Proof It is enough to show that the left-to-right implication holds. Suppose $(\Pi _1,\Pi _2)\approx (\Pi _1',\Pi _2')$ and $\Pi $ is not an isomorphism. First, note that $\Pi $ is bijective. Let $F:=\Pi _*(F_{A'})$ be the push-forward sorting data on A. Since $\Pi $ is not an isomorphism, $F_A\subset F$ . Put $\Pi _1":=\Pi \circ \Pi _1':(G_1,F_1)\rightarrow (A,F)$ and $\Pi _2":=\Pi \circ \Pi _2'$ . Since $\Pi :(A',F_{A'})\rightarrow (A,F)$ is an isomorphism,

$$ \begin{align*}(\Pi_1,\Pi_2)\approx(\Pi_1',\Pi_2')\approx (\Pi_1",\Pi_2").\end{align*} $$

So, there is a morphism $\Pi ':(A,F_A)\rightarrow (A,F)$ such that $\Pi '\circ \Pi _1=\Pi _1$ and $\Pi '\circ \Pi _2=\Pi _2$ , which implies that $\Pi '$ is the identity map. Thus, we have that $F\subseteq F_A$ , a contradiction.

Proof Since $H'\le H$ , by [Reference Haran and Lubotzky7, Lemma 1.7], there are epimorphisms $\Pi _1'$ and $\Pi _2'$ with ${\operatorname {Im}}(\Pi _1')={\operatorname {Im}}(\Pi _2')=:A'$ and an epimorphism $\Pi :A'\rightarrow A$ such that the following diagram is commutative:

where the square is Cartesian. Let $F_{A'}$ be a sorting data on $A'$ such that all epimorphisms $\Pi _1'$ , $\Pi _2'$ , and $\Pi $ are sorted. Let $F'$ be the sorting data on $H'$ such that $(H',F')$ is the fibre product of $(G_1,F_1)$ and $(G_2,F_2)$ over $(A',F_{A'})$ . Then, by the choice of $F'$ and $F_{A'}$ , we have that

$$ \begin{align*}((H',F'),\Pi_1',\Pi_2')\le' ((H,F),\Pi_1,\Pi_2).\\[-36pt]\end{align*} $$

Proof Let $\psi _2:(G',F')\rightarrow (G,F)$ be a SEP-cover. Then, $(B,F_B)$ is in ${\operatorname {SIm}}(G',F')$ . Since $(G',F')$ has SEP, there is a morphism $\psi _1:(G',F')\rightarrow (B,F_B)$ such that $\Pi _1\circ \psi _1=\Pi _2 \circ \psi _2$ . Since H is a fibre product of B and G over A, by Remark 2.15(2), there is a homomorphism $\psi : G'\rightarrow H$ such that the following diagram is commutative:

Since $\psi [G']\le H$ and $((H,F_H),\Pi _1,\Pi _2)$ is minimal, we have that $\psi [G']=H$ so that $\psi $ is an epimorphism. Indeed, if $\psi [G']\lneq H$ , then by Remark 3.6, there is $(\Pi _1',\Pi _2')$ in $\mathcal P$ such that $((\psi [G'],F_{\psi [G']}),\Pi _1',\Pi _2')\lneq ' ((H,F_H),\Pi _1,\Pi _2)$ , where $(\psi [G'],F_{\psi [G']})$ is the fibre product of $(B,F_B)$ and $(G,F)$ along $\Pi _1'$ and $\Pi _2'$ . This contradicts the minimality of $((H,F_H),\Pi _1,\Pi _2)$ .

We will show that $\psi $ is sorted. Since $(H,F_H)$ is the fibre product of $(B,F_B)$ and $(G,F)$ , the sorting data $F_H$ is generated by the following pre-sorting data $F_X$ (see Definition 2.14):

• • $X=\{p_1^{-1}[N_1]:N_1\in {\mathcal {N}}(B)\}\cup \{p_2^{-1}[N_2]:N_2\in {\mathcal {N}}(G)\}$ .

• • For $N_1\in {\mathcal {N}}(B)$ and $N_2\in {\mathcal {N}}(G)$ ,

  $$ \begin{align*}F_X(p_1^{-1}[N_1])=F_1(N_1),\ F_X(p_2^{-1}[N_2])=F_2(N_2).\end{align*} $$

Since $\psi _1$ and $\psi _2$ are sorted, for $N_1\in {\mathcal {N}}(B)$ and $N_2\in {\mathcal {N}}(G)$ ,

$$ \begin{align*}F_1(N_1)\subseteq F'(\psi_1^{-1}[N_1]),\ F_2(N_2)\subseteq F'(\psi_2^{-1}[N_2]).\end{align*} $$

Since $\psi _1=p_1\circ \psi $ and $\psi _2=p_2\circ \psi $ , we have that for $N_1\in {\mathcal {N}}(B)$ and $N_2\in {\mathcal {N}}(G)$ ,

$$ \begin{align*} F'(\psi^{-1}[p_1^{-1}[N_1]])&=F'(\psi_1^{-1}[N_1])\\ &\supseteq F_1(N_1)\\ &=F_H(p_1^{-1}[N_1]), \end{align*} $$

and

$$ \begin{align*} F'(\psi^{-1}[p_2^{-1}[N_2]])&=F'(\psi_2^{-1}[N_2])\\ &\supseteq F_2(N_2)\\ &=F_H(p_2^{-1}[N_2]), \end{align*} $$

which implies $\psi $ is sorted because $F_H$ is generated by $F_X$ . Therefore, we have that $\psi _2=p_2\circ \psi $ for a morphism $\psi $ , and $p_2$ is a q.s.c.

Proof Suppose a sorted profinite group $(G,F)$ does not have SEP. So, there exist:

• • $(A,F_A),(B,F_B)\in {\operatorname {SIm}}(G,F)$ ; and

• • morphisms $\pi _1:(B,F_B)\rightarrow (A,F_A)$ and $\pi _2 :(G,F)\rightarrow (A,F_A)$ ,

such that there is no morphism $p:(G,F)\rightarrow (B,F_B)$ with $\pi _2=\pi _1\circ p$ . By Lemma 3.7, there is a maximal element $(\Pi _1,\Pi _2)\in {\mathcal {P}}((B,F_B),(G,F))$ such that $(\pi _1,\pi _2)\le (\Pi _1,\Pi _2)$ . Then, we have the following diagram:

where $(H,F_H)$ is the fibre product of $(B,F_B)$ and $(G,F)$ over $(A',F_{A'})$ .

Note that $p_2$ is a q.s.c. by Lemma 3.10. Suppose $p_2$ is an isomorphism. Let $p=p_1\circ p_2^{-1}$ . Then, we have that

$$ \begin{align*} \pi_1\circ p&=\pi_1\circ (p_1\circ p_2^{-1})\\ &=\left ((\pi\circ \Pi_1)\circ p_1\right) \circ p_2^{-1}\\ &=\pi\circ (\Pi_2\circ p_2)\circ p_2^{-1}\\ &=\pi\circ \Pi_2\\ &=\pi_2, \end{align*} $$

which is a contradiction. So, $p_2$ is not an isomorphism. If $p_2$ has a non-trivial kernel, then $p_2$ is the desired q.s.c. Suppose $p_2$ has trivial kernel. Consider a sorting data $F'$ on G given as follows: For $N\in {\mathcal {N}}(G)$ ,

$$ \begin{align*}F'(N):=F_H(p_2^{-1}[N]).\end{align*} $$

Then, $p_2 : (H,F_H)\rightarrow (G,F')$ is an isomorphism and a q.s.c. So, by Remark 3.9(1), ${\operatorname {id}}=p_2^{-1}\circ p_2 :(G,F')\rightarrow (H,F_H)\rightarrow (G,F)$ is a q.s.c. with $F\subset F'$ .

Proof If $\alpha $ is a successor ordinal, that is, $\alpha =\alpha '+1$ , then $(G,F)=(G_{\alpha '},F_{\alpha '})$ and we are done. We assume that $\alpha $ is a limit ordinal. Let $p:(G',F')\rightarrow (G_0,F_0)$ be a SEP-cover. To show that $\pi $ is a q.s.c., using transfinite induction, we will construct a sequence $(p_i:(G',F')\rightarrow (G_i,F_i))_{i<\alpha }$ of morphisms such that for each $0\le i<j<\alpha $ , $p_i=\pi _{j,i}\circ p_j$ . Put $p_0:=p$ . Suppose we have constructed $(p_i)_{i< \gamma }$ for some $\gamma <\alpha $ . If $\gamma $ is a limit ordinal, there is a desired morphism $p_{\gamma }:(G',F')\rightarrow (G_{\gamma },F_{\gamma })$ because $(G_{\gamma },F_{\gamma })$ is the inverse limit of $(G_i,F_i)_{i<\gamma }$ . If $\gamma =\gamma '+1$ is a successor ordinal, there is a morphism $r:(G',F')\rightarrow (G_{\gamma },F_{\gamma })$ such that $p_{\gamma '}=\pi _{\gamma ,\gamma '}\circ r$ because $\pi _{\gamma ,\gamma '}$ is a q.s.c. Put $p_{\gamma }:=r$ , which is the desired one.

Since $(G,F)$ is the inverse limit of $(G_i,F_i)_{i<\beta }$ , there is $q:(G',F')\rightarrow (G,F)$ such that for each $i<\alpha $ , $p_i=\pi _i\circ q$ . Thus, we have that $p=\pi _0\circ q$ , and $\pi _0$ is a q.s.c.

Proof If $(G,F)$ has SEP, then ${\operatorname {id}} : (G,F)\rightarrow (G,F)$ is a universal SEP-cover. Suppose $(G,F)$ does not have SEP. Let $\kappa $ be a cardinal such that any q.s.c. $(H,F_H)$ of $(G,F)$ has the cardinality less than $\kappa $ and let $\aleph $ be a cardinal with $2^{\kappa }<\aleph $ .

Using transfinite induction, we will construct an inverse system $((G_i,F_i))_i$ indexed by ordinals $i\le \alpha $ for some ordinal $\alpha <\aleph $ such that:

1. (1) For each $i<j$ , the transition map $\pi _{j,i}$ is a q.s.c. with a non-trivial kernel.

2. (2) For each limit ordinal $\beta $ , $(G_{\beta },F_{\beta })$ is the inverse limit of the inverse system $((G_i,F_i))_{i<\beta }$ with transition maps $\pi _{i,j}$ for $i<j<\beta $ .

3. (3) For each limit ordinal $\beta $ and for $i<\beta $ , the transition map $\pi _{\beta ,i}$ is the natural projection from $(G_{\beta },F_{\beta })$ to $(G_i,F_i)$ coming from the inverse limit construction.

4. (4) Any q.s.c. to $(G_{\alpha },F_{\alpha })$ has trivial kernel.

Put $(G_0,F_0)=(G,F)$ . Suppose we have constructed an inverse system $((G_i,F_i))_{i<\beta }$ for an ordinal $\beta $ , which satisfies $(1)$ – $(3)$ .

Case. $\beta $ is a successor ordinal, that is, $\beta =\beta '+1$ . If $(G_{\beta '},F_{\beta '})$ has SEP, then we stop the process. Suppose $(G_{\beta '},F_{\beta '})$ does not have SEP. By Lemma 3.11, there is a q.s.c. to $(G_{\beta '},F_{\beta '})$ , which is not an isomorphism. If every q.s.c. to $(G_{\beta '},F_{\beta '})$ is injective, we stop the process. So, we may assume that there is a q.s.c. $p:(G',F')\rightarrow (G_{\beta '},F_{\beta '})$ with a non-trivial kernel. Put $(G_{\beta },F_{\beta }):=(G',F')$ and $\pi _{\beta ,\beta '}:=p$ . For each $i<\beta $ , put $\pi _{\beta ,i}:=\pi _{\beta ',i}\circ p$ . By Remark 3.9(1), each $\pi _{\beta ,i}$ is a q.s.c.

Case. $\beta $ is a limit ordinal. Let $(G_{\beta },F_{\beta })$ be the inverse limit of $((G_i,F_i))_{i<\beta }$ . For each $i<\beta $ , let $\pi _{\beta ,i}$ be the natural projection map from $(G_{\beta },F_{\beta })$ to $(G_i,F_i)$ . By Lemma 3.12, each $\pi _{\beta ,i}$ is a q.s.c.

For each $i<j$ , $\pi _{j,i}$ has a non-trivial kernel. Namely, suppose there are $i<j$ such that ${\operatorname {Ker}}(\pi _{j,i})$ is trivial. Since $\pi _{j,i}=\pi _{i+1,i}\circ \pi _{j,i+1}$ , where $\pi _{k,k}={\operatorname {id}}$ for each k, ${\operatorname {Ker}}(\pi _{i+1,i})$ is also trivial, which is a contradiction. In our construction, $\beta $ should be less than $\aleph $ . Suppose $\beta \ge \aleph $ . Without loss of generality, we may assume that $\beta $ is a successor ordinal and write $\beta =\beta '+1$ . We have that $|G_{\beta '}|\le \kappa $ . So, $|{\mathcal {N}}(G_{\beta '})|\le 2^{\kappa }$ . Since $|\beta '|\ge \aleph>2^{\kappa }$ , by the pigeon hole principle, for some $i<j<\beta '$ , ${\operatorname {Ker}}(\pi _{\beta ',j})={\operatorname {Ker}}(\pi _{\beta ',i})$ . Since $\pi _{\beta ',i}=\pi ^j_i\circ \pi _{\beta ',j}$ , we have that $\pi _{j,i}$ is injective, which is a contradiction.

Now, take a maximal inverse system $((G_i,F_i))_{i<\beta }$ satisfying $(1)$ – $(3)$ . Note that $\beta $ is a successor ordinal and write $\beta =\alpha +1$ . Then, for the q.s.c. $\pi _{\alpha ,0}:(G_{\alpha },F_{\alpha })\rightarrow (G,F)$ , we have that every q.s.c. to $(G_{\alpha },F_{\alpha })$ is injective and so $((G_i,F_i))_{i\le \alpha }$ is the desired one.

Let $(E(G),E(F)):=(G_{\alpha },F_{\alpha })$ and let $\pi :=\pi _{\alpha ,0}$ . If $(E(G),E(F))$ has SEP, then $\pi $ is a universal SEP-cover of $(G,F)$ . Suppose $(E(G),E(F))$ does not have SEP. By transfinite induction, we will construct a sequence $(E^i(F))_{i\le \gamma '}$ of sorting data on $E(G)$ such that for $i<j\le \gamma '$ :

• • $E^i(F)\subset E^j(G)$ .

• • ${\operatorname {id}}_{j,i}:={\operatorname {id}}:(E(G),E^j(F))\rightarrow (E(G),E^i(F))$ is a q.s.c.

• • $(E(G),E^{\gamma '}(F))$ has SEP.

Suppose we have constructed such a sequence $(E^i(F))_{i< \alpha '}$ for an ordinal $\alpha '$ . If $\alpha '$ is a limit ordinal, put $E^{\alpha '}(F):=\bigcup _{i<\alpha '}E^i(F)$ . By Lemma 3.12, each ${\operatorname {id}}_{\alpha ',i}:(E(G),E^{\alpha '}(F))\rightarrow (E(G),E^i(F))$ is a q.s.c.

Suppose $\alpha '$ is a successor ordinal, that is, $\alpha '=\alpha "+1$ . If $(E(G),E^{\alpha "}(F))$ has SEP, put $\gamma '=\alpha "$ and stop the process. If $(E(G),E^{\alpha "}(F))$ does not have SEP, then by Lemma 3.11(2), there is a sorting data $F'$ on $E(G)$ such that $E^{\alpha "}(F)\subset F'$ and ${\operatorname {id}}:(E(G),F')\rightarrow (E(G),E^{\alpha "}(F))$ is a q.s.c, and put $E^{\alpha '}(F):=F'$ . Note that we cannot apply Lemma 3.11(1) in this case. Namely, if there is a q.s.c $\pi :(G',F')\rightarrow (E(G),E^{\alpha "}(F))$ whose kernel is non-trivial, then we have a q.s.c. ${\operatorname {id}} \circ \pi :(G',F')\rightarrow (E(G),E^{\alpha "}(F))\rightarrow (G,F)$ whose kernel is non-trivial, which contradicts the choice of $(G,F)$ .

This transfinite inductive process should stop for some $\gamma '<|{\mathcal {J}}|^+$ so that $(E(G),E^{\gamma '}(F))$ has SEP. Thus, a q.s.c.

$$ \begin{align*}\pi_{\alpha,0}\circ {\operatorname{id}}_{\gamma',0}:(E(G),E^{\gamma'}(F))\rightarrow (E(G),E(F))\rightarrow(G,F)\end{align*} $$

is a SEP-cover so that the SEP-cover $\pi _{\alpha ,0}\circ {\operatorname {id}}_{\gamma ',0}$ is a universal SEP-cover.

We now prove the furthermore part. Suppose G is finitely generated. Let $p_i:(G_i,F_i)\rightarrow (G,F)$ be a universal SEP-cover of $(G,F)$ for $i=1,2$ . By Remark 3.3 and [Reference Haran and Lubotzky7, Theorem 1.12], $G_1$ and $G_2$ are finitely generated. By universality, there are morphisms $q:(G_2,F_2)\rightarrow (G_1,F_1)$ and $q':(G_1,F_1)\rightarrow (G_2,F_2)$ such that ${p_2=p_1\circ q}$ and $p_1=p_2\circ q'$ . Since $G_1$ and $G_2$ are finitely generated, by [Reference Ribes13, Proposition 7.6], both q and $q'$ are bijective. We want to show that $q_*(F_2)=F_1$ . Since $q:(G_2,F_2)\rightarrow (G_1,F_1)$ is a morphism, $F_1\subseteq q_*(F_2)$ . Also we have that $q:(G_2,F_2)\cong (G_1,q_*(F_2))$ and $p_2\circ q^{-1}:(G_1,q_*(F_2))\rightarrow (G,F)$ is a universal SEP-cover. So, there is a morphism $r:(G_1,F_1)\rightarrow (G_1,q_*(F_2))$ such that $p_1=(p_2\circ q^{-1})\circ r$ . Since $p_2=p_1\circ q$ , the morphism r should be the identity map on G and $q_*(F_2)\subseteq F_1$ . Thus, $F_1=q_*(F_2)$ .

Fact 3.16 [Reference Chatzidakis1, Theorem 2.7].

A profinite group H has the unique universal embedding cover $E(H)$ and

$$ \begin{align*}{\operatorname{Im}}(E(H))=\bigcup_{A\in {\operatorname{Im}}(H)}{\operatorname{Im}}(E(A)),\end{align*} $$

where $E(A)$ is the unique embedding cover of A.

Proof By Theorem 3.13, each $(A,F_A)\in {\operatorname {FSIm}}(H,F_H)$ has the unique SEP-cover $(E(A),E(F_A))$ . Let

$$ \begin{align*}\Gamma:=\bigcup_{(A,F)\in {\operatorname{FSIm}}(H,F_H)}{\operatorname{FSIm}}(E(A),E(F_A)).\end{align*} $$

Let G be the unique embedding cover of H. Note that

$$ \begin{align*}{\operatorname{Im}}(G)=\{A:(A,F_A)\in \Gamma \mbox{ for some sorting data }F_A\}\end{align*} $$

by Fact 3.16. Define a pre-sorting data $\hat F$ on G as follows: For each $N\in {\mathcal {N}}(G)$ ,

$$ \begin{align*}\hat F(N):=\bigcup_{(A,F_A)\in \Gamma, A\cong G/N}F_A(\{e_A\}).\end{align*} $$

Proof It is enough to show that $\hat F$ satisfies $(3)$ and $(4)$ in Definition 2.2, that is:

1. (3) Suppose $N_1\subseteq N_2$ and $[G:N_1]\le k$ . For $J\in {\mathcal {J}}^{<\mathbb N}$ ,

   $$ \begin{align*}J\in \hat F(N_1)\Rightarrow J^*_{\subseteq}(k,J)\in \hat F(N_2).\end{align*} $$

2. (4) For $J_1\in \hat F(N_1)$ and $J_2\in \hat F(N_2)$ , $J^*_{\cap }(J_1,J_2)\in \hat F(N_1\cap N_2)$ .

$(3)$ Let $N_1\subseteq N_2\in {\mathcal {N}}(G)$ with $[G:N_1]\le k$ and let $J\in {\mathcal {J}}^{<\mathbb N}$ with $J\in \hat F(N_1)$ . By the choice of $\hat F$ , there are:

• • the universal SEP-cover $(E(A),E(F_A))$ of some $(A,F_A)\in {\operatorname {FSIm}}(H,F_H)$ ; and

• • $(G/N_1,F_1)\in {\operatorname {FSIm}}(E(A),E(F_A))$ such that $J\in F_1(\{e_{G/N_1}\})$ .

Then, $J^*_{\subseteq }(k,J)\in F_1(N_2/N_1)$ . Consider a push-forward sorting data $F_2:=\pi _*(F_1)$ on $G/N_2$ where $\pi :G/N_1\rightarrow G/N_2$ is the projection. Then, $(G/N_2,F_2)\in {\operatorname {FSIm}}(E(A),E(F_A))$ and $J^*_{\subseteq }(k,J)\in F_2(\{e_{G/N_2}\})=F_1(N_2/N_1)\subseteq \hat F(N_2)$ . $(4)$ Take $N_1,N_2\in {\mathcal {N}}(G)$ , $J_1\in \hat F(N_1)$ , and $J_2\in \hat F(N_2)$ arbitrary. By the choice of $\hat F$ , for each $i=1,2$ , there are:

• • $\bar N_i\in {\mathcal {N}}(H)$ ;

• • the universal SEP-cover $(E(G/\bar N_i),E(\bar F_i))$ of some $(G{/}\bar N_i,\bar F_i)\kern1.2pt{\in}\kern1.2pt {\operatorname {FSIm}}(H,F_H)$ ; and

• • $(G/N_i,F_i)\in {\operatorname {FSIm}}(E(G/\bar N_i),E(\bar F_i))$ such that $J_i\in F_i(\{e_{G/N_i}\})$ .

Let $\bar F_0$ be a sorting data on $H/(\bar N_1\bar N_2)$ such that the projection from $H/\bar N_i$ to $H/(\bar N_1\bar N_2))$ is sorted for $i=1,2$ . Let $(H/(\bar N_1\cap \bar N_2),\bar F_3)$ be the fibre product of $(H/\bar N_1,\bar F_1)$ and $(H/\bar N_2,\bar F_2)$ over $(H/(\bar N_1\bar N_2), \bar F_0)$ . Let $\left (E(H/(\bar N_1\cap \bar N_2)),E(\bar F_3)\right )$ be the unique SEP-cover of $(H/(\bar N_1\cap \bar N_2),\bar F_3)$ .

Let $F_0$ be a sorting data on $G/(N_1N_2)$ such that the projection from $G/N_i$ to $G/(N_1N_2)$ is sorted for $i=1,2$ . Let $(G/(N_1\cap N_2),F_3)$ be the fibre product of $(G/N_1,F_1)$ and $(G/N_2,F_2)$ over $(G/(N_1N_2),F_0)$ .

Then, there is a morphism from $(E(H/(\bar N_1\cap \bar N_2)),E(\bar F_3))$ to $(E(H/\bar N_i),E(\bar F_i))$ and so there is a morphism from $(E(H/(\bar N_1\cap \bar N_2)),E(\bar F_3))$ to $(G/N_i,F_i)$ for $i=1,2$ . Since $(G/(N_1\cap N_2),F_3)$ is the fibre product of $(G/N_1,F_1)$ and $(G/N_2,F_2)$ over $(G/(N_1N_2),F_0)$ , there is a morphism from $(E(H/(\bar N_1\cap \bar N_2)),E(\bar F_3))$ to $(G/(N_1\cap N_2),F_3)$ . So, we have the following diagram:

Therefore,

$$ \begin{align*}(G/(N_1\cap N_2),F_3)\in {\operatorname{FSIm}}(E(H/(\bar N_1\cap \bar N_2)),E(\bar F_3))\subseteq \Gamma,\end{align*} $$

and so

$$ \begin{align*}J^*_{\cap}(J_1,J_2)\in F_3(\{e_{G/(N_1\cap N_2)}\})\subseteq \hat F(N_1\cap N_2).\\[-35pt]\end{align*} $$

By Theorem 2.22(3), $(G,\hat F)$ has SEP. Now we show that $(G,\hat F)$ is the unique universal SEP-cover of $(H,F_H)$ .

Claim 3.19. For any SEP-cover $(G',F')$ of $(H,F_H)$ , any epimorphism ${\varphi :G'\rightarrow G}$ is sorted, that is, for each $N\in {\mathcal {N}}(G)$ ,

$$ \begin{align*}\hat F(N)\subseteq F'(\varphi^{-1}[N]).\end{align*} $$

Let $(G',F')$ be a SEP-cover of $(H,F_H)$ . Since $G'$ is an embedding cover of H and G is the universal embedding cover of H, there is an epimorphism from $\varphi :G'\rightarrow G$ . By Claim 3.19, the epimorphism $\varphi $ is sorted. This implies that $(G,F)$ is a universal SEP-cover of $(H,F_H)$ .

It remains to show that $(G, \hat F)$ is unique up to isomorphism. Let $(G',F')$ be a universal SEP-cover of $(H,F_H)$ . Since $G'$ is a universal embedding cover of H, there is an isomorphism $\varphi :G'\rightarrow G$ . We show that the isomorphism $\varphi :G'\rightarrow G$ induces an isomorphism $\varphi :(G',F')\rightarrow (G,\hat F)$ . It is enough to show that for any $N\in {\mathcal {N}}(G)$ ,

$$ \begin{align*}\hat F(N)=F'(\varphi^{-1}[N]).\end{align*} $$

Take $N\in {\mathcal {N}}(G)$ arbitrary. By Claim 3.19, $\varphi $ is sorted so that $\hat F(N)\subseteq F'(\varphi ^{-1}[N])$ . Since $(G',F')$ is a universal SEP-cover of $(H,F_H)$ , there is a morphism $\psi :(G,\hat F)\rightarrow (G',F')$ . So, we have that:

• • $G/N\cong G'/\varphi ^{-1}[N]\cong G/\psi ^{-1}[\varphi ^{-1}[N]]$ ; and

• • $\hat F(N)\subseteq F'(\varphi ^{-1}[N])\subseteq \hat F(\psi ^{-1}[\varphi ^{-1}[N]])$ .

Since $G/N\cong G/\psi ^{-1}[\varphi ^{-1}[N]]$ , by Lemma 2.21,

$$ \begin{align*}\hat F(\psi^{-1}[\varphi^{-1}[N]])=\hat F(N)\end{align*} $$

and so $\hat F(N)= F'(\varphi ^{-1}[N])= \hat F(\psi ^{-1}[\varphi ^{-1}[N]])$ .

Proof $(1)$ Suppose $S_1$ is finitely generated by a finite subset X of $S_1$ . We may assume that X is relatively dense. Since X is finite, $\min X\neq \emptyset $ , and since X is relatively dense, $\min X\subset \min S_1$ .

Next, we show that $S_1\vee S_2$ is not empty. Choose $a_0\in \min S_1$ arbitrary. Then, there are only finitely many elements in the set $\{x\in S:a_0\le x\}$ up to $\sim $ and so there is a minimal element $c\in S$ such that $[c]=[a_0]\vee [b]$ for some $b\in S_2$ . So, $c\in S_1\vee S_2$ and $S_1\vee S_2\neq \emptyset $ .

For all $a_0\in \min S_1$ and for all $c\in S_1\vee S_2$ , $a_0\le c$ , and so there are only finitely many elements in $S_1\vee S_2$ up to $\sim $ . Thus, $S_1\vee S_2\neq \emptyset $ .

It remains to show that $\min S_0\sim S_1\vee S_2$ . Since $S_0\subset S_1$ , we have that $\min S_0\neq \emptyset $ . Take $c\in S_1\vee S_2$ and $d\in \min S_0$ arbitrary. Since $d\in S_1\cap S_2$ , it is clear that $c\le d$ . Take $a\in S_1\cap m(k,J)$ and $b\in S_2\cap m(k',J')$ such that $[c]=[a]\vee [b]$ and $c\in m(kk',J_{\subset }^*(k,J))$ . Since $S_1$ and $S_2$ are subsystems, there is $c'\in S_1\cap S_2\cap m(kk',J_{\subset }^*(k,J)^{\frown } JJ')$ such that $c'\sim c$ . Thus, $d\le c'\le c$ .

$(2)$ Fix $a\in \min S'\cap m(k,J)$ for some $(k,J)\in \mathbb N\times {\mathcal {J}}^{<\mathbb N}$ . Since $G(S)$ is the inverse limit of the group $[x]\cap m(k',J')$ of $x\in m(k',J')$ with the transition maps induced from C, $G(S)\cong [a]\cap m(k,J)$ . Also for each $b\in S$ , after putting $C(b,S):=\{x\in S:S\models C(b,x)\}$ , $C(b,S)\cap m(k',J')$ is linearly ordered by $\le $ for each $(k',J')\in \mathbb N\times {\mathcal {J}}^{<\mathbb N}$ . Thus, for each $x\in m(k',J')$ , there is a unique $a_x\in [a]\cap m(k,J)$ such that x is the unique element in S satisfying $S\models C(a_x,x)\wedge x\in m(k',J')$ .

$(3)$ Let $S'\models SCS$ . Suppose $S'$ is finitely generated by a finite presystem X of $S'$ . Since X is relatively dense in $S'$ and X is finite, $\min X\neq \emptyset $ and $\min X\sim \min S'$ . By Remark 4.5(2), $G(S')$ is finite. Also, since $S'=S_{X'}$ and $X'$ is finite, $F(S')$ is finitely generated. So, ${\mathcal {G}}(S')=(G(S'),F(S'))$ is finitely sorted.