearls first.

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 $f(S)=T$.

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.