\documentclass[12pt]{amsart} \headheight=7pt \topmargin=14pt \textheight=574pt \textwidth=432pt \oddsidemargin=18pt \evensidemargin=18pt \usepackage{latexsym} %\usepackage{epsfig} \title[$\mathbf{P}^2$-irreducible open 3-manifolds] {$\mathbf{R}^2$-irreducible universal covering spaces of $\mathbf{P}^2$-irreducible open 3-manifolds} \author{Robert Myers} \address{Department of Mathematics, Oklahoma State University, Stillwater, OK 74078} \email{myersr@math.okstate.edu} \thanks{Research at MSRI is supported in part by NSF grant DMS-9022140.} \subjclass{Primary: 57M10; Secondary: 57N10, 57M60} \keywords{3-manifold, open 3-manifold, $\mathbf{P}^2$-irreducible, $\mathbf{R}^2$-irreducible, universal covering space, contractible open 3-manifold, Whitehead manifold} \renewcommand{\labelenumi}{(\theenumi)} \newtheorem{prop}{Proposition} \newtheorem{thm}{Theorem} \newtheorem{conj}{Conjecture} \newtheorem{lem}{Lemma}[section] \newcommand{\R}{\ensuremath{\mathbf{R}}} \newcommand{\RR}{\ensuremath{\mathbf{R}^2}} \newcommand{\RRR}{\ensuremath{\mathbf{R}^3}} \newcommand{\PP}{\ensuremath{\mathbf{P}^2}} \newcommand{\bd}{\ensuremath{\partial}} \newcommand{\irr}{irreducible} \newcommand{\rirr}{\RR-\irr} \newcommand{\pirr}{\PP-\irr} \newcommand{\birr}{\bd-\irr} \newcommand{\inc}{incompressible} \newcommand{\tm}{3-manifold} \newcommand{\Mt}{\ensuremath{\widetilde{M}}} \newcommand{\Qt}{\ensuremath{\widetilde{Q}}} \newcommand{\hm}{homeomorphic} \newcommand{\8}{\ensuremath{\infty}} \newcommand{\n}{^{-1}} \newcommand{\p}{^{\prime}} \newcommand{\inte}{int \, } \newcommand{\Inte}{Int \, } \newcommand{\halfspace}{\ensuremath{\RR \times [0,\8)}} \newcommand{\prodspace}{\ensuremath{\RR \times [0,1]}} \newcommand{\pf}{\noindent \textbf{Proof:}} \newcommand{\Dn}{\ensuremath{D_n}} \newcommand{\Dnn}{\ensuremath{D_{n+1}}} \newcommand{\Pn}{\ensuremath{P_n}} \newcommand{\Qn}{\ensuremath{Q_n}} \newcommand{\Rn}{\ensuremath{R_n}} \newcommand{\Hn}{\ensuremath{H_n}} 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\newcommand{\alt}{\ensuremath{\alpha_2}} \newcommand{\beo}{\ensuremath{\beta_1}} \newcommand{\bet}{\ensuremath{\beta_2}} \newcommand{\gao}{\ensuremath{\gamma_1}} \newcommand{\gat}{\ensuremath{\gamma_2}} \newcommand{\deo}{\ensuremath{\delta_1}} \newcommand{\detw}{\ensuremath{\delta_2}} \newcommand{\ep}{\ensuremath{\varepsilon}} \newcommand{\tht}{\ensuremath{\theta}} \newcommand{\tho}{\ensuremath{\theta_1}} \newcommand{\thz}{\ensuremath{\theta_0}} \newcommand{\om}{\ensuremath{\omega}} \newcommand{\omj}{\ensuremath{\omega_j}} \newcommand{\pp}{^{\prime\prime}} \newcommand{\lao}{\ensuremath{\lambda_1}} \newcommand{\lat}{\ensuremath{\lambda_2}} \newcommand{\kao}{\ensuremath{\kappa_1}} \newcommand{\kat}{\ensuremath{\kappa_2}} \newcommand{\kad}{\ensuremath{\kappa_3}} \newcommand{\nul}{\emptyset} \newcommand{\pe}{poly-excellent} \newcommand{\Pnnjp}{\ensuremath{P_{n+1,j}^+}} \newcommand{\Simnn}{\ensuremath{\Sigma^m_{n+1}}} \newcommand{\Lamnn}{\ensuremath{\Lambda^m_{n+1}}} \newcommand{\Omn}{\ensuremath{O^m_n}} \newcommand{\Icm}{\ensuremath{I_m}} \newcommand{\Fmp}{\ensuremath{F_m\p}} \newcommand{\Fmpp}{\ensuremath{F_m\pp}} \newcommand{\Fmm}{\ensuremath{F_{m+1}}} \newcommand{\Fm}{\ensuremath{F_m}} \newcommand{\Fmmp}{\ensuremath{F_{m+1}\p}} \newcommand{\Ft}{\ensuremath{\widetilde{F}}} \newcommand{\Ej}{\ensuremath{E_j}} \newcommand{\ali}{\ensuremath{\alpha_i}} \newcommand{\alk}{\ensuremath{\alpha_k}} \newcommand{\bei}{\ensuremath{\beta_i}} \newcommand{\bek}{\ensuremath{\beta_k}} \newcommand{\gai}{\ensuremath{\gamma_i}} \newcommand{\gak}{\ensuremath{\gamma_k}} \newcommand{\dei}{\ensuremath{\delta_i}} \newcommand{\dek}{\ensuremath{\delta_k}} \newcommand{\Pin}{\ensuremath{P^i_n}} \newcommand{\Qin}{\ensuremath{Q^i_n}} \newcommand{\Hin}{\ensuremath{H^i_n}} \newcommand{\lait}{\ensuremath{\lambda^i_t}} \newcommand{\Limun}{\ensuremath{L^{i,\mu}_n}} \newcommand{\Zimun}{\ensuremath{Z^{i,\mu}_n}} \newcommand{\Zimu}{\ensuremath{Z^{i,\mu}}} \newcommand{\Cimum}{\ensuremath{C^{i,\mu,m}}} \newcommand{\Cimumn}{\ensuremath{C^{i,\mu,m}_n}} \newcommand{\Cmmn}{\ensuremath{C^{m+1}_n}} \newcommand{\Cimummn}{\ensuremath{C^{i,\mu,m+1}_n}} \newcommand{\Mnnn}{\ensuremath{M_{n-1}}} \newcommand{\Rnnn}{\ensuremath{R_{n-1}}} \newcommand{\Cmnnn}{\ensuremath{C^m_{n-1}}} \newcommand{\Xns}{\ensuremath{X_{n,s}}} \newcommand{\Ks}{\ensuremath{K_{\sigma}}} \newcommand{\Mtphi}{\ensuremath{\widetilde{M}[\varphi]}} \newcommand{\Mtpsi}{\ensuremath{\widetilde{M}[\psi]}} \newcommand{\Ut}{\ensuremath{\widetilde{U}}} \begin{document} \begin{abstract} An \irr\ open \tm\ $W$ is \rirr\ if it contains no non-trivial planes, i.e. given any proper embedded plane $\Pi$ in $W$ some component of $W-\Pi$ must have closure an embedded halfspace \halfspace. In this paper it is shown that if $M$ is a connected, \pirr, open \tm\ such that $\pi_1(M)$ is finitely generated and the universal covering space \Mt\ of $M$ is \rirr, then either \Mt\ is \hm\ to \RRR\ or $\pi_1(M)$ is a free product of infinite cyclic groups and fundamental groups of closed, connected surfaces other than $S^2$ or \PP. Given any finitely generated group $G$ of this form, uncountably many \pirr, open 3-manifolds $M$ are constructed with $\pi_1(M)\cong G$ such that the universal covering space \Mt\ is \rirr\ and not \hm\ to \RRR; the \Mt\ are pairwise non-\hm. Relations are established between these results and the conjecture that the universal covering space of any \irr, orientable, closed \tm\ with infinite fundamental group must be \hm\ to \RRR. \end{abstract} \maketitle \section{Introduction} Suppose $M$ is a connected, \pirr, open \tm\ with $\pi_1(M)$ finitely generated and non-trivial. It is easy to construct examples of such $M$ for which the universal covering space \Mt\ is not \hm\ to \RRR. Start with any \tm\ $N$ satisfying the given conditions. Let $U$ be a \textbf{Whitehead manifold}, i.e. an \irr, contractible, open \tm\ which is not \hm\ to \RRR (see e.g. \cite{Wh}, \cite{Mc}). Choose end-proper embeddings of $[0,\8)$ in each of $N$ and $U$. (A map between manifolds is \textbf{end-proper} if pre-images of compact sets are compact; it is \textbf{$\bd$-proper} if the pre-image of the boundary is the boundary; it is \textbf{proper} if it has both these properties. These terms are applied to a submanifold if its inclusion map has the corresponding property.) Let $X$ and $Y$ be the exteriors of these rays. (The \textbf{exterior} of a submanifold is the closure of the complement of a regular neighborhood of it.) $\bd X$ and $\bd Y$ are each planes. We identify them to obtain a \pirr\ open \tm\ $M$ with $\pi_1(M) \cong \pi_1(N)$. Let $p:\Mt \rightarrow M$ be the universal covering map. Then \Mt, $p\n (X)$, and $p\n (Y)$ are \pirr\ \cite{MSY}. Each component $\widetilde{Y}$ of $p\n(Y)$ has interior \Ut\ \hm\ to $U$ and so contains a compact, connected subset $J$ which does not lie in a 3-ball in \Ut. If \Mt\ were \hm\ to \RRR\ then $J$ would lie in a 3-ball $B$ in \Mt. Standard general position and minimality arguments applied to $\bd B$ and $\bd \widetilde{Y}$ would then yield a 3-ball $B\p$ in \Ut\ containing $J$, a contradiction. Alternatively, one could use the Tucker Compactification Theorem \cite{Tu} to obtain a compact polyhedron $K$ in \Ut\ such that some component $V$ of $\Ut-K$ has non-finitely generated fundamental group. But this is impossible since the union of $V$ and $\Mt-\Ut$ is a component of $\Mt-K$ whose fundamental group is isomorphic to $\pi_1(V)$. In this example $\bd \widetilde{Y}$ is a \textbf{non-trivial plane} in \Mt, i.e. a proper plane $\Pi$ such that no component of $\Mt-\Pi$ has closure \hm\ to \halfspace\ with $\Pi=\RR \times \{0\}$. This paper shows that it is harder to find examples if one rules out this behavior by requiring that \Mt\ be \textbf{\rirr} in the sense that, in addition to being \irr, it contains no non-trivial planes. Define a \textbf{closed surface group} to be the fundamental group of a closed, connected 2-manifold. \begin{thm} Let $M$ be a connected, \pirr, open \tm\ with $\pi_1(M)$ finitely generated. If the universal covering space \Mt\ of $M$ is \rirr, then either \begin{enumerate} \item \Mt\ is \hm\ to \RRR\ or \item $\pi_1(M)$ is a free product of infinite cyclic groups and infinite closed surface groups. \end{enumerate} \end{thm} The second possibility can be disjoint from the first. \begin{thm} Suppose $G$ is a free product of finitely many infinite cyclic groups and infinite closed surface groups. Then there is a \pirr\ open \tm\ $M$ such that $\pi_1(M)\cong G$ and $\Mt$ is an \rirr\ Whitehead manifold. Moreover, for each given $G$ there are uncountably many such $M$ for which the \Mt\ are pairwise non-\hm. \end{thm} This generalizes an example of Scott and Tucker \cite{Sc-Tu} for which $G$ is infinite cyclic. These results have a bearing on the following well-known problem. \begin{conj}[Universal Covering Conjecture] Let $X$ be a closed, connected, \irr, orientable \tm\ with $\pi_1(X)$ infinite. Then the universal covering space $\widetilde{X}$ of $X$ is \hm\ to \RRR. \end{conj} Since there are only countably many homeomorphism types of closed 3-manifolds Theorem 2 implies that there must exist uncountably many \rirr\ Whitehead manifolds \Mt\ which cover open 3-manifolds $M$ with $\pi_1(M)\cong G$ but cannot cover a closed 3-manifold. This generalizes a result of Tinsley and Wright \cite{Ti-Wr} which shows that there must exist uncountably many non-\rirr\ Whitehead manifolds \Mt\ which cover open 3-manifolds $M$ with $\pi_1(M)$ infinite cyclic but cannot cover a closed \tm. Unfortunately this argument does not provide any \textit{specific} such examples. Specific examples of non-\rirr\ Whitehead manifolds \Mt\ which cover open 3-manifolds $M$ with $\pi_1(M)$ infinite cyclic or, more generally, a countable free group, but cannot cover a closed 3-manifold are given in \cite{My cover} and \cite{My free}, respectively. At the time of this writing the problem of providing specific examples of \rirr\ Whitehead manifolds which non-trivially cover other open 3-manifolds but cannot cover a closed 3-manifold is still open. One can make several conjectures related to Conjecture 1. We consider the selection below. In all of them $G$ is assumed to be a finitely generated group of covering translations acting on a Whitehead manifold $W$ with quotient a 3-manifold $M$. \begin{conj} $G$ is a free product of infinite cyclic groups and fundamental groups of \birr\ Haken manifolds. \end{conj} \begin{conj} $G$ is a free group or contains an infinite closed surface group. \end{conj} \begin{conj} If $W$ is \rirr, then $G$ is a free product of infinite cyclic groups and infinite closed surface groups. \end{conj} A proper plane $\Pi$ in $W$ is \textbf{equivariant} if for each $g \in G$ either $g(\Pi)=\Pi$ or $\Pi \cap g(\Pi)=\nul$. \begin{conj}[Special Equivariant Plane Conjecture] If $G$ is not a free product of infinite cyclic groups and infinite closed surface groups, then $W$ contains a non-trivial equivariant plane. \end{conj} \begin{conj}[Equivariant Plane Conjecture] If $W$ contains a non-trivial plane, then it contains a non-trivial equivariant plane. \end{conj} These conjectures are related as follows. \begin{thm} $(4)\Leftarrow(1)\Leftrightarrow(2)\Leftrightarrow(3) \Leftrightarrow(5)\Leftarrow(4+6)$ \end{thm} Theorems 1 and 3 are proven in section 2. Theorem 2 is proven in sections 3--7. Section 3 presents a modified version of the criterion used by Scott and Tucker \cite{Sc-Tu} for showing that a \tm\ is \rirr. Sections 4 and 5 treat, respectively, the special cases in which $G$ is an infinite cyclic group and an infinite closed surface group. The constructions and notation of these special cases are used in section 6, which treats the general case. Section 7 shows how to get uncountably many $M$ with non-\hm\ \Mt\ for each group $G$. \section{The proofs of Theorems 1 and 3} \begin{lem} Let $M$ be a connected, \pirr, open \tm. Let $Q$ be a compact, connected, 3-dimensional submanifold of $M$ such that $\bd Q$ is \inc\ in $M$ and $\pi_1(Q)$ is not an infinite closed surface group. Let $p:\Mt \rightarrow M$ be the universal covering map and $G$ the group of covering translations. Let \Qt\ be a component of $p\n(Q)$. Then \begin{enumerate} \item Each component of $p\n(\bd Q)$ is a plane. \item There is no component $\Pi$ of $\bd \Qt$ which is invariant under the subgroup $G_0$ of $G$ consisting of those covering translations which leave $\Qt$ invariant. \item If each component of $\bd \Qt$ is a trivial plane, then \Mt\ is \hm\ to \RRR. \end{enumerate} \end{lem} \begin{proof} (1) follows from the incompressibility of $\bd Q$ in $M$. Suppose $S$ is a component of $\bd Q$ and $\Pi$ is a component of $p\n(S)$ which is invariant under $G_0$. Since the restriction of $p$ to \Qt\ is the universal covering space of $Q$ and the restriction of $G_0$ to \Qt\ is the group of covering translations we have that $\pi_1(S) \rightarrow \pi_1(Q)$ is an isomorphism, contradicting our assumption on $\pi_1(Q)$. This establishes (2). We now prove (3). Suppose that each component $\Pi$ of $\bd \Qt$ bounds an end-proper halfspace $H_{\Pi}$ in \Mt. Let $K_{\Pi}$ be the closure of the component of $\Mt-\Pi$ which does not contain $\inte \Qt$. Assume that for all such $\Pi$ we have $H_{\Pi}=K_{\Pi}$. Then \Mt\ is the union of \Qt\ and an open collar attached to $\bd \Qt$, hence \Mt\ is \hm\ to $\inte \Qt$. Since $Q$ is Haken, the Waldhausen Compactification Theorem \cite{Wa} implies that \Qt\ is \hm\ to a closed 3-ball minus a closed subset of its boundary, hence $\inte \Qt$ is \hm\ to \RRR, and we are done. Thus we may assume that for some $\Pi$ we have $H_{\Pi}\neq K_{\Pi}$. Then $H_{\Pi}\cap K_{\Pi}=\Pi$ and $H_{\Pi}\cup K_{\Pi}=\Mt$. Now $G_0$ has an element $g$ such that $g(\Pi)\neq \Pi$. Since $\Qt \subseteq H_{\Pi}$ and $g(\Qt)=\Qt$ we must have $g(K_{\Pi}) \subseteq H_{\Pi}$. Since \halfspace\ is \rirr\ (see e.g. \cite{My endsum}) it follows that $K_{\Pi}$ is \hm\ to \halfspace. Thus \Mt\ is \hm\ to \RRR. \end{proof} \begin{proof}[Proof of Theorem 1] By passing to a covering space of $M$, if necessary, we may assume that $\pi_1(M)$ is indecomposable with respect to free products and is neither an infinite cyclic group nor an infinite closed surface group. Let $C$ be the Scott compact core \cite{Sc} of $M$, i.e. $C$ is a compact, connected, 3-dimensional submanifold of $M$ such that $\pi_1(C) \rightarrow \pi_1(M)$ is an isomorphism. The conditions on $\pi_1(M)$ imply that $\bd C$ is \inc\ in $M$. We thus can apply Lemma 2.1 with $Q=C$ to finish the proof. \end{proof} \begin{proof}[Proof of Theorem 3] We first show that $(1)\Rightarrow(2)\Rightarrow(3)\Rightarrow(1)$. If (1) is true, then $M$ must be non-compact; this follows from the fact that if $M$ were closed and non-orientable, then it would be Haken and so have universal covering space \hm\ to \RRR. Let $C$ be the Scott compact core for $M$. Since $M$ is \irr\ we may assume that no component of $\bd C$ is a 2-sphere; it follows that $C$ is \irr. If $C$ is \birr, then we are done. If $C$ is not \birr, then there is a finite set of compressing disks for $\bd C$ in $C$ which express $C$ as a \bd-connected sum of 3-balls and \birr\ Haken manifolds, thus yielding (2). Clearly $(2)\Rightarrow(3)$. Suppose (3) is true and $M$ is closed. If $G$ is free, then $M$ is by \cite[Theorem 5.2]{He} a connected sum of 2-sphere bundles over $S^1$, hence is not aspherical, hence $W$ is not contractible. If $G$ contains an infinite closed surface group, then by a result of Hass, Rubinstein, and Scott \cite{HRS} $W$ is \hm\ to \RRR. Clearly Theorem 1 and the fact that $M$ cannot be closed and non-orientable show that $(1)\Rightarrow(4)$. We now show that $(1)\Rightarrow(5)$. Let $C$ be the Scott compact core of $M$. Then the assumptions on $G$ imply that there is a set of compressing disks for $\bd C$ in $C$ such that some component $Q$ of $C$ split along this collection of disks satisfies the hypotheses of Lemma 2.1. Thus any component of the pre-image of $\bd Q$ is an equivariant non-trivial plane. We next show that $(5)\Rightarrow(1)$. Assume $M$ is closed. If $\pi_1(M)$ is a free product of infinite cyclic groups and infinite closed surface groups, then we apply (3) to obtain (1). If $\pi_1(M)$ is not such a group, then the existence of an equivariant plane, together with the compactness of $M$, implies that $M$ is Haken, and so (1) follows by Waldhausen \cite{Wa}. Finally we show that $(4+6)\Rightarrow(1)$. If $W$ is \rirr, then (4) implies the hypothesis of (2), hence implies (1). If $W$ is not \rirr, then (6) implies as before that $M$ is Haken, thus (1) holds. \end{proof} \section{Nice quasi-exhaustions and \RR-irreducibility} We shall reformulate a criterion due to Scott and Tucker \cite{Sc-Tu} for a \pirr\ open \tm\ to be \rirr. A proper plane $\Pi$ in an open \tm\ $W$ is \textbf{homotopically trivial} if for any compact subset $C$ of $W$ the inclusion map of $\Pi$ is end-properly homotopic to a map whose image is disjoint from $C$. \begin{lem} Let $W$ be an \irr, open \tm, and let $\Pi$ be a proper plane in $W$. If $\Pi$ is homotopically trivial, then $\Pi$ is trivial. \end{lem} \begin{proof} This is Lemma 4.1 of \cite{Sc-Tu}. \end{proof} \begin{lem} Let $W$ be a connected, \irr, open \tm, and let $\{C_n\}_{n\geq 1}$, be a sequence of compact 3-dimensional submanifolds of $W$ such that $C_n \subseteq \inte C_{n+1}$ and \begin{enumerate} \item each $C_n$ is \irr, \item each $\bd C_n$ is \inc\ in $W-\inte C_n$, \item if $D$ is a proper disk in $C_{n+1}$ which is in general position with respect to $\bd C_n$ such that $\bd D$ is not null-homotopic in $\bd C_{n+1}$, then $D\cap \bd C_n$ has at least two components which are not null-homotopic in $\bd C_n$ and bound disjoint disks in $D$. \end{enumerate} Then any proper plane in $W$ can be end-properly homotoped off $C_n$ for any $n$. \end{lem} \begin{proof} This is Lemma 4.2 of \cite{Sc-Tu}. \end{proof} The precise criterion we shall use is as follows. \begin{lem} Let $W$ be a connected, \irr, open \tm. Suppose that for each compact subset $K$ of $W$ there is a sequence $\{C_n\}_{n\geq 1}$ of compact 3-dimensional submanifolds such that $C_n \subseteq \inte C_{n+1}$ and \begin{enumerate} \item each $C_n$ is \irr, \item each $\bd C_n$ is \inc\ in $W-\inte C_n$ and has positive genus, \item each $C_{n+1}-\inte C_n$ is \irr, \birr, and anannular, \item $K \subseteq C_1$. \end{enumerate} Then $W$ is \rirr. \end{lem} \begin{proof} Let $D$ be a disk as in part (iii) of Lemma 3.2. If every component of $D \cap \bd C_n$ is null-homotopic in $\bd C_n$, then one can isotop $D$ so that $D\cap C_n=\nul$ and hence $\bd C_{n+1}$ is compressible in $C_{n+1}-\inte C_n$. If only one component $\alpha$ of $D \cap \bd C_n$ is not null-homotopic in $\bd C_n$, then $\bd D \cup \alpha$ bounds an annulus $A$ which can be isotoped so that $A \cap \bd C_n=\alpha$, hence $C_{n+1}-\inte C_n$ is not anannular. If no two of the components of $D \cap \bd C_n$ which are not null-homotopic in $\bd C_n$ bound disjoint disks in $D$, then these components must be nested on $D$. We can isotop $D$ to remove null-homotopic components and then intermediate annuli to again get an incompressible annulus joining $\bd C_{n+1}$ to $\bd C_n$. Now apply Lemma 3.2 and then Lemma 3.1. \end{proof} Let $\{C_n\}$ be a sequence of compact, connected 3-dimensional submanifolds of an \irr, open \tm\ $W$ such that $C_n \subseteq \inte C_{n+1}$ such that $W-\inte C_n$ has no compact components. This will be called a \textbf{quasi-exhaustion} for $W$. A quasi-exhaustion for $W$ whose union is $W$ is an \textbf{exhaustion} for $W$. A quasi-exhaustion is \textbf{nice} if it satisfies conditions (1)--(3) of Lemma 3.3. Thus that lemma can be rephrased by saying that if every compact subset of $W$ is contained in the first term of a nice quasi-exhaustion, then $W$ is \rirr. We shall need some tools for constructing Whitehead manifolds with nice quasi-exhaustions. Define a compact, connected \tm\ $Y$ to be \textbf{nice} it is is \pirr, \birr, and anannular, it contains a two-sided proper \inc\ surface, and it is not a 3-ball; define it to be \textbf{excellent} if, in addition, every connected, proper, incompressible surface of zero Euler characteristic in $Y$ is \bd-parallel. So in particular an excellent \tm\ is anannular and atoroidal while a nice \tm\ is anannular but may contain a non-\bd-parallel \inc\ torus. A proper 1-manifold in a compact \tm\ is \textbf{excellent} if its exterior is excellent; it is \textbf{\pe\ } if the union of each non-empty subset of the set of its components is excellent. \begin{lem} Every proper 1-manifold in a compact, connected \tm\ whose boundary contains no 2-spheres or projective planes is homotopic rel \bd\ to an excellent proper 1-manifold. \end{lem} \begin{proof} This is a special case of Theorem 1.1 of \cite{My excel}. \end{proof} Define a \textbf{$k$-tangle} to be a disjoint union of $k$ proper arcs in a 3-ball. \begin{lem} For all $k\geq 1$ \pe\ $k$-tangles exist. \end{lem} \begin{proof} This is Theorem 6.3 of \cite{My attach}. \end{proof} We shall also need the following criterion for gluing together excellent 3-manifolds to get an excellent 3-manifold. \begin{lem} Let $Y$ be a compact, connected \tm. Let $S$ be a compact, proper, two-sided surface in $Y$. Let $Y\p$ be the \tm\ obtained by splitting $Y$ along $S$. Let $S\p$ and $S\pp$ be the two copies of $S$ which are identified to obtain $Y$. If each component of $Y\p$ is excellent, $S\p$, $S\pp$, and $(\bd Y\p)-\inte (S\p \cup S\pp)$ are \inc\ in $Y\p$, and each component of $S$ has negative Euler characteristic, then $Y$ is excellent. \end{lem} \begin{proof} This is Lemma 2.1 of \cite{My excel}. \end{proof} \section{The infinite cyclic case} In \cite{Sc-Tu} Scott and Tucker described an \rirr\ Whitehead manifold which is an infinite cyclic covering space. This section gives a general procedure for constructing such examples. The construction introduced here will be incorporated into that for the general case in section 6. Let $\Pn=\Dn \times [0,1]$, where \Dn\ is the disk of radius $n$. We call \Pn\ a \textbf{pillbox}. Identify $\Dn \times \{0\}$ with $\Dn \times \{1\}$ to obtain a solid torus \Qn. Let \Rn\ be a solid torus and \Hn\ a 1-handle $D \times [0,1]$ joining $\bd \Dn \times (0,1)$ to $\bd \Rn$. Let $\Vn=\Pn \cup \Hn \cup \Rn$ and $\Mn=\Qn \cup \Hn \cup \Rn$. We call \Vn\ an \textbf{eyebolt}. We embed \Mn\ in the interior of \Mnn\ as follows. We choose a collection of arcs \thz, \tho, \alz, \alo, \alt, \beo, \bet, \gao, \gat, \deo, \detw, \ep\ in \Pnn which satisfy certain conditions described below. \thz, \tho, and \alz\ meet in a common endpoint in $\inte \Pnn$ but are otherwise disjoint. The other endpoints of \thz\ and \alz\ lie in $(\inte \Dnn) \times \{0\}$; that of \tho\ lies in $(\inte \Dnn) \times \{1\}$. We let $\tht=\thz \cup \tho$. All the other arcs are proper arcs in \Pnn which are disjoint from each other and from $\tht \cup \alz$. \gao, \bet, and \detw\ run from $(\inte \Dnn)\times \{0\}$ to itself. \gat, \beo, and \deo run from $(\inte \Dnn)\times \{1\}$ to itself. \alo\ runs from $(\inte \Dnn) \times \{0\}$ to $(\inte \Dnn) \times \{1\}$. \alt\ runs from $(\inte \Dnn) \times \{1\}$ to $\inte (\Pnn \cap \Hnn)$. \ep\ runs from $\inte (\Pnn \cap \Hnn)$ to itself. We denote the image in \Qnn\ of an arc by the same symbol, relying on the context to distinguish an arc in \Pnn\ from its image in \Qnn. We require that \tht\ be a simple closed curve in \Qnn\ and that $\alz \cup \beo \cup \gao \cup \deo \cup \alo \cup \bet \cup \gat \cup \detw$ is an arc consisting of subarcs which occur in the given order. We require that any union of these arcs which contains \alz\ and at least one other arc has excellent exterior in \Pnn, and that the same is true for any union of these arcs which contains neither \thz, \tho, nor \alz. This can be achieved as follows. Note that the exterior of \alz\ in \Pnn\ is a 3-ball $B$. Choose a poly-excellent 11-tangle in $B$ and then slide its endpoints so that exactly two of the arcs meet a regular neighborhood of \alz. Extend them to meet \alz\ in the desired configuration. Next let \kao, \kat, and \kad\ be product arcs in \Hnn\ joining $(\inte D) \times \{0\}$ to $(\inte D) \times \{1\}$. Let $\Rn \subseteq \inte \Rnn$ be any null-homotopic embedding. Let \lao\ and \lat\ be disjoint proper arcs in $\Rnn-\inte \Rn$ with \lao\ joining $\inte (\Hnn \cap \Rnn)$ to itself and \lat\ joining $\inte (\Hnn \cap \Rnn)$ to $\bd \Rn$. We require $\lao \cup \lat$ to be excellent in $\Rnn-\inte \Rn$. We also require that these arcs, together with \ep, fit into an arc whose subarcs form the sequence \kao, \lao, \kat, \ep, \kad, \lat\ and that \kao\ meets \alt\ in a common endpoint. Now we embed \Pn\ in \Pnn\ as a regular neighborhood of the arc \tht\ so that the two disks of $\Pn \cap (\Dnn \times \{0,1\})$ are identified to give an embedding of \Qn\ in \Qnn. Note that these embeddings are not consistent with the product structures. From the discussion above we have an arc \om\ in $\Mnn-\inte(\Qn \cup \Rn)$ running from $\bd \Qn$ to $\bd \Rn$. We embed \Hn\ as a regular neighborhood of \om. We change notation slightly by now letting \alz\ be the old \alz\ minus its intersection with the interior of \Qn. We let $M$ be the direct limit of the \Mn\ and let $p:\Mt \rightarrow M$ be the universal covering map. $p\n(\Qn)=p\n(\Pn)$ is the union of pillboxes $\Pnj=\Dn \times [j,j+1]$ meeting along the $\Dn \times \{j\}$ to form $\Dn \times \R$. Note that this embedding is not the product embedding. $p\n(\Rn)$ is a disjoint union of solid tori \Rnj. $p\n(\Hn)$ is a disjoint union of 1-handles \Hnj\ joining $\bd \Dn \times (j,j+1)$ to $\bd \Rnj$; these are regular neighborhoods of lifts \omj\ of \om. $p\n(\Mn)= p\n(\Vn)$ is the union of $p\n(\Pn)$, $p\n(\Hn)$, and $p\n(\Rn)$. It is the union of eyebolts $\Vnj=\Pnj \cup \Hnj \cup \Rnj$ meeting along the $\Dn \times \{j\}$. \Mt\ is the nested union of the $p\n(\Mn)$. Let $\Simn=\cup^m_{j=-m} \Vnj$ and $\Lamn=P_{n,-(m+1)} \cup P_{n,m+1}$. Let $\Phi^m_1=\nul$, and, for $n\geq 2$, let $\Phmn=\cup^{m+n}_{j=m+2} (P_{n,-j} \cup \Pnj)$. Note that \Lamn\ and \Phmn\ (for $n\geq 2$) are each disjoint unions of two 3-balls, $\Lamn \cap \Simn$ is a pair of disjoint disks, and (for $n\geq 2$) so is $\Lamn \cap \Phmn$. Define $\Cmn= \Simn \cup \Lamn \cup \Phmn$. \begin{lem} $\{C^m_m\}$ is an exhaustion for \Mt. Each $C^m$ is a nice quasi-exhaustion. \end{lem} \begin{proof} $\Cmn \subseteq \inte \Cmnn$, and $\Cmn \subseteq C^{m+1}_n$. A given compact subset $K$ of \Mt\ lies in some $p\n(\Mn)$ and thus in a finite union of \Vnj\ and hence in some $\Simn \subseteq \Cmn \subseteq C^q_q$, where $q=\max\{m,n\}$. Thus $\{C^m_m\}$ is an exhaustion for \Mt. \Cmn\ is a cube with $2m+1$ handles. Let $Y=\Cmnn-\inte \Cmn$. We will show that $Y$ is excellent by successive applications of Lemma 3.6. Consider a \Pnnj\ contained in \Cmnn. If $|j|m} V_{n+q,j}$. We have that $C^- \cap C^m_{n+q}$ and $C^+ \cap C^m_{n+q}$ are disjoint disks, while $C^- \cap C^+=\nul$. It follows that $\bd \Cmn$ is \inc\ in $p\n(M_{n+q})-\inte \Cmn$. Since \Mt\ is the nested union of the $p\n(M_{n+q})$ over all $q\geq 1$ we have the desired result. \end{proof} \section{The surface group case} Let $F$ be a closed, connected surface other than $S^2$ or \PP. Let $n\geq 1$. Regard $F$ as being obtained from a $2k$-gon $E$, $k\geq 2$, by identifying sides $s_i$ and $s_i\p$, $1\leq i \leq k$. This induces an identification of the lateral sides $S_i=s_i \times [-n,n]$ and $S_i\p=s_i \times [-n,n]$ of the \textbf{prism} $\Pn=E \times [-n,n]$ which yields $\Qn=F \times [-n,n]$. Let $\Rn$ be a solid torus and $\Hn$ a 1-handle $D \times [0,1]$. Let $\Vn=\Pn \cup \Hn \cup \Rn$, where $\Pn \cap \Rn=D \times \{1\}$ is a disk in $\bd \Rn$, and $\Pn \cap \Rn=\nul$. We again call \Vn\ an \textbf{eyebolt}. It is a solid torus whose image under the identification is $\Mn=\Qn \cup \Hn \cup \Rn$, a space \hm\ to the \bd-connected sum of $F\times [-n,n]$ and a solid torus. We define an open \tm\ $M$ by specifying an embedding of \Mn\ in the interior of \Mnn\ and letting $M$ be the direct limit. The inclusion $[-n,n] \subseteq [-(n+1),n+1]$ induces $\Pn \subseteq \Pnn$ and hence $\Qn \subseteq \Qnn$. We let $\Rn \subseteq \inte \Rnn$ be any null-homotopic embedding. Again the interesting part of the embedding will be that of \Hn\ in \Mnn. It will be the regular neighborhood of a certain arc \om\ in $\Mnn-\inte(\Qn \cup \Rn)$ joining $\bd \Qn$ to $\bd \Rn$. The arc \om\ is the union of $4k+7$ arcs any two of which are either disjoint or have one common endpoint. The $4k+2$ arcs \alz, \ali, \bei, \gai, \dei, $1\leq i \leq k$, and \ep\ lie in $E \times [n,n+1]$ and are identified with their images in \Qnn; the three arcs \kao, \kat, and \kad\ lie in \Hnn, and the two arcs \lao\ and \lat\ lie in \Rnn. These arcs will have special properties to be described later. We first describe their combinatorics. The arcs in \Pnn\ are all proper arcs in $E \times [n,n+1]$. \alz\ runs from $(\inte E) \times \{n\}$ to $\inte S_1$. For $1 \leq i < k$, \ali\ runs from $\inte S_i$ to $\inte S_{i+1}$. \alk\ runs from $\inte S_k\p$ to $\inte (\Pnn \cap \Hnn)$. For $1 \leq i \leq k$, \bei\ and \dei\ each run from $\inte S_i\p$ to itself, while \gai\ runs from $\inte S_i$ to itself. These arcs are chosen so that under the identification their endpoints match up in such a way as to give an arc which follows the sequence \alz, \beo, \gao, \deo, \alo, \ldots, \bek, \gak, \dek, \alk. We require \ep\ to run from $\inte(\Pnn \cap \Hnn)$ to itself. \kao, \kat, and \kad\ are product arcs in \Hnn\ lying in $(\inte D) \times [0,1]$. \lao\ and \lat\ are proper arcs in $\Rnn-\inte \Rn$, with \lao\ running from $\inte (\Hnn \cap \Rnn)$ to itself and \lat\ running from $\inte(\Hnn \cap \Rnn)$ to $\bd \Rn$. These arcs are chosen so as to fit together into the sequence \kao, \lao, \kat, \ep, \kad, \lat\ with the endpoint of \kao\ other than $\kao \cap \lao$ being the same as the endpoint of \alk\ other than $\alk \cap \dek$. This gives \om. We now describe the special properties required of these arcs. We require that $\alz \cup \beo \cup \gao \cup \deo \cup \alo \cup \cdots \cup \bek \cup \gak \cup \dek \cup \ep$ be a \pe\ $(4k+2)$-tangle in $E \times [n,n+1]$ and $\lao \cup \lat$ to be an excellent 1-manifold in $\Rnn-\inte \Rn$. We now consider the universal covering map $p:\Mt \rightarrow M$. Our goal is to construct a sequence $\{C^m\}$ of nice quasi-exhaustions whose diagonal $\{C^m_m\}$ is an exhaustion for \Mt. The universal covering space \Ft\ of $F$ is tesselated by copies \Ej\ of $E$. We fix one such copy $E_1$. We inductively define an exhaustion $\{\Fm\}$ for \Ft\ as follows. $F_1=E_1$. \Fmm\ is the union of \Fm\ and all those \Ej\ which meet it. Each \Fm\ is a disk (which we call a \textbf{star}). The \textbf{inner corona} \Icm\ of \Fm\ is the annulus $\Fmm-\inte \Fm$. Each vertex on $\bd \Fm$ lies in either one or two of those \Ej\ contained in \Fm. Each \Ej\ in \Icm\ meets meets \Fm\ in either an edge or a vertex; in both cases it meets exactly two adjacent $E_\ell$ of $\Icm$, and each of these intersections is an edge. For $n\geq 2$ we define the \textbf{outer $n$-corona} \Omn\ to be the annulus $F_{m+n}-\inte \Fmm$; we define $O^m_1=\nul$. Let $\sigma_2$ be a proper arc in $F_2$ consisting of three edges of the polygons in $F_2$. . Inductively define a proper arc $\sigma_{m+1}$ in \Fmm\ by adjoining to $\sigma_m$ two arcs spanning $\Icm$ which are edges of polygons in \Icm. Thus each $\sigma_m$ is an edge path in \Fm\ splitting it into two unions of polygons \Fmp\ and \Fmpp. We now consider the structure of \Mt. For $n\geq 1$, $p\n(\Qn)=p\n(\Pn)$ is the union of prisms $\Pnj=\Ej \times [-n,n]$ meeting along their lateral sides to form $\Ft \times [-n,n]$. $p\n(\Rn)$ is a disjoint union of solid tori \Rnj. $p\n(\Hn)$ is a disjoint union of 1-handles \Hnj\ running from $\Ej\times \{n\}$ to $\bd \Rnj$; these are regular neighborhoods of lifts \omj\ of \om. Now $p\n(\Mn)=p\n(\Vn)$ is the union of $p\n(\Pn)$, $p\n(\Hn)$, and $p\n(\Rn)$. It can be expressed as the union of the eyebolts $\Vnj= \Pnj \cup \Hnj \cup \Rnj$ meeting along the lateral sides of the \Pnj. Finally \Mt\ is the nested union of the $p\n(\Mn)$. Let \Simn\ be the union of those \Vnj\ such that \Ej\ is in the star \Fm. Let \Lamn\ be the union of those \Pnj\ such that \Ej\ is in the inner corona \Icm. Let \Phmn\ be the union of those \Pnj\ such that \Ej\ is in the outer $n$-corona \Omn. Note that \Lamn\ and \Phmn\ (for $n\geq 2$) are solid tori, $\Lamn \cap \Simn$ is an annulus which goes around $\Lamn$ once longitudinally and consists of those lateral sides of the prisms in \Simn\ which lie on $\bd \Simn$, and (for $n \geq 2$) $\Lamn \cap \Phmn$ is an annulus which goes around each of these solid tori once longitudinally. We now define $\Cmn=\Simn \cup \Lamn \cup \Phmn$. \begin{lem} $\{C^m_m\}$ is an exhaustion for \Mt. Each $C^m$ is a nice quasi-exhaustion. \end{lem} \begin{proof} Note that $\Cmn \subseteq \inte \Cmnn$, and $\Cmn \subseteq \Cmmn$. Suppose $K$ is some compact subset of \Mt. Then $K$ lies in some $p\n(\Mn)$ and thus in a finite union of \Vnj\ and hence in some $\Simn \subseteq \Cmn \subseteq C^q_q$, where $q=\max\{m,n\}$. Thus $\{C^m_m\}$ is an exhaustion for \Mt. Each \Cmn\ is a cube with handles, so is \irr. The number of handles is at least one, so $\bd \Cmn$ has positive genus. Let $Y=\Cmnn-\inte \Cmn$. We will prove that $Y$ is excellent by successive applications of Lemma 3.6. Let $\Pnnjp$ and $P^-_{n+1,j}$ denote, respectively, $\Ej \times [n,n+1]$ and $\Ej \times [-(n+1),-n]$. Consider a \Pnnj\ contained in \Simnn. It meets \Cmn\ in \Pnnj\ together with regular neighborhoods of certain arcs in \Pnnjp. These arcs consist at least of the $j^{th}$ copies of the \ali, the \gai, and \ep\ which are part of the lift \omj\ of \om. If another prism $P_{n+1,\ell}$ in \Simnn\ meets \Pnnj\ in a common lateral side, then either \omj\ or $\om_{\ell}$ will meet this side; in the latter case this contributes a \bei\ and \dei\ to the subsystem of arcs in \Pnnjp. Since the full system of arcs was chosen to be \pe\ this subsystem of arcs is excellent and so has excellent exterior $Y \cap \Pnnjp$. Let $U\p$ be the union of those $Y\cap \Pnnjp$ such that $\Ej \subseteq \Fmp$. This space can be built up inductively by gluing on one $Y \cap \Pnnjp$ at a time, with the gluing being done along either a disk with two holes (when \Pnnj\ is glued along one lateral side) or a disk with four holes (when \Pnnj\ is glued along two adjacent lateral sides). No component of the complement of this surface in the boundary of either manifold is a disk, hence this surface is \inc\ in each manifold. It follows that $U\p$ is excellent. Similar remarks apply to the space $U\pp$ associated with $\Fmpp$. Next consider a \Pnnjp\ contained in \Lamnn. If $\Ej \subseteq \Fmm$ and meets \Fmp\ in an edge of $E_{\ell} \subseteq \Fmp$, then either $\om_{\ell}$ misses \Pnnjp\ or meets it in copies of \bei\ and \dei. Thus enlarging $U\p$ by adding $Y \cap \Pnnjp$ either adds a 3-ball along a disk in its boundary, giving a space \hm\ to $U\p$ or gives a new excellent \tm. We adjoin all such $Y\cap \Pnnjp$ to $U\p$. Then we consider those $\Ej$ which meet \Fmp\ in a vertex. Then $\Pnnjp=Y\cap\Pnnjp$, and one can successively adjoin these 3-balls along disks in their boundaries. We denote the enlargement of $U\p$ from all these additions again by $U\p$. Similar remarks apply to $U\pp$. Now $(F\p_{m+n+1}-\inte \Fmmp) \times [n,n+1]$ is a 3-ball which meets $U\p$ in a disk, so we adjoin it to $U\p$ to get a new $U\p$ \hm\ to the old one. We then adjoin the 3-ball $(F\p_{m+n+1}-\inte F\p_{m+n}) \times [-n,n]$ which meets this space along a disk to obtain our final $U\p$. The same construction gives $U\pp$. Now $U\p$ and $U\pp$ are each excellent. $U\p \cap U\pp$ is an annulus with a positive number of disks removed from its interior corresponding to its intersection with arcs passing from $\Fmp \times [n,n+1]$ to $\Fmpp \times [n,n+1]$. No component of the complement of this surface in $\bd U\p$ or in $\bd U\pp$ is a disk; this corresponds to the fact that $\Fmp \times \{n\}$, $\Fmpp \times \{n\}$, $\Fmp \times \{n+1\}$, and $\Fmpp \times \{n+1\}$ each meet some \omj. Thus this surface is \inc\ in both $U\p$ and $U\pp$, so $U\p \cup U\pp$ is excellent. Finally we add on the $Y\cap\Rnj$ for $\Ej \subseteq \Fm$ to $U\p \cup U\pp$ to conclude that $Y$ is excellent. It remains to show that each $\bd \Cmn$ is \inc\ in $\Mt-\inte \Cmn$. First note that since each $C^m_{n+s+1}-\inte C^m_{n+s}$ is \birr\ we must have that $\bd \Cmn$ is \inc\ in $C^m_{n+q}-\inte \Cmn$ for each $q\geq 1$. Now consider the set \[ \widetilde{M}_{n+q}=p\n(M_{n+q}) \cup (\Ft \times [-(n+q+1),-(n+q)]). \] It can be obtained from $C^m_{n+q}$ as follows. First add the solid tori $R_{n+1,j} \cup H_{n+q,j}$ in $p\n(M_{n+q})$ for which $\Ej \subseteq F_{m+q+n}$; these meet $C^m_{n+q}$ in disks. Then add \[ (F_{m+q+n} \times [-(n+q+1),-(n+q)]) \cup (\Ft-\inte F_{m+q+n})\times[-(n+q+1),n+q]). \] This is a space \hm\ to \prodspace\ which meets $C^m_{n+q}$ in the disk \[ (F_{m+q+n} \times \{-(n+q)\})\cup((\bd F_{m+q+n})\times[-(n+q),n+q]). \] Lastly add all the remaining solid tori $R_{n+q,j} \cup H_{n+q,j}$, where $\Ej \subseteq \Ft-\inte F_{m+q+n}$; these do not meet $C^m_{n+q}$. This description shows that $C^m_{n+q} \cap (\Mt_{n+q}-\inte C^m_{n+q})$ consists of (finitely many) disjoint disks, and therefore $\bd \Cmn$ is \inc\ in $\Mt_{n+q}-\inte \Cmn$. Finally since \Mt\ is the nested union of the $\Mt_{n+q}$ over all $q\geq 1$ we have that $\bd \Cmn$ is \inc\ in $\Mt-\inte \Cmn$. \end{proof} \section{The general case} Suppose $G_1$, $\ldots$, $G_k$ are infinite cyclic groups and infinite closed surface groups. For $i=1, \ldots, k$ let \Pin\ be a pillbox or prism, as appropriate, with quotient \Qin\ a solid torus or product $I$-bundle over a closed surface, respectively. We let \Hin\ be a 1-handle attached to \Pin\ as before. We let \Rn\ be a common solid torus to which we attach the other ends of all the \Hin. The union of the \Qin\ and \Hin\ with \Rn\ is called \Mn. As before we choose arcs in the \Pin, \Hin, and \Rn\ and use them to define an embedding of \Mn\ into the interior of \Mnn. The choice of arcs in $\Rnn-\inte \Rn$, as well as the embedding $\Rn \subseteq \inte \Rnn$, requires some discussion, since we will want this family $\lambda$ of arcs to be \pe. Choose a \pe\ $(2k+2)$-tangle $\lambda^+$ in a 3-ball $B$, with components \lait, $1\leq i\leq k+1$, $t=1,2$. Construct a graph in $B$ by sliding one endpoint of each $\lambda^i_2$, $1\leq i \leq k$, so that it lies on $\inte \lambda^{k+1}_2$. Thus these $\lambda^i_2$ now join $\bd B$ to distinct points on $\inte \lambda^{k+1}_2$; all the other \lait\ still join $\bd B$ to itself. Now choose disjoint disks $E_1$ and $E_2$ in $\bd B$ such that $E_t$ meets the graph in $\bd \lambda^{k+1}_t \cap \inte E_t$. Glue $E_1$ to $E_2$ so that $B$ becomes a solid torus \Rnn\ and $\lambda^{k+1}_1 \cup \lambda^{k+1}_2$ becomes a simple closed curve. The regular neighborhood of this simple closed curve is our embedding of \Rn\ in the interior of \Rnn. Clearly \Rn\ is null-homotopic in \Rnn. By Lemma 3.6 its exterior is excellent as is the exterior of the union of \Rn\ with any of the \lait, $1\leq i\leq k$, $t=1,2$. Let $p:\Mt\rightarrow M$ be the universal covering map. Then $p\n(\Rn)$ consists of disjoint solid tori whose union separates $p\n(\Mn)$ into components with closures \Limun, where \Limun\ is a component of $p\n(\Qin \cup \Hin)$. Let \Zimun\ be the union of \Limun\ and all those components of $p\n(\Rn)$ which meet it. Then $\Zimu=\cup_{n\geq 1}\Zimun$ is an open subset of \Mt\ which has a family $\{\Cimum\}$ of quasi-exhaustions as previously described. We will develop from these families an appropriate family $\{C^m\}$ of quasi-exhaustions of $\Mt$. We start by choosing a component $\widehat{R}_1$ of $p\n(R_1)$. For each $n$ there is then a unique component $\widehat{R}_n$ of $p\n(\Rn)$ which contains $\widehat{R}_1$. We define $C^1_n$ to be the union of $\widehat{R}_n$ and the (finitely many) $C^{i,\mu,1}_n$ which contain it; this space is a solid torus which meets each of these $C^{i,\mu,1}_n$ in a 3-ball which is either a pillbox and a 1-handle or a prism and a 1-handle. Suppose \Cmn\ has been defined and that it is the union of the \Cimumn\ for which $\Cmn \cap \Limun \neq \nul$. We define \Cmmn\ in two steps. We first take the union $C\p$ of all the $C^{i,\mu,m}_n$ such that $\Cimumn \subseteq \Cmn$. This is just the union of the $n^{th}$ elements of the $(m+1)^{st}$ quasi-exhaustions for those \Zimun\ such that $\{i,\mu\}$ is in the current index set. The second step is to enlarge the index set by adding those $\{i,\nu\}$ for which $C\p \cap L^{i,\nu}_n \neq \nul$ and then adjoin the $C^{i,\nu,m+1}_n$ to $C\p$ in order to obtain \Cmmn. One can observe that the \Limun\ and $p\n(\Rn)$ give $p\n(\Mn)$ a tree-like structure and that the passage from \Cmn\ to \Cmmn\ goes out further along this tree. \begin{lem} $\{C^m_m\}$ is an exhaustion for \Mt. $C^m$ is a nice quasi-exhaustion. \end{lem} \begin{proof} Again we have $\Cmn \subseteq \inte \Cmnn$ and $\Cmn \subseteq \Cmmn$ with the result that $\{C^m_m\}$ is an exhaustion for \Mt. As regards the excellence of $\Cmnn-\inte \Cmn$ we note that the only thing new takes place in those components of $p\n(\Rnn)$ contained in \Cmnn. Instead of two arcs $\lambda_1$ and $\lambda_2$ as before we have $\lambda^i_1$ and $\lambda^i_2$ as $i$ ranges over some non-empty subset of $\{1, \ldots, k\}$ We then apply the poly-excellence of the full set of \lait. The incompressibility of $\bd \Cmn$ in $\Mt-\inte \Cmn$ follows as before. We first note that $\bd \Cmn$ is \inc\ in $C^m_{n+q}-\inte \Cmn$ for each $q\geq 1$. Now define $\Mt_{n+q}$ to be the union of $p\n(M_{n+q})$ and, for each of the surface group factors $G_i$ of $G$, the copy $\Ft^{i,\mu} \times [-(n+q+1),-(n+q)]$ of $\Ft^i \times [-(n+q+1),-(n+q)]$ contained in \Zimu, where $\Ft^i$ is the universal covering space of the surface $F^i$ with $\pi_1(F^i)\cong G_i$. Then the exterior of $C^m_{n+q}$ in $\Mt_{n+q}$ meets it in a collection of disjoint disks, from which it follows that $\bd \Cmn$ is \inc\ in $\Mt_{n+q}-\inte \Cmn$, thus is \inc\ in $\Mt-\inte \Cmn$. \end{proof} \section{Uncountably many examples} We now describe how to get uncountably many examples for a given group $G$. We will use a trick introduced in \cite{My attach}. Let $\{X_{n,s}\}$ be a family of exteriors of non-trivial knots in $S^3$ indexed by $n\geq 2$ and $s\in \{0,1\}$; they are chosen to be anannular, atoroidal, and pairwise non-\hm. (One such family is that of non-trivial, non-trefoil twist knots.) One chooses a function $\varphi(n)$ with values in $\{0,1\}$, i.e. a sequence of 0's and 1's indexed by $n$, and constructs a \tm\ $M[\varphi]$ by embedding $X_{n,\varphi(n)}$ in $\Mn-\inte \Mnnn$ so that $\bd X_{n,\varphi(n)}$ in \inc\ in $\Mn-\inte \Mnnn$ (but is compressible in \Mn). The idea is to do this in such a way that for ``large'' compact sets $C$ in \Mtphi\ one has components of $p\n(X_{n,\varphi(n)})$ which lie in $\Mt-C$ and have \inc\ boundary in $\Mt-C$ for ``large'' values of $n$; moreover, every knot exterior having these properties should be \hm\ to some $X_{n,\varphi(n)}$. Thus if \Mtphi\ and \Mtpsi\ are \hm\ one must have $\varphi(n)=\psi(n)$ for ``large'' $n$. One then notes that there are uncountably many functions which are pairwise inequivalent under this relation. We proceed to the details. First assume $\varphi$ is fixed, so we can write $s=\varphi(n)$. The most innocuous place to embed $\Xns$ is in $\Rn-\inte \Rnnn$ since this space is common to all our constructions. Recall that this space contains arcs $\lambda_1$, $\lambda_2$ or, if $G$ is a non-trivial free product, arcs $\lambda^i_1$, $\lambda^i_2$, $1\leq i\leq k$; call this collection of arcs $\lambda$. We wish \Xns\ to lie in the complement of $\lambda$ in such a way that it is \pe\ in $\Rn-\inte(\Rnnn \cup \Xns)$. We revise the construction of $\lambda$ from section 6 as follows. Let $B_0$ and $B_1$ be 3-balls. Choose disjoint disks $D_r$ and $D_r\p$ in $\bd B_r$. Let $\zeta_r$ be a simple closed curve in $\bd B_r-(D_r \cup D_r\p)$ which separates $D_r$ from $D_r\p$. Let $A_r$ and $A_r\p$ be the annuli into which $\zeta_r$ splits the annulus $\bd B_r-\inte (D_r \cup D_r\p)$, with the notation chosen so that $A_r \cap D_r=\nul$. Let $\tau_r$ be a poly-excellent $(4k+4)$-tangle in $B_r$ which is the union of $(2k+2)$-tangles $\rho_r$ and $\rho_r\p$ satisfying the following conditions. Each component of $\rho_0$ runs from $\inte D_0$ to $\inte A_0\p$. Each component of $\rho_1$ runs from $\inte A_1\p$ to $\inte D_1\p$. Each component of $\rho_0\p$ runs from $\inte D_0\p$ to itself. Each component of $\rho_1\p$ runs from $\inte D_1\p$ to $\inte D_1$. We then glue $A_0\p$ to $A_1\p$ and $D_0\p$ to $D_1\p$ so as to obtain a space \hm\ to a 3-ball minus the interior of an unknotted solid torus contained in the interior of the 3-ball. The 2-sphere boundary component is $D_0 \cup D_1$; the torus boundary component is $A_0 \cup A_1$. The gluing is done so that the endpoints of the arcs match up to give a system $\lambda^+$ of $2k+2$ arcs. Each arc in this system consists of an arc of $\rho_0$ followed by an arc of $\rho_1$ followed by an arc of $\rho_0\p$ followed by an arc of $\rho_1\p$. We then glue \Xns\ to this space along their torus boundaries so as to obtain a 3-ball $B$. We then apply the construction of section 6 to $\lambda^+$ to get a \pe\ system $\lambda$ of arcs in $\Rn-\inte \Rnnn$. It is easily seen that this \tm\ is nice and that $\bd \Xns$ is, up to isotopy, the unique \inc\ non-\bd-parallel torus in it; $\bd \Xns$ is also, up to isotopy, the unique \inc\ torus in the exterior \Ks\ of any non-empty union $\sigma$ of components of $\lambda$. \begin{lem} If \Mtphi\ and \Mtpsi\ are \hm\ then there is an index $N$ such that $\varphi(n)=\psi(n)$ for all $n\geq N$. \end{lem} \begin{proof} Consider \Mt. $Y=\Cmn-\inte \Cmnnn$ contains copies of \Ks\ for various choices of $\sigma$. The closure of the complement in $Y$ of these copies consists of excellent 3-manifolds which meet the copies along \inc\ planar surfaces. It follows that the various copies of $\bd \Xns$ in $Y$ are, up to isotopy and for $n\ge3$, the unique \inc\ tori in $Y$. The incompressibility of $\bd \Cmn$ in $\Mt-\inte \Cmn$ implies that these tori are also \inc\ in $\Mt-\inte \Cmnnn$. Suppose $T$ is an \inc\ torus in $\Mt-\inte \Cmnnn$. Then $T$ lies in $\Mt_{n+q}$ for some $q \geq 0$. The exterior of $C^m_{n+q}$ in $\Mt_{n+q}$ consists of copies of $D \times \R$ and \prodspace\ to which disjoint 1-handles have been attached. It meets $C^m_{n+q}$ in a set of disjoint disks. It follows that $T$ can be isotoped into $C^m_{n+q}-\inte \Cmnnn$. Since $\bd C^m_{n+u}$ for $1\leq u\ell$. Since the knot exteriors are pairwise non-\hm\ we must have $n=v$ and $\varphi(n)=\psi(v)= \psi(n)$. \end{proof} \begin{thebibliography}{99} \bibitem{HRS} J. Hass, H. Rubinstein, and P. 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