Arbitrary Stage Spectra for Raikov Remainders Along Quotient Towers

1. Introduction

All topological groups are Hausdorff. The empty space is regarded as pseudocompact. Each group is viewed as a dense subgroup of its Raikov completion via the canonical embedding.

For a Hausdorff topological group G, let $\rho G$ denote its Raikov completion and put

$$r_{\rho }\left(G\right)=\rho G\setminus G.$$

This completion remainder is called the Raikov remainder of G. Work on group remainders includes Arhangel’skii’s pseudocompact–Lindelöf dichotomy for arbitrary compactification remainders and subsequent results on remainder duality and completeness-type phenomena [1,2,6]. Arhangel’skii and Choban studied Raikov remainders in the broader setting of group extensions. In particular, they showed that every nonempty group-remainder is dense in its ambient group [5, Proposition 1.2].

Let N be a closed normal subgroup of G, let $q:G\to G/N$ be the quotient homomorphism, and let $\widehat{q}:\rho G\to \rho (G/N)$ be its continuous homomorphic extension [8, Corollary 3.6.17]. The behavior of the source remainder under $\widehat{q}$ is governed by two completion facts. The kernel of $\widehat{q}$ is the closure of N in $\rho G$, and $\widehat{q}$ is open onto its image [8, Theorem 3.6.19]. If N is precompact, then $\widehat{q}$ is onto [8, Proposition 3.7.19]. Surjectivity need not hold in general. Leischner gave an example of an incomplete quotient of a complete topological group and later studied quotients of Raikov-complete groups [18,19].

For a Čech-complete invariant subgroup H of X, Morales and Tkachenko prove that X is strongly Dieudonné complete if and only if $X/H$ is strongly Dieudonné complete when X is $\omega$-balanced, and that X is strongly realcompact if and only if $X/H$ is strongly realcompact when X is $\omega$-narrow [20, Theorem 4.5]. In the Abelian setting, Bello, Chasco, Domínguez, and Tkachenko show that completion is compatible with quotients under a Čech-completeness hypothesis and obtain an exact completed extension under the corresponding assumptions [11, Propositions 3.9 and 3.10]. Recent work of Arnautov and Ermakova gives further quotient-completeness results for discrete, compact, and locally compact subgroups [9,10]. Braunling and Ren prove an exactness theorem for completion on locally precompact Hausdorff Abelian groups [12, Proposition 6.5]. Together, these results concern completeness properties of quotients or exactness of completion rather than first-entry strata of remainders.

The remainder identities rest on these completion facts. With $K={\overline{N}}^{\rho G}$, the kernel-closure theorem gives the basic fiber description for $\widehat{q}{|}_{r_{\rho }\left(G\right)}$. The group K is a Raikov completion of N. If $gN\in G/N$, then the part of the fiber over $gN$ lying in the source remainder is $g(K\setminus N)$, a translated copy of $r_{\rho }\left(N\right)$. If the target point lies in $\widehat{q}\left(\rho G\right)\setminus (G/N)$, its whole $\widehat{q}$-fiber lies in the source remainder and is a copy of $\rho N$. Thus a Raikov-complete kernel contributes no source-remainder fiber over old quotient points, whereas a kernel that is not Raikov complete contributes one over every old quotient point.

For a precompact kernel, the completion extension is automatically onto. The fiber dichotomy then gives a more detailed decomposition. The source remainder splits into points lying over the old quotient group and points lying over the quotient remainder. The first part has fibers homeomorphic to $r_{\rho }\left(N\right)$, while the second part has fibers homeomorphic to $\rho N$. When neither N nor $G/N$ is Raikov complete, both parts are dense in $\rho G$ by the classical density theorem for nonempty group-remainders [5, Proposition 1.2]. The quotient map therefore gives a canonical decomposition of the source remainder into two dense pieces.

The quotient-induced decomposition is compatible with further quotients. For nested closed normal subgroups $N\subseteq M$, the old-point defects defined below satisfy an exact two-stage law obtained from functoriality of the completion extensions and the preimage formula in Theorem 1. The old-point defect for the direct quotient by M splits into the defect already visible after quotienting by N and the pullback of the old-point defect for the second quotient $M/N$. Iterating this identity along a normal quotient tower starting from the trivial subgroup gives a transfinite first-entry stratification. Successor strata are pullbacks of the old-point defects of the successive quotients. At continuous limit stages, the current kernel closure is obtained by taking the closure in $\rho G$ of the union of the earlier kernel closures. The corresponding limit stratum consists precisely of the points that first enter at that stage. The two-stage identity describes remainder fibers under successive quotients and is not an exactness theorem for Raikov completion.

Every subset of the nonzero stages occurs as the stage spectrum of a continuous tower in a pseudocompact Boolean group, with every member of the tower pseudocompact. The construction also determines exactly which successive factors are compact. Peng studied a different, cardinal-valued quotient-weight spectrum for precompact Abelian groups and showed that such spectra can be arbitrary subsets of a cardinal interval, with pseudocompact realizations in part of that range [21]. Peng and Zhou prove that a free pro-p group of uncountable rank has no nontrivial metrizable closed normal subgroup [22].

Dense subgroups of compact Boolean products are standard ingredients in constructions of pseudocompact groups. A closely related classical construction is due to Comfort and Soundararajan. For a prime p and an uncountable cardinal $\alpha$, they obtain a dense pseudocompact subgroup of ${(\mathbb{Z}/p\mathbb{Z})}^{\alpha }$ of index p by starting from the countable-support subgroup $\Sigma$ [14, Lemma 4.2]. See also the related pseudocompact Abelian example of Leiderman and Tkachenko [17, Theorem 3.4]. Gabriyelyan, van Mill, and Reznichenko likewise use $\Sigma$-product methods in pseudocompact-group constructions [16, Theorem 5.1]. They also state the standard Boolean-group fact that maps from independent subsets to $\mathbb{Z}/2\mathbb{Z}$ extend to homomorphisms [16, Fact 4.2]. Theorem 5 uses these standard ingredients to realize arbitrary Raikov first-entry stage spectra under pseudocompactness.

In the compact-kernel case the old-point part disappears. The restricted remainder map is then open and perfect, so pseudocompactness passes in both directions. The perfectness of quotient maps by compact subgroups is classical in the theory of coset spaces [3]. Arhangel’skii and Choban also use open perfect restrictions arising from compact-subgroup quotients in their study of group-remainders [5]. Their later work treats properties of group-remainders and their perfect images in a broader mapping setting [7].

If $r_{\rho }\left(G\right)$ is nonempty and pseudocompact, its density in $\rho G$ forces $\rho G$ to be pseudocompact. The compactness theorem for Raikov completions of pseudocompact groups then implies that $\rho G$ is compact [8, Corollary 3.7.18]. Hence G is precompact, and the precompact-kernel dichotomy applies to every closed normal subgroup of G.

Examples 1 and 2 show that $\widehat{q}\left(r_{\rho }\left(G\right)\right)=\rho (G/N)$ can occur with either a pseudocompact or a non-pseudocompact quotient remainder. They use the Bohr topology on $\mathbb{Z}$ and a result of Arhangel’skii and Bella stated in the abstract of their paper [4]. In that abstract, a group is called pseudocompact at infinity when every compactification remainder is pseudocompact.

2. Quotient Fibers and Canonical Decomposition

2.1. Fiber Structure and the Quotient Image Dichotomy

For the quotient homomorphism $q:G\to G/N$, let

$$\widehat{q}:\rho G⟶\rho (G/N)$$

be its extension, and put

$$K={\overline{N}}^{\rho G}\mathrm{and}I=\widehat{q}\left(\rho G\right).$$

By the classical completion theorem, $\widehat{q}$ is open onto I and $ker\widehat{q}=K$ [8, Theorem 3.6.19]. Since K is a closed subgroup of the Raikov-complete group $\rho G$, it is Raikov complete [8, Exercise 3.6.m]. The subgroup N is dense in K, and the subgroup-uniformity and uniqueness results show that this inclusion realizes K as a Raikov completion of N [8, Proposition 1.8.4 and Theorem 3.6.14].

Theorem 1.

Let N be a closed normal subgroup of a topological group G. With q, K, I, and $\widehat{q}$ as above,

$${\widehat{q}}^{-1}(G/N)=GK.$$

The restriction of $\widehat{q}$ to $r_{\rho }\left(G\right)$ has the following fiber structure.

1. 

If $y=q\left(g\right)\in G/N$, then

$${\left(\widehat{q}{|}_{r_{\rho }\left(G\right)}\right)}^{-1}\left(y\right)=g(K\setminus N),$$

which is homeomorphic to $r_{\rho }\left(N\right)$.

2. 

If $y\in I\setminus (G/N)$ and $\widehat{q}\left(x\right)=y$, then

$${\left(\widehat{q}{|}_{r_{\rho }\left(G\right)}\right)}^{-1}\left(y\right)=xK,$$

which is homeomorphic to $\rho N$.

Consequently, the image is given by the following two cases.

1. 

If N is Raikov complete, then

$$\widehat{q}\left(r_{\rho }\left(G\right)\right)=I\setminus (G/N).$$

2. 

If N is not Raikov complete, then

$$\widehat{q}\left(r_{\rho }\left(G\right)\right)=I.$$

In particular,

$$r_{\rho }(G/N)=\widehat{q}\left(r_{\rho }\left(G\right)\right)$$

holds if and only if N is Raikov complete and $\widehat{q}$ is onto.

Proof. 

By the classical facts above, $ker\widehat{q}=K$ and K is a Raikov completion of N. If $x\in {\widehat{q}}^{-1}(G/N)$, choose $g\in G$ such that $\widehat{q}\left(x\right)=q\left(g\right)$. Then $g^{-1}x\in K$, so $x\in GK$. The reverse inclusion follows from $\widehat{q}\left(K\right)=\left\{e\right\}$. Hence

$${\widehat{q}}^{-1}(G/N)=GK.$$

Because N is closed in G,

$$K\cap G=N.$$

Fix $y=q\left(g\right)\in G/N$. The full $\widehat{q}$-fiber over y is $gK$. Its intersection with G is $gN$. Therefore

$${\left(\widehat{q}{|}_{r_{\rho }\left(G\right)}\right)}^{-1}\left(y\right)=gK\setminus gN=g(K\setminus N).$$

Left translation by g is a homeomorphism, and under the completion isomorphism $K\cong \rho N$ fixing N, the set $K\setminus N$ corresponds to $r_{\rho }\left(N\right)$.

Now let $y\in I\setminus (G/N)$ and choose $x\in \rho G$ with $\widehat{q}\left(x\right)=y$. The full fiber is $xK$. If $xk\in G$ for some $k\in K$, then

$$y=\widehat{q}\left(xk\right)\in G/N,$$

a contradiction. Hence $xK\subseteq r_{\rho }\left(G\right)$, and the restricted fiber equals $xK$, which is homeomorphic to K and therefore to $\rho N$.

If N is Raikov complete, then $K=N$ and the fibers over points of $G/N$ are empty. The image of the source remainder is therefore $I\setminus (G/N)$. If N is not Raikov complete, then $K\setminus N$ is nonempty. Every point of $G/N$ is then hit by the source remainder, and every point of $I\setminus (G/N)$ is hit by the second fiber formula. Hence the image is all of I.

The final equivalence follows from the two cases. When N is Raikov complete, the image equals the target remainder exactly when $I=\rho (G/N)$. When N is not Raikov complete, the image contains $G/N$, so it cannot equal the target remainder. □

If N is precompact, then $\widehat{q}$ is onto [8, Proposition 3.7.19], and a precompact group is Raikov complete exactly when it is compact [8, Theorem 3.7.15].

Corollary 1.

Let N be a closed normal precompact subgroup of a topological group G, and let $\widehat{q}:\rho G\to \rho (G/N)$ be the completion extension of the quotient homomorphism. Then

$$\widehat{q}\left(r_{\rho }\left(G\right)\right)=\left\{\begin{array}{cc}{r}_{\rho }(G/N),\hfill & \mathit{if}N\mathit{is}\mathit{compact},\hfill \\ \rho (G/N),\hfill & \mathit{if}N\mathit{is}\mathit{not}\mathit{compact}.\hfill \end{array}\right.$$

In particular,

$$r_{\rho }(G/N)=\widehat{q}\left(r_{\rho }\left(G\right)\right)$$

if and only if N is compact.

Proof. 

Apply Theorem 1 using the two classical facts above. □

2.2. Canonical Decomposition for Precompact Kernels

For a precompact kernel, the fiber theorem determines both the image of the source remainder and a canonical intermediate group between G and its Raikov completion.

Let N be a closed normal precompact subgroup of G, put

$$Q=G/N,K={\overline{N}}^{\rho G},$$

and let $q:G\to Q$ be the quotient homomorphism and $\widehat{q}:\rho G\to \rho Q$ its completion extension. The map $\widehat{q}$ is onto [8, Proposition 3.7.19]. Since K is a Raikov completion of the precompact group N, it is compact [8, Theorem 3.7.15]. Define

$$H_N={\widehat{q}}^{-1}\left(Q\right)=GK.$$

Set

$$L_N=H_N\setminus G\mathrm{and}{T}_N=\rho G\setminus H_N.$$

Thus $L_N$ is the part of the source remainder lying over old quotient points, while $T_N$ is the part lying over the quotient remainder.

Theorem 2.

Let N be a closed normal precompact subgroup of a topological group G. With the notation above, the following assertions hold.

1. 

The group $H_N$ is a dense intermediate subgroup

$$G\subseteq H_N\subseteq \rho G,$$

and $\rho G$ is a Raikov completion of $H_N$. With this choice of completion, $\rho H_N=\rho G$. The compact group K is normal in $H_N$, and

$$H_N/K\cong Q.$$

2. 

The source remainder has the canonical decomposition

$$r_{\rho }\left(G\right)=L_N\dot{\cup }{T}_N,$$

where

$$L_N=H_N\setminus G=GK\setminus G$$

and

$$T_N=\rho G\setminus H_N=r_{\rho }\left(H_N\right).$$

3. 

One has $H_N=G$ if and only if N is compact. One has $H_N=\rho G$ if and only if Q is Raikov complete.

4. 

If N is not compact, then

$$\widehat{q}{|}_{L_N}:L_N⟶Q$$

is a continuous open surjection. Every fiber is homeomorphic to $r_{\rho }\left(N\right)$.

5. 

If Q is not Raikov complete, then

$$\widehat{q}{|}_{T_N}:T_N⟶r_{\rho }\left(Q\right)$$

is an open perfect surjection. Every fiber is homeomorphic to $\rho N$.

Proof. 

The equality $H_N=GK$ follows from Theorem 1. Since $G\subseteq H_N\subseteq \rho G$ and G is dense in $\rho G$, the group $H_N$ is dense in $\rho G$. The two-sided uniformity on a subgroup is the restriction of the ambient two-sided uniformity [8, Proposition 1.8.4]. Hence $\rho G$ is a Raikov completion of $H_N$ [8, Theorem 3.6.14]. With this choice, $\rho H_N=\rho G$.

The kernel of $\widehat{q}{|}_{H_N}$ is K, and the restriction is an open surjection from $H_N$ onto Q. Thus $H_N/K\cong Q$. The definitions now give

$$L_N=H_N\setminus G$$

and

$$T_N=\rho G\setminus H_N=r_{\rho }\left(H_N\right),$$

which proves the decomposition of $r_{\rho }\left(G\right)$.

By Theorem 1, the set $L_N$ is empty exactly when the precompact group N is Raikov complete. This is equivalent to compactness of N. The set $T_N$ is empty exactly when $r_{\rho }\left(Q\right)$ is empty, which is equivalent to Raikov completeness of Q. This proves the two endpoint characterizations.

Suppose that N is not compact. The old-point fiber formula in Theorem 1 gives

$${\left(\widehat{q}{|}_{L_N}\right)}^{-1}\left(q\left(g\right)\right)=g(K\setminus N)\cong r_{\rho }\left(N\right).$$

It remains to check openness. Let $U=O\cap L_N$, where O is open in $\rho G$. If $y\in \widehat{q}\left(O\right)\cap Q$, then the full fiber over y is a coset $gK$ meeting O. The group N is a proper dense subgroup of K. A proper dense subgroup has empty interior, so $K\setminus N$ is dense in K. Hence the nonempty open set $O\cap gK$ meets $g(K\setminus N)$. Therefore

$$\widehat{q}\left(U\right)=\widehat{q}\left(O\right)\cap Q,$$

which is open in Q.

Assume that Q is not Raikov complete. The open surjection $\widehat{q}$ has compact kernel K, so the induced topological isomorphism $\left(\rho G\right)/K\cong \rho Q$ and the compact-subgroup quotient theorem show that $\widehat{q}$ is perfect [8, Theorem 1.5.7]. The set $T_N$ is the full inverse image of $r_{\rho }\left(Q\right)$. The restriction of an open perfect map to the full inverse image of any subspace is again open and perfect. Its fibers are cosets of $K\cong \rho N$. □

If N is not compact, then $L_N=H_N\setminus G$ is a nonempty group-remainder of G in $H_N$ and is therefore dense in $H_N$ [5, Proposition 1.2]. Since $H_N$ is dense in $\rho G$, the set $L_N$ is dense in $\rho G$. If Q is not Raikov complete, then $T_N=r_{\rho }\left(H_N\right)$ is a nonempty group-remainder and is dense in $\rho H_N=\rho G$ by the same theorem.

For precompact N, these conclusions give four cases. If N is compact and Q is Raikov complete, then $r_{\rho }\left(G\right)$ is empty. If exactly one of these two conditions fails, the source remainder is precisely one of $L_N$ and $T_N$. If N is not compact and Q is not Raikov complete, then

$$r_{\rho }\left(G\right)=L_N\dot{\cup }{T}_N$$

with both pieces dense in $\rho G$ and hence in $r_{\rho }\left(G\right)$. In the last case, the source remainder is therefore resolvable by the classical density theorem for nonempty group-remainders. The quotient map canonically selects the two dense pieces, and its restrictions to them have different fiber types.

3. Quotient Towers and Stage Spectra

In this section, an increasing chain ${\left(N_{\alpha }\right)}_{\alpha \le \theta }$ of closed subgroups of a topological group G is continuous at a nonzero limit ordinal $\lambda \le \theta$ if

$$N_{\lambda }={\overline{\bigcup _{\alpha <\lambda }{N}_{\alpha }}}^G.$$

The chain is continuous if this condition holds at every nonzero limit ordinal $\lambda \le \theta$.

3.1. Two-Stage Defect Law

For a closed normal subgroup N of G, let $q_N:G\to G/N$ be the quotient homomorphism and let

$${\widehat{q}}_N:\rho G⟶\rho (G/N)$$

be its extension. Write

$$K_N={\overline{N}}^{\rho G}$$

and define the old-point defect by

$$D_G\left(N\right)={\widehat{q}}_N^{-1}(G/N)\setminus G=G{K}_N\setminus G.$$

Thus $D_G\left(N\right)$ is exactly the part of $r_{\rho }\left(G\right)$ that is sent back into the embedded quotient group. By Theorem 1, the set $D_G\left(N\right)$ is empty if and only if N is Raikov complete.

Theorem 3.

Let $N\subseteq M$ be closed normal subgroups of a topological group G. Use the canonical topological isomorphism

$$(G/N)/(M/N)\cong G/M$$

and regard the second quotient as $G/M$. Let

$$q_{M/N}:G/N⟶G/M$$

be the induced quotient homomorphism and put

$$K_{M/N}={\overline{M/N}}^{\rho (G/N)}.$$

Let

$$a={\widehat{q}}_N:\rho G\to \rho (G/N)$$

and let

$$b={\widehat{q}}_{M/N}:\rho (G/N)\to \rho (G/M)$$

be the completion extension of $q_{M/N}$. Then the extension of the direct quotient by M is $b\circ a$. Also,

$$a^{-1}\left(K_{M/N}\right)=K_M$$

and

$$a^{-1}\left(D_{G/N}(M/N)\right)=G{K}_M\setminus G{K}_N.$$

Consequently,

$$\begin{array}{|c|}\hline D_G\left(M\right)=D_G\left(N\right)\dot{\cup }{a}^{-1}\left(D_{G/N}(M/N)\right).\\ \hline\end{array}$$

Proof. 

The quotient homomorphisms satisfy

$$q_M=q_{M/N}\circ q_N.$$

Both ${\widehat{q}}_M$ and $b\circ a$ are continuous homomorphisms from $\rho G$ to the Hausdorff group $\rho (G/M)$, and each restricts to $q_M$ on the dense subgroup G. Hence they coincide, so

$${\widehat{q}}_M=b\circ a.$$

The kernel-closure theorem gives

$$a^{-1}\left(K_{M/N}\right)=a^{-1}(kerb)=ker(b\circ a)=ker{\widehat{q}}_M=K_M.$$

Applying Theorem 1 to the second quotient gives

$$b^{-1}(G/M)=(G/N)K_{M/N}.$$

Taking the inverse image under a and using $b\circ a={\widehat{q}}_M$ gives

$$a^{-1}\left((G/N)K_{M/N}\right)={\widehat{q}}_M^{-1}(G/M)=G{K}_M.$$

For the first quotient, Theorem 1 gives

$$a^{-1}(G/N)=G{K}_N.$$

Since inverse images commute with set difference,

$$\begin{array}{cc}\hfill a^{-1}\left(D_{G/N}(M/N)\right)& =a^{-1}\left((G/N)K_{M/N}\setminus (G/N)\right)\hfill \\ & =G{K}_M\setminus G{K}_N.\hfill \end{array}$$

The remaining set difference decomposes as

$$G{K}_M\setminus G=\left(G{K}_N\setminus G\right)\dot{\cup }\left(G{K}_M\setminus G{K}_N\right),$$

which proves the asserted two-stage identity. □

The two-stage identity separates points according to the first quotient stage at which they become old quotient points. A point in $D_G\left(N\right)$ has already become an old quotient point after the first quotient by N. A point in the second term remains outside $G/N$ after the first quotient, but its image becomes an old point when the second quotient by $M/N$ is applied.

3.2. Transfinite Stratification

Theorem 4.

Let θ be an ordinal, let G be a topological group, and let ${\left(N_{\alpha }\right)}_{\alpha \le \theta }$ be an increasing chain of closed normal subgroups of G, where $N_0=\left\{e\right\}$. Put

$$K_{\alpha }={\overline{N_{\alpha }}}^{\rho G},H_{\alpha }=G{K}_{\alpha }.$$

For $0<\gamma \le \theta$, define the γ-th first-entry stratum by

$$S_{\gamma }=\left\{\begin{array}{cc}{H}_{\delta +1}\setminus H_{\delta },\hfill & \mathit{if}\gamma =\delta +1,\hfill \\ {\displaystyle H_{\gamma }\setminus \bigcup _{\alpha <\gamma }{H}_{\alpha },}\hfill & \mathit{if}\gamma \mathit{is}a\mathit{limit}\mathit{ordinal}.\hfill \end{array}\right.$$

Then, for every $\beta \le \theta$,

$$D_G\left(N_{\beta }\right)={\dot{\bigcup }}_{0<\gamma \le \beta }{S}_{\gamma }.$$

At a successor stage $\gamma =\delta +1$, if

$$a_{\delta }:\rho G⟶\rho (G/N_{\delta })$$

denotes the completion extension, then

$$S_{\delta +1}=a_{\delta }^{-1}\left(D_{G/N_{\delta }}(N_{\delta +1}/N_{\delta })\right).$$

If the chain is continuous at a nonzero limit ordinal $\lambda \le \theta$, then

$$K_{\lambda }={\overline{\bigcup _{\alpha <\lambda }{K}_{\alpha }}}^{\rho G}.$$

Thus $S_{\lambda }$ consists precisely of the points that belong to $G{K}_{\lambda }$ but to no $G{K}_{\alpha }$ with $\alpha <\lambda$, where $K_{\lambda }$ is given by the displayed closure formula.

If, in addition, $N_{\theta }$ is precompact, then for every $\delta <\theta$,

$$S_{\delta +1}=⌀⟺N_{\delta +1}/N_{\delta }\mathit{is}\mathit{compact}.$$

Proof. 

The groups $H_{\alpha }$ form an increasing chain and $H_0=G$. Fix $x\in D_G\left(N_{\beta }\right)=H_{\beta }\setminus G$. The set of ordinals $\gamma \le \beta$ for which $x\in H_{\gamma }$ has a least member, say ${\gamma }_x>0$. If ${\gamma }_x=\delta +1$, then $x\in H_{\delta +1}\setminus H_{\delta }=S_{{\gamma }_x}$. If ${\gamma }_x$ is a limit ordinal, then x belongs to $H_{{\gamma }_x}$ and to none of the preceding groups, so $x\in S_{{\gamma }_x}$. This gives the disjoint stratification.

The successor formula is Theorem 3 applied to $N_{\delta }\subseteq N_{\delta +1}$. At a continuous limit stage,

$$\begin{array}{cc}\hfill K_{\lambda }& ={\overline{N_{\lambda }}}^{\rho G}={\overline{\bigcup _{\alpha <\lambda }{N}_{\alpha }}}^{\rho G}={\overline{\bigcup _{\alpha <\lambda }{K}_{\alpha }}}^{\rho G},\hfill \end{array}$$

which proves the limit-stage assertion.

Suppose in addition that $N_{\theta }$ is precompact, and fix $\delta <\theta$. Then $N_{\delta }$ and $N_{\delta +1}/N_{\delta }$ are precompact. Hence $a_{\delta }$ is onto [8, Proposition 3.7.19]. By the successor formula,

$$S_{\delta +1}=⌀⟺D_{G/N_{\delta }}(N_{\delta +1}/N_{\delta })=⌀.$$

By Theorem 1, the latter condition is equivalent to Raikov completeness of $N_{\delta +1}/N_{\delta }$, and precompactness makes this equivalent to compactness [8, Theorem 3.7.15]. This proves the final assertion. □

For $0<\gamma \le \theta$, put $P_{\gamma }={\bigcup }_{\alpha <\gamma }{H}_{\alpha }$. Because $G\subseteq P_{\gamma }\subseteq H_{\gamma }\subseteq \rho G$ and G is dense in $\rho G$, the group $P_{\gamma }$ is dense in $H_{\gamma }$ and $H_{\gamma }$ is dense in $\rho G$. Moreover,

$$S_{\gamma }=H_{\gamma }\setminus P_{\gamma }.$$

Hence each nonempty first-entry stratum $S_{\gamma }$ is a group-remainder of $P_{\gamma }$ in $H_{\gamma }$. By the general density theorem for nonempty group-remainders [5, Proposition 1.2], it is dense in $H_{\gamma }$, and therefore in $\rho G$.

Assume again that $N_{\theta }$ is precompact. For every $\delta <\theta$, the open surjection $a_{\delta }$ has compact kernel $K_{\delta }$. The induced topological isomorphism $\left(\rho G\right)/K_{\delta }\cong \rho (G/N_{\delta })$ and the compact-subgroup quotient theorem show that $a_{\delta }$ is perfect [8, Theorem 1.5.7]. By the successor formula, $S_{\delta +1}$ is the full inverse image of $D_{G/N_{\delta }}(N_{\delta +1}/N_{\delta })$ under $a_{\delta }$. Hence the restriction

$$a_{\delta }{|}_{S_{\delta +1}}:S_{\delta +1}⟶D_{G/N_{\delta }}(N_{\delta +1}/N_{\delta })$$

is an open perfect surjection whose fibers are homeomorphic to the compact group $K_{\delta }\cong \rho N_{\delta }$. The stratification specifies the quotient stage at which a point first becomes an old quotient point. By contrast, limit strata need not be determined by the compactness of the successive factors.

3.3. Arbitrary Stage Spectra

For a tower as in Theorem 4, define its stage spectrum by

$$\Sigma \left(N_{•}\right)=\{\gamma \in (0,\theta ]:S_{\gamma }\ne ⌀\}.$$

Under the precompactness hypothesis in Theorem 4, for each $\delta <\theta$, the stratum $S_{\delta +1}$ is empty exactly when $N_{\delta +1}/N_{\delta }$ is compact. At a nonzero limit stage $\lambda$, by contrast, the closure relation imposed by continuity does not determine whether $S_{\lambda }$ is empty.

In this subsection, Boolean groups are written additively.

Theorem 5.

Let θ be a nonzero ordinal and let $S\subseteq (0,\theta ]$. There exist a Hausdorff pseudocompact Boolean group G and a continuous increasing chain ${\left(N_{\alpha }\right)}_{\alpha \le \theta }$ of closed normal subgroups of G, with $N_0=\left\{0\right\}$, such that every $N_{\alpha }$ is pseudocompact and

$$\Sigma \left(N_{•}\right)=S.$$

For this tower,

$$r_{\rho }\left(G\right)={\dot{\bigcup }}_{\gamma \in S}{S}_{\gamma }.$$

The construction can be arranged so that

$$G/N_{\theta }\cong \mathbb{Z}/2\mathbb{Z}.$$

For every $\delta <\theta$,

$$N_{\delta +1}/N_{\delta }\mathit{is}\mathit{compact}⟺\delta +1\notin S.$$

If $S\ne ⌀$, then G and $N_{\theta }$ are not Raikov complete.

Proof. 

If $S=⌀$, take $G=\mathbb{Z}/2\mathbb{Z}$ and $N_{\alpha }=\left\{0\right\}$ for all $\alpha \le \theta$. Then $r_{\rho }\left(G\right)=⌀$, and all assertions follow. Hence assume $S\ne ⌀$.

Let

$$I=\theta \times S\times {\omega }_1,K=2^I,$$

where $2=\mathbb{Z}/2\mathbb{Z}$, and equip K with the compact product topology. For $\alpha \le \theta$, put

$$C_{\alpha }=\{x\in K:x(\xi ,s,\eta )=0\mathrm{whenever}\xi \ge \alpha \}.$$

Then $C_0=\left\{0\right\}$, $C_{\theta }=K$, and for every nonzero limit $\lambda \le \theta$,

$$C_{\lambda }={\overline{\bigcup _{\alpha <\lambda }{C}_{\alpha }}}^K.$$

Let

$$D=\{x\in K:|supp\left(x\right)|\le \omega \}.$$

Thus D is the countable-support subgroup of K. It is $G_{\delta }$-dense in K. Indeed, let $Z\subseteq K$ be a nonempty $G_{\delta }$ set and write $Z={\bigcap }_{n<\omega }{O}_n$, where each $O_n$ is open in K. Choose $z\in Z$. For every n, choose a finite set $J_n\subseteq I$ such that the cylinder of all points agreeing with z on $J_n$ is contained in $O_n$. If $J={\bigcup }_{n<\omega }{J}_n$, the point that agrees with z on J and is zero off J belongs to $D\cap Z$. By the Comfort–Ross characterization of pseudocompact totally bounded groups, D is pseudocompact [13, Theorem 1.2].

Regard $K/D$ as a vector space over ${\mathbb{F}}_2$, write $q:K\to K/D$ for the algebraic quotient map, and put $U_{\alpha }=q\left(C_{\alpha }\right)$. For each $\gamma \in S$, choose $x_{\gamma }\in C_{\gamma }$ as follows. If $\gamma =\delta +1$, let

$$supp\left(x_{\gamma }\right)=\{(\delta ,\gamma ,\eta ):\eta <{\omega }_1\}.$$

If $\gamma$ is a nonzero limit ordinal, let

$$supp\left(x_{\gamma }\right)=\{(\xi ,\gamma ,\eta ):\xi <\gamma ,\eta <{\omega }_1\}.$$

Write ${\overline{x}}_{\gamma }=q\left(x_{\gamma }\right)$. For every $\alpha <\gamma$ and every $y\in C_{\alpha }$, the difference $x_{\gamma }-y$ has uncountably many nonzero coordinates whose first coordinate is at least $\alpha$. Thus $x_{\gamma }-y\notin D$, so

$${\overline{x}}_{\gamma }\in U_{\gamma }\setminus \bigcup _{\alpha <\gamma }{U}_{\alpha }.$$

(3.1)

The supports of the $x_{\gamma }$ are pairwise disjoint in their second coordinate. Thus every nonempty finite sum of them has uncountable support. Consequently

$$\{{\overline{x}}_{\gamma }:\gamma \in S\}$$

is linearly independent in $K/D$.

By transfinite recursion, choose bases ${\mathcal{B}}_{\alpha }$ of $U_{\alpha }$ such that

$${\mathcal{B}}_{\alpha }\subseteq {\mathcal{B}}_{\beta }(\alpha \le \beta \le \theta )$$

and

$${\overline{x}}_{\gamma }\in {\mathcal{B}}_{\alpha }\mathrm{whenever}\gamma \in S\mathrm{and}\gamma \le \alpha .$$

(3.2)

At a successor stage, this follows from (3.1) and basis extension. At a limit stage, the union of the earlier bases spans ${\bigcup }_{\beta <\alpha }{U}_{\beta }$ and remains independent. If $\alpha \in S$, equation (3.1) allows ${\overline{x}}_{\alpha }$ to be adjoined, after which the resulting independent set can be extended to a basis of $U_{\alpha }$.

Let

$$V=\underset{\gamma \in S}{⨁}{\mathbb{F}}_2{e}_{\gamma }.$$

Define a linear map $\pi :K/D\to V$ on the basis ${\mathcal{B}}_{\theta }$ by

$$\pi \left({\overline{x}}_{\gamma }\right)=e_{\gamma }(\gamma \in S)$$

and by $\pi \left(b\right)=0$ for every other $b\in {\mathcal{B}}_{\theta }$. Put

$$\varphi =\pi \circ q:K⟶V,G_0=ker\varphi .$$

Only the algebraic homomorphism property of $\varphi$ is needed. Since $D\subseteq G_0$ and D is $G_{\delta }$-dense in K, the group $G_0$ is $G_{\delta }$-dense in K and hence pseudocompact [13, Theorem 1.2]. In particular, $G_0$ is dense in K. The subgroup uniformity is the restriction of the ambient two-sided uniformity, so K is a Raikov completion of $G_0$ [8, Proposition 1.8.4 and Theorem 3.6.14]. Under the completion isomorphism extending the inclusion $G_0\subseteq K$, write $\rho G_0=K$.

For $\alpha \le \theta$, set

$$M_{\alpha }=G_0\cap C_{\alpha }.$$

The subgroup $M_{\alpha }$ is closed in $G_0$. The subgroup $D\cap C_{\alpha }$ is $G_{\delta }$-dense in $C_{\alpha }$ by the same cylinder argument used for D. Since it is contained in $M_{\alpha }$, the subgroup $M_{\alpha }$ is $G_{\delta }$-dense in $C_{\alpha }$ and is pseudocompact by [13, Theorem 1.2]. In particular,

$${\overline{M_{\alpha }}}^K=C_{\alpha }.$$

(3.3)

If $\lambda \le \theta$ is a nonzero limit ordinal, every finite-support point of $C_{\lambda }$ belongs to some $C_{\alpha }$ with $\alpha <\lambda$. Such points lie in ${\bigcup }_{\alpha <\lambda }{M}_{\alpha }$ and are dense in $C_{\lambda }$. Since $M_{\lambda }=G_0\cap C_{\lambda }$ is closed in $G_0$,

$$M_{\lambda }={\overline{\bigcup _{\alpha <\lambda }{M}_{\alpha }}}^{G_0}.$$

(3.4)

Hence ${\left(M_{\alpha }\right)}_{\alpha \le \theta }$ is continuous.

By (3.1)–(3.2) and the definition of $\varphi$,

$$\varphi \left(C_{\alpha }\right)=W_{\alpha }:=span\{e_{\gamma }:\gamma \in S,\gamma \le \alpha \}.$$

(3.5)

Equation (3.5) and $G_0=ker\varphi$ give

$$G_0+C_{\alpha }={\varphi }^{-1}\left(W_{\alpha }\right).$$

(3.6)

Write $H_{\alpha }^0=G_0+C_{\alpha }$. If $\gamma =\delta +1$, then (3.5)–(3.6) give

$$H_{\gamma }^0\setminus H_{\delta }^0\ne ⌀⟺W_{\gamma }\ne W_{\delta }⟺\gamma \in S.$$

(3.7)

If $\gamma$ is a nonzero limit ordinal, every vector in V has finite support, so

$$\bigcup _{\alpha <\gamma }{W}_{\alpha }=span\{e_s:s\in S,s<\gamma \}.$$

It follows that

$$H_{\gamma }^0\setminus \bigcup _{\alpha <\gamma }{H}_{\alpha }^0\ne ⌀⟺\gamma \in S.$$

(3.8)

To obtain the stated terminal quotient, let $L=\mathbb{Z}/2\mathbb{Z}$ with its discrete topology and put

$$G=G_0\times L,N_{\alpha }=M_{\alpha }\times \left\{0\right\}.$$

Since $G_0$ is a subgroup of the Boolean group K and L is Boolean, the group G is Boolean. It is Hausdorff as a subgroup of the compact Hausdorff group $K\times L$, and it is pseudocompact because it is a finite union of copies of the pseudocompact group $G_0$. Each $N_{\alpha }$ is closed in G because $M_{\alpha }$ is closed in $G_0$, normal because G is Abelian, and pseudocompact because it is homeomorphic to $M_{\alpha }$. Since $C_0=\left\{0\right\}$, one has $N_0=\left\{0\right\}$. Moreover, G is dense in $K\times L$. The two-sided uniformity on G is the restriction of that on $K\times L$, so $K\times L$ is a Raikov completion of G [8, Proposition 1.8.4 and Theorem 3.6.14]. Thus $\rho G$ may be taken to be $K\times L$, and (3.3) gives

$${\overline{N_{\alpha }}}^{\rho G}={\overline{M_{\alpha }}}^K\times \left\{0\right\}=C_{\alpha }\times \left\{0\right\}.$$

The families ${\left(C_{\alpha }\right)}_{\alpha \le \theta }$, ${\left(M_{\alpha }\right)}_{\alpha \le \theta }$, and ${\left(N_{\alpha }\right)}_{\alpha \le \theta }$ are increasing, and (3.4) shows that the tower ${\left(N_{\alpha }\right)}_{\alpha \le \theta }$ is continuous. For this tower, the intermediate groups from Theorem 4 are

$$H_{\alpha }=G+(C_{\alpha }\times \left\{0\right\})=(G_0+C_{\alpha })\times L=H_{\alpha }^0\times L.$$

Hence (3.7)–(3.8) show that $\Sigma \left(N_{•}\right)=S$. Since $M_{\theta }=G_0$,

$$G/N_{\theta }\cong L\cong \mathbb{Z}/2\mathbb{Z}.$$

Since $C_{\theta }=K$, the formula for $H_{\alpha }$ gives $H_{\theta }=\rho G$. Hence $D_G\left(N_{\theta }\right)=r_{\rho }\left(G\right)$, and Theorem 4 together with $\Sigma \left(N_{•}\right)=S$ gives

$$r_{\rho }\left(G\right)={\dot{\bigcup }}_{\gamma \in S}{S}_{\gamma }.$$

In particular, $N_{\theta }=M_{\theta }\times \left\{0\right\}=G_0\times \left\{0\right\}$ is a subgroup of the compact group $K\times L$, so it is precompact. The compactness criterion for successor strata in Theorem 4 now gives

$$N_{\delta +1}/N_{\delta }\mathrm{is}\mathrm{compact}⟺\delta +1\notin S$$

for every $\delta <\theta$. When $S\ne ⌀$, the map $\varphi$ is nonzero, so $G_0$ is a proper dense subgroup of K. Hence $G_0$, $N_{\theta }\cong G_0$, and $G=G_0\times L$ are not Raikov complete. □

As an extreme case, let $\lambda \le \theta$ be a nonzero limit ordinal and take $S=\left\{\lambda \right\}$ in Theorem 5. Then G and every $N_{\alpha }$ are pseudocompact, and every successive factor is compact. The unique nonempty stratum is

$$S_{\lambda }=r_{\rho }\left(G\right)\ne ⌀,$$

and it is dense in $\rho G$. Hence compactness of all successive factors, even together with pseudocompactness of G and every $N_{\alpha }$, does not force all limit strata to be empty.

4. Pseudocompactness and Examples

4.1. Pseudocompactness Consequences

For a closed normal subgroup N of G, write $\widehat{q}:\rho G\to \rho (G/N)$ for the completion extension of the quotient homomorphism.

By the locally finite characterization of pseudocompactness [15, Theorem 3.10.22], an open perfect surjection $f:X\to Y$ between Tychonoff spaces preserves pseudocompactness in both directions. The forward implication follows from preservation under continuous images. For the reverse implication, assume that Y is pseudocompact and let $\mathcal{U}$ be a locally finite family of nonempty open subsets of X. Since f is open, each $f\left(U\right)$ is open. Fix $y\in Y$. Compactness of $f^{-1}\left(y\right)$ gives an open neighborhood W of the fiber that meets only finitely many members of $\mathcal{U}$. Since f is closed, there is an open neighborhood V of y with $f^{-1}\left(V\right)\subseteq W$. If $V\cap f\left(U\right)\ne ⌀$, then $f^{-1}\left(V\right)\cap U\ne ⌀$, so U meets W. Thus there are only finitely many $U\in \mathcal{U}$ for which $V\cap f\left(U\right)\ne ⌀$. Moreover, only finitely many members of $\mathcal{U}$ can have the same image under f. Otherwise, a point in that common image would have a compact fiber meeting infinitely many members of the locally finite family $\mathcal{U}$. Therefore an infinite $\mathcal{U}$ would produce an infinite locally finite family of nonempty open subsets of Y, contradicting pseudocompactness. Hence $\mathcal{U}$ is finite, and X is pseudocompact.

The perfectness of the natural quotient map by a compact subgroup is classical [8, Theorem 1.5.7]. Combining this fact with Corollary 1 gives the following corollary.

Corollary 2.

Let N be a compact normal subgroup of a topological group G, and let $\widehat{q}:\rho G\to \rho (G/N)$ be the completion extension of the quotient homomorphism. Then

$$\widehat{q}{|}_{r_{\rho }\left(G\right)}:r_{\rho }\left(G\right)⟶r_{\rho }(G/N)$$

is an open perfect surjection. Every fiber is homeomorphic to N. Consequently,

$$r_{\rho }\left(G\right)\mathit{is}\mathit{pseudocompact}⟺r_{\rho }(G/N)\mathit{is}\mathit{pseudocompact}.$$

Proof. 

Since N is compact, $\widehat{q}$ is onto by [8, Proposition 3.7.19]. Moreover, N is compact and therefore closed in the Hausdorff group $\rho G$, so ${\overline{N}}^{\rho G}=N$. Theorem 1 now gives

$${\widehat{q}}^{-1}(G/N)=G.$$

Therefore

$$r_{\rho }\left(G\right)={\widehat{q}}^{-1}\left(r_{\rho }(G/N)\right).$$

The map $\widehat{q}$ is open by the classical completion theorem [8, Theorem 3.6.19]. Since it is onto with compact kernel N, the induced topological isomorphism $\left(\rho G\right)/N\cong \rho (G/N)$ and the compact-subgroup quotient theorem show that $\widehat{q}$ is perfect [8, Theorem 1.5.7]. Its restriction to this full inverse image remains open and perfect. The fiber formula in Theorem 1 shows that every fiber is a coset of N. The final equivalence follows because open perfect maps preserve pseudocompactness in both directions. □

The classical density theorem for nonempty group-remainders [5, Proposition 1.2], together with the compactness theorem for Raikov completions of pseudocompact groups [8, Corollary 3.7.18], gives the following consequence.

Corollary 3.

Let G be a topological group. If $r_{\rho }\left(G\right)$ is nonempty and pseudocompact, then $\rho G$ is compact and G is precompact. Consequently, if N is any closed normal subgroup of G and $\widehat{q}:\rho G\to \rho (G/N)$ is the completion extension of the quotient homomorphism, then

$$\widehat{q}\left(r_{\rho }\left(G\right)\right)=\left\{\begin{array}{cc}{r}_{\rho }(G/N),\hfill & \mathit{if}N\mathit{is}\mathit{compact},\hfill \\ \rho (G/N),\hfill & \mathit{if}N\mathit{is}\mathit{not}\mathit{compact}.\hfill \end{array}\right.$$

If N is compact, then Corollary 2 gives the stronger equivalence between pseudocompactness of the source and quotient remainders.

Proof. 

The space $r_{\rho }\left(G\right)$ is dense in $\rho G$ by [5, Proposition 1.2]. Let $f:\rho G\to \mathbb{R}$ be continuous. Its restriction to $r_{\rho }\left(G\right)$ is bounded. Density gives

$$f\left(\rho G\right)\subseteq \overline{f\left(r_{\rho }\left(G\right)\right)},$$

so $\rho G$ is pseudocompact. The Raikov completion of every pseudocompact topological group is compact [8, Corollary 3.7.18]. Since $\rho G$ is already Raikov complete, it is compact. Therefore G is precompact. Every subgroup of a precompact group is precompact, so Corollary 1 gives the displayed dichotomy. The compact-kernel assertion follows from Corollary 2. □

If the kernel is noncompact in Corollary 3, the source remainder maps onto the entire quotient completion, including the embedded quotient group.

For small precompact groups, the image dichotomy does not determine whether the quotient remainder itself is pseudocompact when the kernel is not Raikov complete. Arhangel’skii and Bella state in their abstract that every nonmetrizable topological group of cardinality at most ${\omega }_1$ is pseudocompact at infinity [4]. Combining this statement with the elementary metrizable case gives the following criterion for small precompact groups.

Proposition 1.

Let H be a precompact topological group with $\left|H\right|\le {\omega }_1$. Then $r_{\rho }\left(H\right)$ is pseudocompact if and only if H is nonmetrizable or Raikov complete.

Proof. 

If H is nonmetrizable, then the cited result of Arhangel’skii and Bella shows that H is pseudocompact at infinity. Since $\rho H$ is a compactification of the precompact group H, the remainder $r_{\rho }\left(H\right)$ is pseudocompact.

If H is Raikov complete, then precompactness makes H compact, so $r_{\rho }\left(H\right)=⌀$.

Suppose that H is metrizable and not Raikov complete. The two-sided uniformity of a metrizable topological group is metrizable, so its completion $\rho H$ is a compact metrizable group. Moreover, H is a proper dense subgroup. The remainder $\rho H\setminus H$ is metrizable and noncompact. Indeed, compactness would make it closed in $\rho H$, forcing H to be open and hence closed, contrary to proper density. A pseudocompact metrizable space is compact. Hence $r_{\rho }\left(H\right)$ is not pseudocompact. □

4.2. Countable Abelian Examples

For a discrete Abelian group D, let $D^{\#}$ denote D equipped with its Bohr topology. Then $D^{\#}$ is Hausdorff and precompact [8, Corollary 9.9.6]. Put

$$A={\mathbb{Z}}^{\#}.$$

Every compact subset of A is finite [8, Theorem 9.9.30]. Since A is infinite and precompact, it is not discrete. If A were metrizable, non-discreteness and first countability at the identity would give a sequence of distinct points converging to the identity. That sequence together with its limit would be an infinite compact subset of A, a contradiction. Hence A is nonmetrizable. If A were Raikov complete, precompactness would make it compact [8, Theorem 3.7.15], again contradicting the finiteness of compact subsets. Thus A is not Raikov complete.

Choose an irrational number t and let

$$B=\{e^{2\pi int}:n\in \mathbb{Z}\}$$

be equipped with the subspace topology inherited from the circle group $\mathbb{T}$. Since t is irrational, B is dense in $\mathbb{T}$. Thus B is countable, metrizable, precompact, and not Raikov complete, and $\mathbb{T}$ is a Raikov completion of B. Write $\rho A$ for the Raikov completion of A, and use additive notation for A and multiplicative notation for B.

Example 1.

Let

$$G=A\times A,N=A\times \left\{0\right\}.$$

Then G is countable, precompact, Abelian, nonmetrizable, and not Raikov complete. Hence $r_{\rho }\left(G\right)\ne ⌀$, and Proposition 1 shows that it is pseudocompact. The subgroup N is closed and normal, precompact, and noncompact. Corollary 1 gives

$$\widehat{q}\left(r_{\rho }\left(G\right)\right)=\rho (G/N).$$

On the other hand,

$$G/N\cong A,$$

so $r_{\rho }(G/N)$ is pseudocompact by Proposition 1. Thus the equality $\widehat{q}\left(r_{\rho }\left(G\right)\right)=\rho (G/N)$ is compatible with pseudocompactness of the quotient remainder. The decomposition in Theorem 2 becomes

$$L_N=(\rho A\setminus A)\times A\mathit{and}{T}_N=\rho A\times (\rho A\setminus A).$$

Both sets are dense in $\rho A\times \rho A$ [5, Proposition 1.2], and Theorem 2(5) shows that $T_N$ maps onto $r_{\rho }\left(A\right)$ by an open perfect map.

Example 2.

Let

$$G=A\times B,N=A\times \left\{1\right\}.$$

Again G is countable, precompact, Abelian, nonmetrizable, and not Raikov complete. Hence $r_{\rho }\left(G\right)\ne ⌀$, and Proposition 1 shows that it is pseudocompact. The subgroup N is closed and normal, precompact, and noncompact, so

$$\widehat{q}\left(r_{\rho }\left(G\right)\right)=\rho (G/N)$$

by Corollary 1. Since

$$G/N\cong B,$$

the quotient is metrizable and not Raikov complete. Proposition 1, applied to $G/N$, shows that $r_{\rho }(G/N)$ is not pseudocompact.

Equivalently,

$$r_{\rho }(G/N)\cong \mathbb{T}\setminus B.$$

This space is metrizable and noncompact. If it were compact, it would be closed in $\mathbb{T}$, so the proper dense subgroup B would be open and therefore closed, a contradiction. Here the canonical decomposition is

$$L_N=(\rho A\setminus A)\times B\mathit{and}{T}_N=\rho A\times (\mathbb{T}\setminus B).$$

Again both parts are dense in the source completion [5, Proposition 1.2]. By Theorem 2(5), the natural map from $T_N$ onto $\mathbb{T}\setminus B$ is open and perfect. Hence $T_N$ is not pseudocompact.

In both examples, the kernel is precompact and not Raikov complete, so the image of the source remainder is the entire quotient completion. Neither quotient group is Raikov complete. The quotient in Example 1 is nonmetrizable, whereas the quotient in Example 2 is metrizable.

Example 3.

Let $A={\mathbb{Z}}^{\#}$ and let B be the dense cyclic subgroup of $\mathbb{T}$ used above. Put

$$G=A\times A\times B,$$

and consider the closed normal chain

$$N_0=\left\{0\right\}\times \left\{0\right\}\times \left\{1\right\}\subseteq N_1=A\times \left\{0\right\}\times \left\{1\right\}\subseteq N_2=A\times A\times \left\{1\right\}.$$

The terminal subgroup $N_2\cong A\times A$ is precompact, and both successive factors $N_1/N_0\cong A$ and $N_2/N_1\cong A$ are noncompact. The compactness criterion in Theorem 4 therefore gives two nonempty disjoint first-entry strata

$$S_1=G{K}_1\setminus G,S_2=G{K}_2\setminus G{K}_1,$$

and

$$D_G\left(N_2\right)=S_1\dot{\cup }{S}_2.$$

Each $S_i$ is dense in $\rho G$ [5, Proposition 1.2]. The final quotient satisfies

$$G/N_2\cong B,$$

so its Raikov remainder is not pseudocompact. Meanwhile, G is countable, precompact, nonmetrizable, and not Raikov complete. Hence $r_{\rho }\left(G\right)\ne ⌀$, and Proposition 1 shows that it is pseudocompact. Thus the old-point defect $D_G\left(N_2\right)$ splits into two distinct first-entry strata, while the final quotient remainder is not pseudocompact.

If the source remainder is allowed to be empty, a degenerate Abelian example shows that the quotient remainder need not be pseudocompact. In Problem 7.9.B, Arhangel’skii and Tkachenko ask the reader to show that every Abelian topological group is a quotient of a Raikov-complete Abelian topological group [8, Problem 7.9.B]. Applying this statement to $\mathbb{Q}$ with its usual topology gives a Raikov-complete Abelian group H and a closed normal subgroup N such that $H/N\cong \mathbb{Q}$. Since $\mathbb{R}$ is a Raikov completion of $\mathbb{Q}$, the source remainder is empty while the quotient remainder is homeomorphic to $\mathbb{R}\setminus \mathbb{Q}$, which is not pseudocompact. Example 2 removes this degeneracy.

For a closed normal subgroup, the closure of the kernel in the Raikov completion determines exactly the source-remainder fibers over the embedded quotient and over the points of the quotient remainder that lie in the image of the completion extension. If the kernel is precompact, the source remainder has the canonical decomposition $r_{\rho }\left(G\right)=L_N\dot{\cup }{T}_N$. For nested kernels, the two-stage defect law separates what is already visible after the first quotient from what first becomes an old quotient point after the second, and its transfinite iteration gives the first-entry stratification along normal towers. When the terminal subgroup is precompact, compactness of a successive factor is exactly the criterion for the corresponding successor stratum to vanish. In contrast, a continuous tower can have a nonempty limit stratum even when every successive factor is compact. More generally, the realization theorem shows that every subset of the nonzero stages occurs as the stage spectrum of a continuous tower in a pseudocompact Boolean group, with every member of the tower pseudocompact and with terminal quotient $\mathbb{Z}/2\mathbb{Z}$. For compact kernels, pseudocompactness of the source and quotient remainders is equivalent. For noncompact precompact kernels, the source remainder maps onto the entire quotient completion. The countable Abelian examples show that the quotient remainder can nevertheless be pseudocompact or non-pseudocompact.

5. Further Questions

The results above lead to three questions.

1.

Let N be a closed normal precompact noncompact subgroup of G. Find useful conditions under which the old-point part

$$L_N={\widehat{q}}^{-1}(G/N)\setminus G$$

is pseudocompact or countably compact. In particular, determine to what extent these properties can be inferred from $G/N$ and $r_{\rho }\left(N\right)$. The restriction $\widehat{q}{|}_{L_N}:L_N\to G/N$ is a continuous open surjection, and every fiber is homeomorphic to $r_{\rho }\left(N\right)$. Since N is precompact and noncompact, $\rho N$ is compact and N is a proper dense subgroup. Thus $r_{\rho }\left(N\right)$ is noncompact. If it were compact, it would be closed in $\rho N$, making N an open subgroup and hence closed, contrary to proper density. Therefore the map is not perfect.

2.

Characterize continuous chains of closed normal subgroups for which every limit stratum in Theorem 4 is empty. Theorem 5 shows that compactness of all successive factors is not sufficient. Such a criterion should distinguish continuity of the subgroup chain inside G from the additional closure step inside $\rho G$.

3.

Does there exist a topological group G and a closed normal subgroup N such that $r_{\rho }\left(G\right)$ is nonempty and countably compact while $r_{\rho }(G/N)$ is not countably compact? Any such example must have N noncompact. Indeed, countable compactness of $r_{\rho }\left(G\right)$ implies pseudocompactness, so Corollary 3 makes G precompact. If N were compact, Corollary 2 would make $r_{\rho }(G/N)$ a continuous image of $r_{\rho }\left(G\right)$, and hence countably compact. Consequently, $G/N$ is precompact and $\rho (G/N)$ is compact in any such example. Arhangel’skii and Bella state in their abstract that pseudocompactness at infinity need not imply countable compactness at infinity even for countable groups [4]. The quotient version asks for such a separation within the fixed Raikov completions.