Proof

1. 1) This is by compactness, in detail if $\{C_i\}_{i\in I}$ is a family of closed subsets of $X/E$ such that $\{C_i\cap \pi (Y)\}_{i \in I}$ has the finite intersection property then $\{\pi ^{-1}(C_i)\cap Y\}_{i\in I}$ are a bounded family of type-definable sets with the finite intersection property, and so it has non-empty intersection.

2. 2) Take Z a type-definable set. Then $\pi ^{-1}\pi (Z\cap Y)=Y\cap \{x\in X\mid \text {there exists }x'\in Z, (x,x')\in E\}$ , as Z and E are type-definable $\pi ^{-1}\pi (Y\cap Z)$ is type-definable relative to Y and $\pi (Y\cap Z)$ is closed as required.

   Now to see that $Y/E_Y$ is Hausdorff take $x,y\in Y$ such that $\pi (x)\neq \pi (y)$ . Then for $E_z=\{x'\in X\mid (z,x')\in E\}$ we have $E_x\cap E_y=\emptyset $ . As $E_x$ and $E_y$ are type-definable then there are definable sets $D_1,D_2$ such that $E_x\subset D_1$ $E_y\subset D_2$ and $D_1\cap D_2=\emptyset $ . Then by the previous paragraph there are open sets $U_1,U_2\subset X/E$ such that $\pi ^{-1}(U_i)\subseteq D_i$ .

3. 3) By 2) this space is Hausdorff. Denote $\pi :H\to H/K$ . If C is type-definable and $C\cap H$ is $\pi $ -saturated, then $(C\cap H)^{-1}=C^{-1}\cap H$ is type-definable relative to H and $\pi $ -saturated, so inversion is continuous in $H/K$ . Similarly one obtains that left and right translations are continuous.

   Finally one has to see that the product is continuous at the identity, let $1\subset U\subset H/K$ be an open set, and let $Z\subset G$ be type-definable such that $H\setminus Z=\pi ^{-1}(U)$ . Then $K\cap Z=\emptyset $ . As $K=K^2$ one also has $K^2\cap Z=\emptyset $ . Now by compactness there is $K\subset V$ definable such that $(V\cap G)^2\cap Z=\emptyset $ . Now by 2) there exists $1\subset V'\subset \pi (V\cap H)$ open such that $\pi ^{-1}(V')\subset V\cap H$ , so $(V')^{2}\subset U$ as required. ⊣

Proof Let $o\subset U\subset O$ be an $L'$ -definable symmetric set such that the restriction of $f:O\to G$ to U is injective. Then $f(o)\subset f(U)$ , is such that $f(U)$ is $L'$ -definable and $f(o)$ is L-type-definable. By compactness there is $f(o)\subset R\subset f(U)$ such that R is L-definable. Replacing U by $f^{-1}(R)\cap U$ we may assume $f(U)$ is L-definable. Now take V a symmetric $L'$ -definable set such that $o\subset V$ and $V^4\subset U$ . Similarly replacing V by $f^{-1}(R')\cap V$ for a suitable $R'$ , we may further assume that $f(V)$ is L-definable. A bounded number of translates of V cover O, so by compactness, a denumerable number also cover it say $O=\bigcup _{i<\omega }a_iV$ .

Take $X=\bigsqcup _{i<\omega }f(V)$ , with canonical injections $\tau _i:f(V)\to X$ . Define a map $g:X\to O$ by $g\tau _if(x)=a_ix$ for $x\in V$ . This is a surjective $L'$ -definable map. Define $W_i=f(V)\setminus (g\tau _i)^{-1}\bigcup _{r<i}(g\tau _r)f(V)$ . Then $W_i$ is L-definable. We have $O'=\bigsqcup _{i<\omega } W_i$ a disjoint union of L-definable sets and the restriction of g to $O'$ is a bijective $L'$ -definable map, denoted also by g, $g:O'\to O$ . We also have that $fg\tau _i(W_i)$ is L-definable, and the restriction of $fg\tau _i$ to $W_i$ is translation by $f(a_i)$ which is L-definable. We conclude that $fg$ is L-definable.

Now we transfer the group structure of O to $O'$ using the bijection g. Finally we have to prove that this group structure on $O'$ is L-definable. Now $g(W_i)^{-1}$ is contained in a finite number of translates $a_jV$ so $W_i^{-1}$ is contained in a finite number of $W_j$ . Similarly $W_iW_{i'}$ is contained in a finite number of $W_{j}$ . So it is enough to see that product and inverse are relatively definable. We do the product as the inverse is similar. Suppose $(x,y,z)\in W_i\times W_j\times W_r$ , then $\tau _r(z)=\tau _i(x)\tau _j(y)$ in $O'$ if and only if $a_rz'=a_ix'a_jy'$ in O for $x',y',z'\in V$ which have images under f equal to $x,y$ , and z. In this case $a_ra_i^{-1}a_j^{-j}\in V^3$ (here we use the hypothesis of commutativity), so as f is injective in $V^4$ we get that this happens if and only if $f(a_ra_i^{-1}a_j^{-1})z=xy$ in G. This last condition is L-definable. ⊣

Proof First we note that $o_Z(a)$ is a convex subgroup of the ordered group $O_Z(a)$ , so $O_Z(a)/o_Z(a)$ is an ordered abelian group.

Define a function $\mathbb {Q}\to O_Z(a)$ by $\frac {n}{m}\mapsto \frac {n}{m}(a-r_m)$ , where $n,m\in \mathbb {Z}$ , $(n,m)=1$ , $m>0$ , $r_m\in \mathbb {N}$ is such that $0\leq r_m<m$ , and $a\equiv r_m \mod mZ$ . A straightforward check shows that the composition of this function with the canonical projection $O_Z(a)\to O_Z(a)/o_Z(a)$ is an ordered group morphism, and that this morphism extends in a unique way to an ordered group morphism $\mathbb {R}\to O_Z(a)/o_Z(a)$ . This morphism $\mathbb {R}\to O_Z(a)/o_Z(a)$ is an isomorphism of topological groups. Indeed if we define $F:O_Z(a)\to \mathbb {R}$ as $F(b)=\text {Sup}\{\frac {n}{m}\mid n,m\in \mathbb {Z}, m>0, na<mb\}$ , then F factors through $O_Z(a)/o_Z(a)$ as $\bar {F}:O_Z(a)/o_Z(a)\to \mathbb {R}$ and $\bar {F}$ is the inverse to the map $\mathbb {R}\to O_Z(a)/o_Z(a)$ described before.

It is easy to verify that $\bar {F}$ is continuous if we give $O_Z(a)/o_Z(a)$ the logic topology and $\mathbb {R}$ the usual topology. Now notice that $o_Z(a)$ is type-definable, so $O_Z(a)/o_Z(a)$ is a Hausdorff topological group. Also, $O_Z(a)$ is a denumerable union of definable sets, so $O_Z(a)/o_Z(a)$ is $\sigma $ -compact. As $\mathbb {R}$ is a locally compact Hausdorff topological group, $\bar {F}$ is open by Proposition 2.3 as required. ⊣

Proof Let $G=(a_1+S_1)\cup \cdots \cup (a_n+S_n)$ be a cell decomposition of G, where for each $i S_i=0$ or $S_i=(-b_i,c_i)\cap m_iZ$ for $b_i,c_i\in Z_{>0}$ and $m_i\in \mathbb {Z}$ or $S_i=m_iZ$ or $S_i=(-b_i,\infty )\cap m_iZ$ or $S_i=(-\infty ,c_i)\cap m_iZ$ . Note that if $S_i$ is unbounded, then $\langle a_i +S_i\rangle =a_i\mathbb {Z} +m_iZ$ so $m_iZ\subset G$ . We note that there is at least one index r such that $S_r$ is unbounded, otherwise G is bounded and so as it is definable it has a maximum, but G is a group so this cannot happen unless G is trivial. Then $m_rZ\subset G$ and as $Z/m_rZ\cong \mathbb {Z}/m_r\mathbb {Z}$ is finite cyclic we see that any intermediate group is of the form $nZ$ for an $n|m_r$ . ⊣

Proof Let $H\subset D$ , with D a definable set. We need to see that there is $a\in Z$ such that $H\subset o_Z(a)\subset D$ . By compactness there is $H\subset S\subset D$ such that S is definable symmetric and convex. If $S=H$ then $H=Z$ or $H=0$ by Lemma 3.3. Assume otherwise and take $a\in S\setminus H$ positive. As H is convex $h<a$ for all $h\in H$ . As H is a group, $H\subset o_Z(a)\subset [-a,a]\subset D$ , as required. ⊣

Proof Take C the convex hull and $H\subset D$ a definable set. We have to show that there is a natural number k such that $H\subset kZ\cap C\subset D$ . Indeed because H is a bounded intersection of definable sets this would imply that H is an intersection of groups of the form $kZ\cap C$ . If n is the supernatural number which is the smallest common multiple of all the natural numbers k such that $H\subset kZ$ , then $H=nZ\cap C$ .

Take $H\subset D_2\subset D_1$ symmetric sets such that $D_1+D_1\subset D$ , and $D_2+D_2\subset D_1$ . Take a cell decomposition of $D_2$ , say $D_2=E_1\cup \cdots \cup E_n$ . Then, as in the proof of Lemma 3.3 we conclude that if H is nontrivial then for one of the cells, say $E_1$ , $E_1\cap H$ is unbounded above in H, and for this cell $E_1$ we have that $E_1-E_1$ contains a group of the form $mZ\cap C$ , so that in particular $mZ\cap C\subset D_1$ ; and finally we also have that $H+mZ\cap C=kZ\cap C$ for some natural number k, because it is an intermediate group of $mZ\cap C$ and C which have quotient $C/mZ\cap C\cong \mathbb {Z}/m\mathbb {Z}$ which is cyclic. As $H+mZ\cap C=kZ\cap C\subset D_1+D_1\subset D$ we conclude. ⊣

Proof Take the polynomial $P(T)=T^n-ba^{-1}$ . If R is the valuation ring of K, we have that $P(T)\in R[T]$ . We also have that $v(P(1))>2v(P'(1))$ . We conclude by Hensel’s lemma that there is a unique solution $x\in R$ of $P(T)$ such that $v(x-1)>v(n)$ , as required. ⊣

Proof Note that by Hensel’s lemma $D_{v(b)+2v(n)+1}\subset S$ , see Lemma 4.3. If $\alpha =v(b)+N$ for some $N\in \mathbb {Z}$ then we are done. Now take the set $M=v(S\setminus \{0\})\subset Z$ . This is a definable subset of Z, bounded below by $\alpha $ , so it has a minimum $\beta $ . Now it is clear that we may replace $\alpha $ by $\beta $ and assume $\alpha <v(b)+m$ for all $m\in \mathbb {Z}$ . Now let $a\in S$ be such that $v(a)=\alpha $ . Then by Lemma 4.3 $\{c\in Q_p\mid v(c-a)>\alpha +2v(n)\}\subset S$ . But then $D_{\alpha +2v(n)+1}\subset S-S$ which finishes the proof. ⊣

Proof Take $M=v(S\setminus \{0\})$ . If this set is not bounded below, then the argument of the previous proof shows $D_{\beta +2v(n)+1}\subset S-S$ for all $\beta \in M$ so $S-S=Q_p$ . If M is bounded below, then we are in the situation of the previous Lemma. ⊣

Proof S decomposes as a finite union of cells $a_i+S_i$ where $S_i$ are as in the previous two Lemmas or are $0$ . If all the $S_i$ are $0$ or there is i such that $Q_p=S_i-S_i$ then we are done. Otherwise take $\alpha _i$ as in the previous lemmas for every i such that $S_i\neq 0$ . Then for $\alpha =\text {min}_i\alpha _i$ we obtain the result. ⊣

Proof Let L be a type-definable subgroup of $Q_p$ , and $L\subset S\subset Q_p$ a definable set. We prove that either $Q_p=S$ or $L=0$ or there exists $\alpha \in Z$ such that $L\subset D_{\alpha }\subset S$ . Let T be a symmetric definable set with $L\subset T$ and $3T\subset S$ . If T is finite then L is a finite group, and as $Q_p$ is torsion free, it is trivial. If $T-T=Q_p$ then $S=Q_p$ as required. So assume by the previous Lemma that $T\subset D_{\alpha }+X$ and $D_{\alpha +n}\subset 2T$ . Now $L+D_{\alpha +n}$ is a type-definable group and satisfies $L+D_{\alpha +n}\subset D_{\alpha }+X$ and $L+D_{\alpha +n}\subset S$ .

Now we claim that the group generated by $D_{\alpha }+X$ is of the form $D_{\alpha +m}\oplus \oplus _i \mathbb {Z}a_i$ where every nonzero element in $\Sigma _i \mathbb {Z}a_i$ has valuation $<\alpha +n$ for all $n\in \mathbb {Z}$ . Indeed if this group is denoted A then for $B=A\cap (\bigcup _{n\in \mathbb {Z}} D_{\alpha +n})$ we have that $A/B$ is finitely generated and torsion free so it is finite free and $0\to B\to A\to A/B\to 0$ is split exact. This then implies that $0\to B/D_{\alpha }\to A/D_{\alpha }\to A/B\to 0$ is split exact so $B/D_{\alpha }$ is finitely generated. This implies that $B\subset D_{\alpha -M}$ for an $M\in \mathbb {Z}$ and because $D_{\alpha -M}/D_{\alpha }\cong \mathbb {Z}/p^M\mathbb {Z}$ is cyclic every intermediate group is of the form $D_{\alpha +m}$ , as required.

By compactness a type-definable subgroup of $D_{\alpha }+X$ is then a subgroup of $D_{\alpha +m}$ and as $D_{\alpha +m}/D_{\alpha +n}\cong \mathbb {Z}/p^{n-m}\mathbb {Z}$ we see that every intermediate subgroup is of the form $D_{\alpha +k}$ , so $L+D_{\alpha +n}=D_{\alpha +k}$ as required. ⊣

Proof By compactness there exists a definable sets $L\subset U_1\subset U_0$ such that $U_1-U_1\subset U_0$ , and a definable extension of $\phi $ to $\phi :U_0\to Q_p$ , such that $\phi $ is injective and such that if $a,b\in U_1$ then $\phi (a-b)=\phi (a)-\phi (b)$ . Now by the previous proposition $\phi (L)$ is trivial or an intersection of balls around the origin, so by compactness there is a definable group $\phi (L)\subset A\subset \phi (U_1)$ , with $A=0, D_{\alpha }$ or $Q_p$ . Then $H=\phi ^{-1}(A)\cap U_1$ works. ⊣

Proof Let C be the convex hull of $v(L)$ . Let $L\subset S$ be a definable set. Let $L\subset T$ be a symmetric definable set such that $T^2\subset S$ . Let $T=S_0\cup \cdots \cup S_m$ be a cell decomposition of T. Then there exists one of them, say $S_0$ , such that $v(S_0)\cap C$ is unbounded below in C. If $S_0=\{b\in Q_p\mid \alpha \leq v(b-a)<\beta , b-a\in cQ_p^n\}$ we distinguish two cases, so assume first that $v(a)<\tau $ for all $\tau \in C$ .

If $b\in S_0\cap v^{-1}(C)$ then $v(b)\in C$ , so $v(a)<v(b)$ and $v(b-a)=v(a)$ . Now if $b'\in v^{-1}(C)$ , then $v((b'-a)-(b-a))=v(b'-b)\geq \text {min}(v(b'), v(b))> v(a)+2v(n)=v(b-a)+2v(n)$ , so by Lemma 4.3 we get $b'-a\in cQ_p^{n}$ , and we see then that $b'\in S_0$ and $v^{-1}(C)\subset S_0$ , and this case is done.

Assume now that $\tau \leq v(a)$ for one $\tau \in C$ . Then by compactness we may obtain $\tau \in C$ such that $\tau <v(a)+m$ for all $m\in \mathbb {Z}$ . Now we choose $b\in S_0\cap v^{-1}(C)$ such that $v(b)<\tau $ . Now if $d\in U_{2v(n)+1}$ , then $v((bd-a)-(b-a))=v(b(d-1))=v(b)+v(d-1) > v(b)+2v(n)=v(b-a)+2v(n)$ so $bd\in S_0$ . Then $d\in T^2\subset S$ , as required. ⊣

Proof Embed $Q_p\to Q_p^{*}$ , where $Q_p^{*}$ is a monster model of $\mathbb {Q}_p$ as a valued field together with the exponentiation map $\mathbb {Z}\to \mathbb {Q}_p$ , $x\mapsto p^x$ ; the map $Q_p\to Q_p^{*}$ is taken to be elementary from $Q_p$ into the valued field reduct of $Q_p^*$ . Replacing $Q_p$ by $Q_p^*$ we may assume that there exists a map $Z\to Q_p^{\times }$ elementary equivalent to $x\mapsto p^x$ in $\mathbb {Q}_p$ . We define then ${\textrm {ac}} x=xp^{-v(x)}$ the projection $Q_p^{\times }\to Z_p^{\times }$ associated with the short exact sequence $1\to Z_p^{\times }\to Q_p^{\times }\to Z\to 0$ and the section $Z\to Q_p^{\times }, x\mapsto p^x$ . In the rest of the proof we shall have definable mean definable in the valued field language, and say ${\textrm {ac}}$ -definable to mean definable in the language including ${\textrm {ac}}$ .

Assume first that $v(L)$ is nontrivial and take C and $G_n$ as in the statement. Let $L\subset S$ be a definable set. Let $L\subset T$ , with T a symmetric definable set such that $T^{3}\subset S$ . By Lemma 5.1 we get that there exists n such that $U_n \subset T$ . Define $A=(L\cap Z_p^{\times })U_n$ , this is a definable group because it contains the group $U_n$ which is of finite index in $Z_p^{\times }$ . Now take the $v(L)\to Z_p^{\times }/A$ the composition of the connecting homomorphism $v(L)\to Z_p^{\times }/L\cap Z_p^{\times }$ with the canonical projection. This is an ac-type-definable morphism. By compactness there exists $r\in \mathbb {Z}$ such that $v(L)\subset C\cap rZ$ and an ac-type-definable group morphism extension $C\cap rZ\to Z_p^{\times }/A$ , see Proposition 3.5. As the codomain is finite we see $m(C\cap rZ)=mrC=C\cap mrZ$ is a subgroup of the kernel of $C\cap rZ\to Z_p^{\times }/A$ , (here $m=\text {Card}(Z_p/A)$ ). So if we denote $L_1=v^{-1}(mrZ)\cap LU_n$ we get that the connecting homomorphism in the short exact sequence $1\to A\to L_1\to v(L)\cap mrZ\to 0$ is trivial, so the sequence is split with splitting given by the restriction of the section $Z\to Q_p^{\times }$ to $v(L)\cap mrZ$ . This is to say $x\in L_1$ if an only if $v(x)\in v(L)\cap mrZ$ and $xp^{-v(x)}\in A$ . By compactness there is $s\in \mathbb {Z}$ such that if $v(x)\in C\cap sZ$ and $xp^{-v(x)}\in A$ then $x\in T^2$ . But the set of such x forms the kernel of a group morphism $v^{-1}(C)\to Z/sZ\times Z_p/A$ with finite codomain, so it contains $v^{-1}(C)^k=(Q_p^{\times })^k\cap v^{-1}(C)$ , so S contains $G_k\cap v^{-1}(C)$ as required.

Assume now that $v(L)$ is trivial. Then $L\subset Z_p^{\times }$ . If $L'=L\cap U_1$ then we get a short exact sequence $1\to L'\to L\to \pi (L)\to 1$ . Where $\pi :Z_p^{\times }\to \mathbb {F}_p^{\times }\cong \mathbb {Z}/(p-1)\mathbb {Z}$ . Recall that above a primitive root in $\mathbb {F}_p^{\times }$ lies a primitive $p-1$ th root of $Q_p$ (by Hensel’s lemma), so the sequence $1\to U_1\to Z_p^{\times }\to \mathbb {F}_p^{\times }\to 1$ is split. Then we get the connecting homomorphism $\pi (L)\to U_1/L'$ and from the description of $L'$ in Lemma 5.2 we see that $U_1/L'$ only has $p^n$ torsion, so as it has no $p-1$ -torsion we conclude that $\pi (L)\to U_1/L'$ is trivial, so L is as required. ⊣

Proof Consider the type-definable group $\phi (L)\subset Q_p^{\times }$ , it is of the form described in the statement of Proposition 5.4.

By compactness $\psi =\phi ^{-1}:\phi (L)\to L$ extends to a type-definable injective group homomorphism $\psi :L'\to G$ and after a restriction to a finite index subgroup $L'$ is either $ U_{\alpha }$ or $1$ or $(Q_p^{\times })^n$ or $o(a)^n$ (here we use Lemma 3.4), so without loss of generality $\phi (L)=L'$ . In the first three cases $L'$ is definable of the kind required, the image is a definable bounded index group, so it is finite index by compactness so we are done. Assume now the remaining case.

If we denote $r_m\in \mathbb {Z}$ the element such that $0\leq r_m<m$ and $v(a)-r_m\in mZ$ , and $I_m=v^{-1}([-\frac {1}{m}(a-r_m),\frac {1}{m}(a-r_m)])$ then $\cap _mI_m\cap (Q_p^{\times })^n=o(a)^n$ , so $\psi $ extends to an injective map $\psi _1:I_m\cap O(a)^n\to G$ . We also may find an $m'>m$ such that $\psi _1(ab^{-1})=\psi _1(a)\psi _1(b)^{-1}$ for all $a,b\in I_{m'}\cap O(a)^n$ . Remembering that $O(a)/o(a)\cong (\mathbb {R},+)$ , from which we obtain $O(a)^n/o(a)^n\cong \mathbb {R}$ , we see that the map $\psi _1$ restricted to $I_{m'}\cap O(a)^n$ extends to a definable group homomorphism $\psi _2:O(a)^n\to G$ . The set $A=\psi _2(I_{m'}\cap O(a)^n)$ is definable and a bounded number of translates cover G, so a finite number of translates cover G, so the group generated by it can be generated in a finite number of steps and is definable, indeed the number of translates of A in $A^r$ stabilizes, and for such an r, $A^r$ is the group generated by A. We see then that $H=\psi _2(O(a)^n)$ is a definable subgroup of G of finite index. The kernel of $\psi _2$ is a definable subgroup K of $O(a)^n$ such that $K\cap o(a)^n=1$ . The image of K to $\mathbb {R}\cong O(a)^n/o(a)^n$ is a subgroup $f(K)$ such that $f(K)\cap [-1,1]$ is compact and $f(K)\cap [-\epsilon ,\epsilon ]=0$ . The subgroups of $\mathbb {R}$ are either dense or of the form $f(K)=f(b)\mathbb {Z}$ , or trivial. K cannot be trivial as in this case $\psi _2$ is injective to a definable set, which cannot happen by compactness. So $K=\langle b\rangle $ and we are done. ⊣

Proof Observe that $F_2(\mathbb {Q}_p)\subset {\textrm {GL}}_2(\mathbb {Z}_p)$ has the topology given by a p-valuation in the group $G(\mathbb {Q}_p)$ . See [Reference Schneider11, Example 23.2]. As this is a one-dimensional compact Lie group then this p-valuation has rank 1, see Theorem 27.1 and (the proof of) Proposition 26.15 of [Reference Schneider11]. Then by [Reference Schneider11, Proposition 26.6] this group is isomorphic to $(p^2\mathbb {Z}_p,+)$ . By [Reference Schneider11, Theorem 29.8] this isomorphism is locally analytic. So this isomorphism extends to an isomorphism $p^2Z_p\to F_2$ in $Q_p^{an}$ . From the definition of the morphism it follows that it transforms the filtration $D_{\alpha }=\{x\in Z_p\mid v(x)\geq \alpha \}$ into $F_{\alpha }$ . So Proposition 4.7 finishes the proof. ⊣

Proof $\phi (L)$ is a type-definable subgroup of E, so after restricting to a bounded index type-definable subgroup $\phi (L)\subset o_E(a)$ . So as in the proof of Proposition 5.5 we see that the restriction of the inverse of $\phi $ to a finite index type-definable subgroup extends to a definable group morphism $O_E(b)^n\to H$ with kernel $\langle b^n\rangle $ or $E_{1,\alpha }\to H$ with trivial kernel, onto a finite index subgroup H definable subgroup of G. ⊣

Proof By Proposition 1.2, we may assume G is commutative, and Theorem 1.1 applies. So we get a type-definable bounded index group $T\subset G$ and an algebraic group H and a type-definable group morphism $T\to H$ with finite kernel. Replacing H by the Zariski closure of the image of $T\to H$ we may assume H is a one-dimensional algebraic group. Replacing H by the connected component of the identity (which is a finite index algebraic subgroup) we assume that H is a connected algebraic group. Then H is isomorphic as an algebraic group to the additive group or the multiplicative group or the one dimensional twisted torus or an elliptic curve. These cases are dealt with in Propositions 4.8 5.5, and 6.2, Section 7, and Proposition 7.1 for the Tate curve case. ⊣

Proof Assume 1). Suppose that $0\to K\to G\to G/K\to 0$ splits. Then tensoring by $\mathbb {Z}/n\mathbb {Z}$ remains exact, which is exactly 2).

Now assume 1) and 2). Choose a set theoretic section $\phi :G/K\to G$ . Then the set of all such sections is in bijective correspondence with the maps $K^{G/K}$ , and the set of group sections is closed in the product topology. As $K^{G/K}$ is a compact topological space it is then enough to show that for all finitely generated subgroups $A\subset G/K$ the map $\pi ^{-1}A\to A$ splits, that is, without loss of generality $G/K$ is finitely generated. Take $\mathbb {Z}^n\to G/K$ is a surjective group homomorphism and $\alpha :\mathbb {Z}^n\to G$ is a lift. Then after a base change one may assume that the kernel of $\mathbb {Z}^n\to G/K$ is $T=n_1\mathbb {Z}\times \cdots \times n_r\mathbb {Z}$ . From the assumption we conclude that there exists $\beta :\mathbb {Z}^n\to K$ such that $\alpha -\beta $ has kernel T. That is, $\alpha -\beta $ factors as a section $G/K\to G$ as required. ⊣

Proof As A is the torsion part, $L/A$ is torsion free. So by Proposition 8.2 we obtain that $A\to L$ is split injective. If $L\to A$ is a retraction then it factors through $nL$ for $n={\textrm {Card}}A$ , so it is definable. ⊣