Inverse Systems Profinite Groups and Completion

The identity tuple $(e_i{)}_{i\in I}$ is compatible, because every transition map is a homomorphism and therefore sends $e_j$ to $e_i$. So the inverse limit is nonempty.

If $(g_i)$ and $(h_i)$ are compatible, then for every $i\le j$ one has ${\phi }_{ij}(g_j{h}_j)={\phi }_{ij}(g_j){\phi }_{ij}(h_j)=g_i{h}_i\mathrm{.}$ Likewise ${\phi }_{ij}(g_j^{-1})=g_i^{-1}$. Hence coordinatewise products and coordinatewise inverses remain compatible.

By [F1], step 1.1 and step 1.2 prove that the compatible tuples form a subgroup of the product group.

Define by $u(h):=(f_i(h){)}_{i\in I}\mathrm{.}$ The cone identities make this tuple compatible, so [L1] shows that $u$ lands in the inverse limit.

For $h,h^{\mathrm{\prime }}\in H$ one has $u(h{h}^{\mathrm{\prime }})=(f_i(h{h}^{\mathrm{\prime }}){)}_i=(f_i(h)f_i(h^{\mathrm{\prime }}){)}_i=u(h)u(h^{\mathrm{\prime }}),$ and similarly $u(e_H)=e$. Hence $u$ is a group homomorphism by [F1]. Also ${\pi }_i\circ u=f_i$ for every $i$.

If is another homomorphism with ${\pi }_i\circ v=f_i$ for every $i$, then every coordinate of $v(h)$ agrees with the corresponding coordinate of $u(h)$. Tuples are equal exactly when all coordinates are equal, so $v=u$.

Steps 1.1 through 3.1 give existence and uniqueness of the mediating homomorphism.

For each $i\le j$, define ${\mathrm{\Psi }}_{ij}:{\prod }_k{G}_k\to G_i\times G_i,{\mathrm{\Psi }}_{ij}((g_k))=(g_i,{\phi }_{ij}(g_j))\mathrm{.}$ Because every factor is discrete, $G_i\times G_i$ is discrete, and the diagonal ${\mathrm{\Delta }}_i=\{(x,x):x\in G_i\}$ is closed. A tuple is compatible at the pair $(i,j)$ exactly when it lies in ${\mathrm{\Psi }}_{ij}^{-1}({\mathrm{\Delta }}_i)$. Therefore $L={\bigcap }_{i\le j}{\mathrm{\Psi }}_{ij}^{-1}({\mathrm{\Delta }}_i)$ is closed in the ambient product.

By [L1], $L$ is a subgroup of the product group. The product group operations are continuous coordinatewise, and restricting continuous maps to a subspace preserves continuity. Hence $L$ is a topological subgroup in the sense of [F1].

So $L$ is a closed topological subgroup of the product of the finite discrete groups.

Every finite discrete group $G_i$ is Hausdorff because distinct points have disjoint singleton neighbourhoods. It is compact because an open cover has one member containing each point and [F3] makes those finitely many choices; the selected family is a finite subcover.

By [L2] and step 1.1, the full product is Hausdorff. Intersecting two disjoint ambient neighbourhoods with $L$ shows directly that the subspace $L$ is Hausdorff. Under the Axiom of Choice, [F1] and step 1.1 make the product compact; since $L$ is closed by [L1], [L3] makes $L$ compact.

By [L4], fixing finitely many coordinates gives a neighbourhood basis in $L$. Because every factor is discrete, each such cylinder trace is clopen. In particular, for $x\in L$ and an index $i$, the set $C_i(x):=\{z\in L:z_i=x_i\}$ is clopen.

If a connected subset $C\subseteq L$ contained distinct points $x\mathrm{\ne }y$, they would differ in some coordinate $i$. Then $C\cap C_i(x)$ and $C\setminus C_i(x)$ would be disjoint nonempty subsets that are both open and closed in $C$, a contradiction. Hence every nonempty connected subset of $L$ is a singleton, so $L$ is totally disconnected.

Each kernel $\ker {\pi }_i$ is a normal subgroup because ${\pi }_i$ is a homomorphism, and it is open because $\{e_i\}\subseteq G_i$ is open in the discrete factor and ${\pi }_i^{-1}(\{e_i\})=\ker {\pi }_i$.

Let $U$ be any neighbourhood of the identity in $L$. By [F1], some basic cylinder fixing finitely many coordinates, say those in a finite set $F\subseteq I$, is contained in $U$. Choose $k\in I$ dominating every index in $F$ by directedness. If $x\in \ker {\pi }_k$, then compatibility gives ${\pi }_i(x)={\phi }_{ik}(e_k)=e_i$ for every $i\in F$, so $x$ lies in that cylinder. Hence $\ker {\pi }_k\subseteq U$.

Therefore the family $(\ker {\pi }_i)$ is a neighbourhood basis of open normal subgroups at the identity.

If $f$ is continuous, then every composite ${\pi }_i\circ f$ is continuous because it is the composite of two continuous maps, namely $f$ and the coordinate map [L1].

Conversely, suppose every ${\pi }_i\circ f$ is continuous. Regard $f$ as a map into the ambient product. By the product-topology clause of [F1], those coordinatewise continuities imply that the product-valued map is continuous. Since the image already lies in the subspace $L$, the same map is continuous as a map into $L$.

Steps 1.1 and 1.2 prove the equivalence.

Restricting a compatible tuple on $I$ to its $J$-coordinates gives a homomorphism Compatibility is preserved because every relation used in the $J$-limit is already one of the relations used in the $I$-limit.

For a compatible tuple $(x_j{)}_{j\in J}$ and $i\in I$, let $A_i:=\{{\phi }_{ij}(x_j):j\in J,i\le j\}\mathrm{.}$ Cofinality makes $A_i$ nonempty. If $j,j^{\mathrm{\prime }}$ both dominate $i$, choose $k\in J$ dominating both; compatibility gives ${\phi }_{ij}(x_j)={\phi }_{ik}(x_k)={\phi }_{i{j}^{\mathrm{\prime }}}(x_{j^{\mathrm{\prime }}})\mathrm{.}$ Thus $A_i$ is a singleton. Define $\sigma ((x_j){)}_i$ to be its unique member. This makes no simultaneous choice, and the transition identities show that the resulting tuple is compatible on $I$.

By construction, $\rho \circ \sigma$ and $\sigma \circ \rho$ fix every coordinate, hence are identity maps on the corresponding inverse limits. Therefore $\rho$ and $\sigma$ are inverse isomorphisms.

Assume first that $X$ is totally separated. Let $C\subseteq X$ be connected and choose distinct points $x,y\in C$ if possible. By [L2] there is a clopen set $U$ with $x\in U$ and $y\mathrm{\notin }U$. Then $C\cap U$ and $C\setminus U$ are disjoint nonempty clopen subsets of $C$, contradicting connectedness. Hence every connected component is a singleton, so $X$ is totally disconnected.

Assume now that $X$ is totally disconnected. By [L1], quasicomponents also are singletons. For distinct points $x\mathrm{\ne }y$, the definition of quasicomponent therefore supplies a clopen set containing $x$ but not $y$. This is exactly total separatedness by [L2].

Steps 1.1 and 1.2 prove the equivalence for compact Hausdorff spaces.

If $G$ is profinite, then by [F1] it is topologically isomorphic to an inverse limit of finite discrete groups. Therefore [L1] gives that $G$ is compact, Hausdorff, and totally disconnected. The same presentation gives the open normal basis in [L2]; each quotient by a coordinate kernel is isomorphic to the image of a projection and is therefore finite.

Conversely, assume that $G$ is compact, Hausdorff, and totally disconnected. By [L3], $G$ is totally separated. Let $U$ be an identity neighbourhood and choose an open identity neighbourhood $O\subseteq U$. For each $x\in G\setminus O$, total separatedness gives a clopen set $A_x$ containing $1$ but not $x$. The open complements $G\setminus A_x$ cover the compact set $G\setminus O$, so finitely many suffice. Their complementary intersection is a clopen identity neighbourhood $C\subseteq O\subseteq U$.

Put $K:=(G\setminus C)\cap C^2$. The set $C$ is compact by [L7], so $C\times C$ is compact by [L6], its product image $C^2$ is compact by [L5], and $K$ is compact by [L7]. Since $G$ is Hausdorff, [L5] makes $K$ closed. For each $h\in C$, continuity of multiplication at $(h,1)$ and openness of $C$ give open neighbourhoods $W_h\subseteq C$ of $h$ and $X_h\subseteq C$ of $1$ such that $W_h{X}_h\subseteq (G\setminus K)\cap C^2\subseteq C\mathrm{.}$ Compactness of $C$ gives finitely many $W_{h_j}$ covering it.

Let $X:={\bigcap }_j{X}_{h_j}$ and $Y:=X\cap X^{-1}$. Then $Y$ is a symmetric open identity neighbourhood and $CY\subseteq C$. Hence $H_0:={\bigcup }_{n\ge 1}{Y}^n$ is an open subgroup contained in $C$. Its open cosets cover compact $G$, so $H_0$ has finite index. The normalizer of $H_0$ contains $H_0$, so $H_0$ has only finitely many distinct conjugates. Their intersection $N$ is therefore an open normal finite-index subgroup with $N\subseteq H_0\subseteq C\subseteq U$. Since $U$ was arbitrary, these $N$ form an identity-neighbourhood basis.

Now let $G$ be any compact Hausdorff topological group having an identity-neighbourhood basis of open normal finite-index subgroups; this includes the situation produced in step 3.1. Let $\mathcal{N}$ be the family of all open normal finite-index subgroups. It still contains such a basis, and it is directed under reverse inclusion because it is closed under finite intersections. The quotient maps define a homomorphism Every coordinate map is continuous because its fibres are open cosets of $N$, so [L4] makes $\eta$ continuous.

The basis separates points: if $g\mathrm{\ne }1$, Hausdorffness gives an identity neighbourhood omitting $g$, and some $N\in \mathcal{N}$ lies inside it. Hence $\eta$ is injective. To prove density, let a nonempty basic cylinder in the inverse limit restrict the finitely many coordinates $N_1,\dots ,N_k$. If $k=0$, it contains $\eta (1)$. Otherwise put $M:=N_1\cap \cdots \cap N_k\in \mathcal{N}$, choose a compatible tuple in the cylinder, and choose a representative $g\in G$ of its $M$-coordinate. Compatibility makes the image of $g$ equal to the prescribed value in every restricted coordinate, so $\eta (g)$ lies in the cylinder. Thus $\eta [G]$ is dense.

By compactness of $G$ and [L5], the continuous image $\eta [G]$ is compact. The target is Hausdorff by [L1], so [L5] makes the image closed. It is both closed and dense by step 5.1, hence is the whole target. Thus $\eta$ is a continuous bijection from compact to Hausdorff and is a homeomorphism by [L5]. It is a group isomorphism by construction, so [F1] makes $G$ profinite.

Step 1.1 proves both forward implications. For a compact Hausdorff totally disconnected group, steps 1.2-3.1 construct the required basis and steps 4.1-6.1 make $G$ profinite. For a compact Hausdorff group already having the stated basis, steps 4.1-6.1 apply directly. This establishes both equivalences exactly as stated.

An element $g\in G$ lies in $\ker {\iota }_G$ exactly when every coordinate $gN$ is the identity coset in $G\mathrm{/}N$. That is equivalent to $g\in N$ for every finite-index normal subgroup $N$. By [L1], this means precisely $g\in R_f(G)$.

By [L2], let $U$ be a nonempty basic open set of $\widehat{G}$. Then $U$ fixes finitely many coordinates, say at $N_1,\dots ,N_m$, to compatible cosets $g_k{N}_k$. Let $M:=N_1\cap \cdots \cap N_m$, which is again finite-index normal. Compatibility means exactly that these finitely many coordinates come from one coset $gM$ in $G\mathrm{/}M$. Since $M\subseteq N_k$, the image of $gM$ in $G\mathrm{/}{N}_k$ is the prescribed coset $g_k{N}_k$ for each $k$. Therefore ${\iota }_G(g)\in U$.

Step 1.2 says that every nonempty basic open set meets ${\iota }_G[G]$, so the image is dense. Together with step 1.1, this proves the theorem.

If ${\iota }_G$ is injective, then its kernel is trivial. By [L1], the finite residual is therefore trivial, and [F1] says that $G$ is residually finite.

If $G$ is residually finite, then [F1] gives $R_f(G)=\{1\}$. By [L1], this is exactly the statement that $\ker {\iota }_G=\{1\}$, so ${\iota }_G$ is injective.

The two implications prove the equivalence.

For each coordinate $q_i:P\to P_i$ of [L1], put $N_i:=\ker (q_i\circ f)$. This is normal, and $G\mathrm{/}{N}_i$ is isomorphic to the image of $q_i\circ f$ in the finite group $P_i$, so $N_i$ has finite index. Let ${\bar{f}}_i:G\mathrm{/}{N}_i\to P_i$ be the induced homomorphism.

Let ${\pi }_{N_i}:\widehat{G}\to G\mathrm{/}{N}_i$ be the completion coordinate and define ${\widehat{f}}_i:={\bar{f}}_i\circ {\pi }_{N_i}$. It is continuous because both finite quotients are discrete and ${\pi }_{N_i}$ is a coordinate projection for the topology in [F2]. Moreover, [F2] gives ${\widehat{f}}_i\circ {\iota }_G=q_i\circ f\mathrm{.}$

If $i\le j$ and ${\phi }_{ij}:P_j\to P_i$ is the transition map, then $q_i={\phi }_{ij}\circ q_j$, so $N_j\subseteq N_i$. Let ${\psi }_{ij}:G\mathrm{/}{N}_j\to G\mathrm{/}{N}_i$ be the natural quotient map. The two induced maps satisfy ${\phi }_{ij}\circ {\bar{f}}_j={\bar{f}}_i\circ {\psi }_{ij},$ and the completion coordinates satisfy ${\psi }_{ij}\circ {\pi }_{N_j}={\pi }_{N_i}$. Hence ${\phi }_{ij}\circ {\widehat{f}}_j={\widehat{f}}_i$, so $({\widehat{f}}_i{)}_i$ is a compatible cone.

By [F1], the compatible cone from step 3.1 induces a homomorphism with coordinate maps ${\widehat{f}}_i$. The coordinate identities in step 2.1 give $h\circ {\iota }_G=\theta \circ f$. Define $\widehat{f}:={\theta }^{-1}\circ h$; then $\widehat{f}\circ {\iota }_G=f$.

Every coordinate composite of $h$ is the continuous map ${\widehat{f}}_i$, so [L3] makes $h$ continuous. The inverse ${\theta }^{-1}$ is continuous because $\theta$ is a topological group isomorphism, hence $\widehat{f}$ is continuous.

For uniqueness, let $u,v:\widehat{G}\to P$ be continuous homomorphisms with $u\circ {\iota }_G=v\circ {\iota }_G=f$. By [L1] and [L4], $P$ is Hausdorff. The equalizer of $u$ and $v$ is therefore closed, while [L2] says it contains the dense subset ${\iota }_G[G]$. Thus the equalizer is all of $\widehat{G}$, and $u=v$.

Steps 4.1, 5.1, and 5.2 give the unique continuous homomorphism $\widehat{f}:\widehat{G}\to P$ extending $f$. Therefore $\widehat{G}$ is initial among continuous homomorphisms from $G$ to profinite groups.

By [F1], $\widehat{H}$ is a profinite topological group.

Fix a finite-index normal subgroup $N⊴H$. The kernel of the coordinate map $G\stackrel{f}{\to }H\to H\mathrm{/}N$ is $f^{-1}(N)$, and the induced map $G\mathrm{/}{f}^{-1}(N)\to H\mathrm{/}N$ is injective. Thus $f^{-1}(N)$ is finite-index normal in $G$. Every fibre of the coordinate map is a coset of this open subgroup, so the coordinate map to the finite discrete group $H\mathrm{/}N$ is continuous. Since this holds for every coordinate, [L2] makes ${\iota }_H\circ f:G\to \widehat{H}$ continuous for the profinite topology on $G$.

Apply [L1] to the continuous homomorphism from step 2.1. It gives a unique continuous homomorphism $\widehat{f}:\widehat{G}\to \widehat{H}$ satisfying $\widehat{f}\circ {\iota }_G={\iota }_H\circ f$.

If $f={\mathrm{id}}_G$, then both $\widehat{f}$ and ${\mathrm{id}}_{\widehat{G}}$ compose with ${\iota }_G$ to give ${\iota }_G$. Uniqueness in [L1] therefore forces $\widehat{{\mathrm{id}}_G}={\mathrm{id}}_{\widehat{G}}$.

For homomorphisms $G\stackrel{f}{\to }H\stackrel{g}{\to }K$, both $\widehat{g}\circ \widehat{f}$ and $\widehat{g\circ f}$ compose with ${\iota }_G$ to give ${\iota }_K\circ g\circ f$. Uniqueness in [L1] again yields $\widehat{g\circ f}=\widehat{g}\circ \widehat{f}\mathrm{.}$ So profinite completion is functorial.

By [L1], replace $g$ by a conjugate and assume it is represented by a nontrivial cyclically reduced word $w=x_1\cdots x_n$ in finitely many letters. Build a finite pointed labelled graph with distinct vertices $v_0,\dots ,v_n$ and a distinguished path from $v_{k-1}$ to $v_k$ labelled $x_k$. Because $w$ is reduced, these prescribed edges define partial injective transitions. Complete each partial transition to a permutation of the finite vertex set, so every vertex has exactly one incoming and one outgoing edge of each used label.

Reading words from $v_0$ gives an action of the free group on the finite vertex set. The word $w$ sends $v_0$ to the distinct vertex $v_n$. Therefore the induced homomorphism $F(X)\to \mathrm{Sym}(V)$ does not kill $g$.

The kernel of that finite permutation action has finite index and omits $g$. Since $g\mathrm{\ne }1$ was arbitrary, [F1] shows that the finite residual is trivial. Hence free groups are residually finite.

The present hypotheses are exactly the subgroup-separability conclusion of [F1]. Therefore there exists a finite-index subgroup of $F$ that contains $H$ but not $g$. Since $g\in F\setminus H$ was arbitrary, every finitely generated subgroup of $F$ is closed in the profinite topology.

This is precisely the definition of subgroup separability from A subgroup is separable when it is closed in the profinite topology, and a group is LERF when every finitely generated subgroup is separable. Hence every finitely generated free group is subgroup separable.

The identity tuple of the product is compatible, because every transition homomorphism preserves identity.

Therefore the inverse limit contains at least that identity tuple. So by [L1] it is never empty.

This refutes the statement.

The statement is a global set-theoretic assertion about all surjective inverse systems, not a theorem of ZF proved by [L1]. In models of ZF without sufficient choice, there are inverse systems of finite groups with surjective bonding maps for which a prescribed coordinate value has no compatible lift.

In such a model, the corresponding coordinate projection from the inverse limit fails to be surjective. Therefore the universal claim in the Statement is false in ZF.

So surjectivity of all coordinate projections requires extra choice beyond bare ZF. This refutes the statement.

Because $G$ is not residually finite, the finite residual $R_f(G)$ contains some nonidentity element.

By [L1], the finite residual is exactly the kernel of the canonical map. Hence that nonidentity element lies in the kernel, so the canonical map is not injective.

Therefore the statement is false.

The circle group $S^1$ is compact and Hausdorff in its Euclidean subspace topology, and it is a topological group under multiplication.

The circle group is connected, so it is not totally disconnected. Therefore [L1] shows that it is not profinite.

This gives a compact Hausdorff group that is not profinite, so the statement is false.

Every homomorphism from the divisible group $\mathbb{Q}$ to a finite group is trivial: if $F$ is finite of order $m$ and $\varphi :\mathbb{Q}\to F$, then for any $q\in \mathbb{Q}$ one has $q=mr$ for some $r\in \mathbb{Q}$, so $\varphi (q)=m\varphi (r)=0$. Therefore every finite quotient of $H=\mathbb{Z}\oplus \mathbb{Q}$ factors through the projection onto $\mathbb{Z}$.

The finite quotients of both $G$ and $H$ are therefore exactly the finite cyclic groups $\mathbb{Z}\mathrm{/}n\mathbb{Z}$, with the same transition maps. By [L1], both profinite completions are the inverse limit of that same system, namely $\widehat{\mathbb{Z}}$. But $G$ and $H$ are not isomorphic because $H$ contains a nontrivial divisible subgroup and $G$ does not.

Thus nonisomorphic groups can have isomorphic profinite completions. The statement is false.