%this is a LaTeX file of ``A Standard Form for Incompressible Surfaces in a % Handlebody'' by Linda Green \documentclass[12pt]{amsart} \textheight=574pt \textwidth=432pt \oddsidemargin=18pt \evensidemargin=18pt \topmargin=14pt \usepackage{epsf} \newcommand{\pfpf}[1]{\medskip\par\noindent{\bf {#1}}} \newcommand{\clcl}[1]{\medskip\par\noindent{\it {#1}:}} \newcommand{\pfpfend}{$\bullet$ \medskip} \newcommand{\caca}[1]{\medskip\par\noindent{\bf {#1}:}} \newcommand{\stst}[1]{\medskip\par\noindent{\bf {#1}:}} \newcommand{\defglo}[1]{{\em {#1}}} \newcommand{\mvmv}[1]{\medskip\par\noindent{\textsc {{#1}}:}} \newcommand{\de}{\partial} \newcommand{\sm}{\setminus} \newcommand{\interior}{\mbox{int}} \newcommand{\dv}{\de_v} \newcommand{\F}{\Sigma} \newcommand{\X}{X} \newcommand{\Si}{\Sigma} \newcommand{\Z}{Z} \newcommand{\R}{I \! \! R} \newtheorem{thm}{Theorem} \newtheorem{fact}{Fact} \begin{document} \title{A Standard Form for Incompressible Surfaces in a Handlebody} \author{Linda Green} \address{MSRI \\ 1000 Centennial Drive \\ Berkeley \\ CA 94720} \email{lindag@msri.org} \thanks{Research at MSRI is supported in part by NSF grant DMS-9022140} \begin{abstract} Let $\F$ be a compact surface and let $I$ be the unit interval. This paper gives a standard form for all 2-sided incompressible surfaces in the 3-manifold $\F \times I$. Since $\F \times I$ is a handlebody when $\F$ has boundary, this standard form applies to incompressible surfaces in a handlebody. \end{abstract} \maketitle \section{Introduction} Let $M$ be a $3$-dimensional manifold and let $X \subset M$ be a properly embedded surface. A \defglo{compression disk} for $\X \subset M$ is an embedded disk $D \subset M$ such that $\de D \subset \X$, $\interior(D) \subset (M \sm \X)$, and $\de D$ is an essential loop in $\X$. The surface $\X \subset M$ is \defglo{incompressible} if there are no compression disks for $\X \subset M$ and no component of $\X$ is a sphere that bounds a ball. If $\X \subset M$ is connected and 2-sided, then $\X$ is incompressible if and only if the induced map $\pi_1(\X) \rightarrow \pi_1(M)$ is injective and $\X$ is not a sphere that bounds a ball. See, for example, \cite[Chapter 6]{Hempel}. Let $\F$ be a compact surface and let $I$ be the unit interval $[0, 1]$. The manifold $\Sigma \times I$ is foliated by copies of $I$, which can be thought of as vertical flow lines. This paper shows that every properly embedded, 2-sided, incompressible surface in $\F \times I$ can be isotoped to a standard form, called ``near-horizontal position''. A surface in near-horizontal position is transverse to the flow on $\Sigma \times I$ in the $I$ direction, except at isolated intervals, where it coincides with flow lines. Near each of these intervals, $\F$ looks like a tiny piece of a helix, with the interval as its core. A surface in near-horizontal position can be described combinatorially by listing its boundary curves and the number of times it crosses each line in a certain finite collection of flow lines. When $\Sigma$ is a compact surface with boundary, then $\Sigma \times I$ is a handlebody. So this paper applies, in particular, to incompressible surfaces in a handlebody. \section{Notation} Throughout this paper, if $E$ is a topological space, then $|E|$ denotes the number of components of $E$. The symbol $I$ denotes the unit interval $[0, 1]$. If $M$ is a manifold, then $\de M$ refers to the boundary of $M$ and $\interior(M)$ refers to its interior. The symbol $\Sigma$ refers to a compact surface, and $p$ denotes the projection map $\Sigma \times I \rightarrow \Sigma$. The surface $\X \subset M$ is a proper embedding if $\interior(\X) \subset \interior(M)$, $\de \X \subset \de M$, and the intersection of $X$ with a compact subset of $M$ is a compact subset of $X$. The map $G: \X \times I \rightarrow M$ is a proper isotopy between $G|_{\X \times 0}$ and $G|_{\X \times 1}$ if for all $t \in I$, $G|_{\X \times t}$ is a proper embedding. Unless otherwise stated, all surfaces contained in 3-manifolds are properly embedded and all isotopies are proper isotopies. A connected surface $\X \subset M$ is \defglo{boundary parallel} if $\X$ separates $M$ and there is a component $K$ of $M \sm \X$ such that $(K, \X)$ is homeomorphic to $(\X \times I, \X \times 0)$. \section{Definition of near-horizontal position} Let $\X$ be a surface in $\F \times I$. Let $C \subset \F$ be the union of loops and arcs $p(\X \cap (\F \times 1))$, let $C' = p(\X \cap (\F \times 0))$, and let $B = \X \cap (\de \F \times I)$. $\X$ is in \defglo{near-horizontal position} if \begin{enumerate} \item $C$ and $C'$ intersect transversely, \item $p|_B: B \rightarrow \de \F$ is injective, \item $p |_{\X \sm p^{-1}(C \cap C')}$ is a local homeomorphism, and \item for any point $z \in {C \cap C'}$, there is a neighborhood $U \subset \F$ of $z$ such that $p^{-1}(U) \cap \X$ either looks like the region \par\noindent % % \begin{figure}[h!tbp] \epsfysize=1.0in \centerline{ \epsfbox{neartwist.eps}} \end{figure} % % \par\noindent or else looks like a union of regions of the form % % \begin{figure}[h!tbp] \epsfysize=0.7in \centerline{ \epsfbox{nearother.eps}} \end{figure} % \end{enumerate} \par\noindent Here $p^{-1}(C) \cap \X$ is drawn with thick lines and $p^{-1}(C') \cap \X$ is drawn with thin lines. Pieces of the boundaries of neighborhoods that are not part of $p^{-1}(C) \cap \X$ or $p^{-1}(C') \cap \X$ are drawn with dotted lines. I will call the vertical line of $p^{-1}(C \cap C')$ in the first picture a \defglo{vertical twist line}. \medskip \label{section-twistsection} \section{Examples of surfaces in near-horizontal position} A surface $X$ in near-horizontal position determines two unions of properly embedded arcs and loops in $\F$: $C$ and $C'$ defined above, and a union of disjoint arcs of $\de \F$: $p(B)$. By definition of near-horizontal position, $C$, $C'$, and $p(B)$ have the following properties: \begin{enumerate} \item $C$ and $C'$ intersect transversely. \item $\de (p(B)) = \de C \cup \de C'$ \end{enumerate} Furthermore, a surface $X$ in near-horizontal position determines a function \[N: \mbox{ components of $\X \sm (C \cup C')$} \longrightarrow \mbox{ non-negative integers}\] such that for each component $r$ of $\F \sm (C \cup C')$ and each point $y \in \interior(r)$, $N(r)$ counts the number of times $X$ intersects $y \times I$. It is easy to check that $N$ has the following properties: \begin{enumerate} \setcounter{enumi}{2} \item If $r$ is a component of $\F \sm (C \cup C')$ with an edge in $p(B)$, then $N(r) = 1$. \item If $r_1$ and $r_2$ are two components which meet along an arc of $C$ or an arc of $C'$, then $|N(r_1) - N(r_2)| = 1$. \item If $r_1$, $r_2$, $r_3$, and $r_4$ meet at a common vertex, then either the set $\{N(r_1), N(r_2), N(r_3), N(r_4)\}$ contains $3$ distinct numbers, or else \linebreak[3] $\{N(r_1), N(r_2), N(r_3), N(r_4)\} = \{0, 1\}$. \end{enumerate} The second possibility occurs if and only if the common vertex is the projection of a vertical twist line. \medskip Conversely, suppose that $C$ and $C'$ are unions of arcs and loops in $\F$, that $p(B)$ is a union of disjoint arcs of $\de \F$, and that \[N: \mbox{ components of $\F \sm (C \cup C')$} \longrightarrow \mbox{ non-negative integers}\] is a numbering scheme satisfying conditions (1) - (5) above. Then the information $(C, C', p(B), N)$ determines a unique surface in near-horizontal position. Figure~\ref{fig-example} depicts a genus $1$ surface with $4$ boundary loops, in near-horizontal position in $\R^2 \times I$. The projection of the surface to $\R^2$ is drawn at left. Vertical twist lines, which project to points, are marked with dots. Loops of $C$ are drawn with thick lines, and loops of $C'$ are drawn with thin lines. The corresponding combinatorial description is given at right. \begin{figure}[htbp] \epsfysize=4in \centerline{\epsfbox{example.eps}} \label{fig-example} \caption{An example of a surface in near-horizontal position.} \end{figure} \section{Putting incompressible surfaces in near-horizontal position} \stst{Theorem} {\it Suppose that $\X \subset \F \times I$ is a properly embedded, $2$-sided, incompressible surface. Then $\X$ is isotopic to a surface in near-horizontal position. } \pfpf{Proof of Theorem:} Notice that it is possible to isotope $\X$ (leaving it fixed outside a neighborhood of $\de \X$), so that $\de \X$ satisfies requirements 1 and 2 in the definition of near-horizontal position. Therefore, I will assume that $C$ intersects $C'$ transversely and that $p|_B$ is an embedding, where $C$, $C'$, and $B$ are defined as above. Suppose that $\X$ contains components $\X_1, \X_2, \ldots \X_n$ that are boundary parallel disks. Each $X_i$ can be isotoped so that $p|_{\X_i}$ is an embedding and so that $p(\de \X_i)$ is disjoint from $p(\de X \sm \de \X_i)$. Therefore, if $\X \sm (\X_1 \cup \X_2 \cup \cdots \cup \X_n)$ can be isotoped to near-horizontal position, so can $\X$. So without loss of generality, I can assume that $\X$ has no components that are boundary parallel disks. Since $\X$ is incompressible, this assumption insures that all loops in $C$ and $C'$ are essential in $\F$. All isotopies in the rest of the proof will leave $\de \X$ fixed. Let $A$ be the union of vertical strips and annuli $C \times [0, 1]$ and let $A' = C' \times [0, 1]$. First, I define a notion of ``pseudo-transverse'' and a measure of complexity for surfaces that are pseudo-transverse to $A \cup A'$. Then I describe three moves which decrease this complexity. In Step 1, I isotope $\X$ so that it is pseudo-transverse to $A$ and $A'$. In Step 2, I isotope $\X$ using the three moves as many times as possible. I then verify six claims about the position of $\X$. In Step 3, I isotope $\X$ so that the projection map $p$ is injective when restricted to any arc or loop of $\X \cap (A \sm A')$ and $\X \cap (A' \sm A)$. In Step 4, I complete the proof of the Theorem by isotoping $\X$ so that $p$ is locally injective everywhere except at twist lines. I will say that $\X \subset \F \times I$ is \defglo{pseudo-transverse} to $A \cup A'$ if the following three conditions hold: \begin{enumerate} \item For any point $z \in (C - C')$ there is a neighborhood $U \subset \F$ of $z$ such that $p^{-1}(U) \cap \X$ is a disjoint union of regions of the form % % \begin{figure}[h!tbp] \epsfysize=0.7in \centerline{ \epsfbox{paperpseudoc.eps}} \end{figure} % In the last two pictures above, the dashed lines represent pieces of $B$. One of these two pictures occurs if and only if $z \in \de C$. % \item For any point $z \in (C' - C)$, there is a neighborhood U of $z$ such that $p^{-1}(U) \cap \X$ is a disjoint union of regions of the form % % \begin{figure}[h!tbp] \epsfysize=0.7in \centerline{\epsfbox{paperpseudocprime.eps}} \end{figure} % One of the last two pictures occurs if and only if $z \in \de C'$. % \item For any point $z \in (C \cap C')$, there is a neighborhood $U$ of $z$ such that $p^{-1}(U) \cap \X$ is either a single region of the form % % % \begin{figure}[h!tbp] \epsfysize=1.0in \centerline{ \epsfbox{neartwist.eps}} \end{figure} % \par\noindent or else a disjoint union of regions of the form % % % \begin{figure}[h!tbp] \epsfysize=0.7in \centerline{ \epsfbox{nearother.eps}} \end{figure} % % \end{enumerate} \medskip Suppose $X$ is pseudo-transverse to $A \cup A'$. Define the complexity of $\X$ by \[\zeta(\X) = (|\X \cap (A \cap A')| \mbox{, rank } H_1(\X \cap A) + \mbox{ rank } H_1(\X \cap A')) \] ordered lexicographically. Consider the following three moves. \mvmv{Move 1} Suppose $D$ is a disk of $A$ such that $D \cap \X = \de D$ and $\partial D \cap A' = \emptyset$. Then $D$ can be used to isotope $\X$ and decrease $\zeta(\X)$. The isotopy leaves $\X$ in the class of pseudo-transverse embeddings. If the roles of $A$ and $A'$ are interchanged, an analogous move is possible. \mvmv{Explanation of Move 1} Since $\X$ is incompressible in $\F \times I$, $\de D$ bounds a disk $D'$ in $\X$. The set $D \cup D'$ forms a sphere in $\F \times I$, which must be embedded since $\interior(D) \cap \X = \emptyset$. Since $\F \times I$ is irreducible, the sphere bounds a ball, which can be used to isotope $\X$ relative to $\de \X$. If $\de D \cap \de \X = \emptyset$, then $D'$ can be pushed entirely off of $\de D$, and one component of $\X \cap A$ is eliminated. Components of $\X \cap (A \cap A')$, components of $\X \cap A'$, and additional components of $\X \cap A$ may also be removed if $\interior(D') \cap (A \cup A') \neq \emptyset$, but no new components of any kind are added. Therefore, $\zeta(\X)$ goes down. If, instead, $\de D \cap \de \X \neq \emptyset$, then the isotopy of $X$ relative to $\de X$ must leave $\de D \cap \de \X$ fixed. But this isotopy still decreases the rank of $H_1(\X \cap A)$ without increasing the rank of $H_1(\X \cap A')$ or the number of components of $\X \cap (A \cap A')$. The explanation is analogous if the roles of $A$ and $A'$ are interchanged. \smallskip \begin{figure}[htbp] \epsfysize=2.0in \centerline{ \epsfbox{move1.eps}} \caption{Move 1 disks.} \label{fig-move1} \end{figure} \mvmv{Move 2} Suppose E is a disk in $\F \times I$ whose boundary consists of two arcs $\alpha$ and $\sigma$. Suppose that $E \cap \X = \sigma$ and that $E \cap A = E \cap A' = \alpha$. Then $E$ can be used to isotope $\X$ relative to $\de \X$ and decrease $\zeta(\X)$. The isotopy will keep $\X$ in the class of pseudo-transverse embeddings. \smallskip \begin{figure}[htbp] \epsfysize=4.0in \centerline{ \epsfbox{move2.eps}} \caption{Move 2 disks.} \label{fig-move2} \end{figure} \begin{figure}[ptbh] \epsfysize=7.0in \centerline{ \epsfbox{move3.eps}} \caption{Using move 2 disks to isotope $\X$.} \label{fig-move3} \end{figure} \mvmv{Explanation of Move 2} Consider three cases depending on how many points of $\de \sigma$ lie in $\de \X$. See Figures~\ref{fig-move2} and~\ref{fig-move3}. \caca{Case 1} Both endpoints of $\sigma$ lie in $\interior(\X)$. Then $\X$ can be isotoped in a neighborhood of $\sigma$ so that it moves through $E$ and slips entirely off of $\alpha$. This isotopy decreases by two the number of components of $\X \cap (A \cap A')$. \caca{Case 2} One endpoint of $\sigma$ lies in $\interior(\X)$ and one endpoint lies in $\de \X$. Now $\X$ cannot be isotoped relative to $\de \X$ entirely off $\alpha$ since the endpoint of $\de \sigma$ in $\de \X$ must remain fixed. But $\X$ can still be pushed off $\interior(\alpha)$ and off the free endpoint, lowering the number of components of $\X \cap (A \cap A') $ by one. \caca{Case 3} Both endpoints of $\sigma$ lie in $\de \X$. Notice that one endpoint must lie in $\X \cap (\F \times 1)$ and one must lie in $\X \cap (\F \times 0)$, since $\alpha$ is a vertical line connecting them. In this case $\X$ can be isotoped relative to $\de \X$ in a neighborhood of $\sigma$ to move $\sigma$ directly onto the vertical line $\alpha$ and produce a twist around this vertical line resembling picture (3)(i). This isotopy decreases the number of components of $\X \cap (A \cap A')$ by one, since it transforms the two endpoints of $\de \sigma$ into a single vertical line. \mvmv{Move 3} Suppose $W$ is an annulus of $A$ such that $W \cap A' = \emptyset$ and $\de W$ is the boundary of an annulus of $\X \sm A$. Then $W$ can be used to isotope $\X$ and decrease $\zeta(\X)$. The isotopy leaves $\X$ in the class of pseudo-transverse embeddings. A similar move is possible if the roles of $A$ and $A'$ are interchanged. \begin{figure}[htbp] \epsfysize=2.0in \centerline{\epsfbox{move3really.eps}} \caption{Move 3 annuli.} \label{fig-move3really} \end{figure} \mvmv{Explanation of Move 3} Let $K$ be the annulus of $\X \sm A$ such that $\de K = \de W$, and let $L$ be the closure of the component of $\F \times I \sm A$ that contains $K$. $W \cup K$ forms a torus, which is embedded in $L$ since $\interior(K) \cap A = \emptyset$. The torus $W \cup K$ compresses in $L$, because $L$ is homotopic to a surface with boundary and therefore $\pi_1(L)$ cannot contain a $\Z \times \Z$ subgroup. Since $L$ is irreducible and $W$ lies on the boundary of $L$, $W \cup K$ actually bounds a solid torus in $L$, which can be used to isotope $\X$ relative to $\partial \X$ by pushing $K$ through $W$. Notice that $\partial W$ and $\de \X$ share at most one component. If $\de W$ and $\de X$ are disjoint, then this isotopy removes at least two components of $\X \cap A$. If $\de W$ and $\de \X$ share a component, the the isotopy removes at least one component of $\X \cap A$. In either case, the isotopy decreases the rank of $H_1(\X \cap A)$ without increasing the rank of $H_1(\X \cap A')$ or the number of components of $\X \cap (A \cap A')$. The explanation is analogous if the roles of $A$ and $A'$ are interchanged. \medskip I will refer to the type of disk used in Move 1 as a move 1 disk, the type of disk used in Move 2 as a move 2 disk, and the type of annulus used in Move 3 as a move 3 annulus. \stst{Step 1} Isotop $\X$ relative to $\de \X$ so that it is pseudo-transverse to $A \cup A'$. This can be accomplished, for example, by making $\X$ honestly transverse to $A \cup A'$. Then only pictures (1)(i), (1)(ii), (1)(iv), (2)(i), (2)(ii), (2)(iv), (3)(ii), (3)(iii), and (3)(iv) in the definition of pseudo-transverse can occur. \smallskip \stst{Step 2} Suppose $\F \times I$ contains a move 1 disk, a move 2 disk, or a move 3 annulus. Use it to isotope $\X$. Repeat this step as often as necessary, until there are no more such disks or annuli. The process must terminate after finitely many moves, since each move decreases $\zeta(\X)$. At this stage, $\X$ already has a neat posture with respect to $A \cup A'$. In particular, the following claims hold, where $K$ is any component of $\X \sm (A \cup A')$ and $L$ is the component of $(\F \times I) \sm (A \cup A')$ that contains $K$. \clcl{Claim 1} Suppose that $\mu$ is an arc contained in $\X \cap (A \sm A')$ with both endpoints in $A \cap A'$. Then either the endpoints of $\mu$ go to distinct vertical lines of $(A \cap A')$ or else $\mu$ wraps all the way around an annulus of $A$. Likewise for arcs of $\X \cap (A' \sm A)$. See Figure~\ref{fig-distinct}. \begin{figure}[htbp] \epsfxsize=3.0in \centerline{\epsfbox{distinct.eps}} \caption{Endpoints of $\mu$ go to distinct vertical lines of $A \cap A'$.} \label{fig-distinct} \end{figure} \clcl{Claim 2} For any point $z \epsilon (C \sm C')$, there is a neighborhood $U$ of $z$ such that $p^{-1}(U) \cap \X$ is a disjoint union of neighborhoods of the form: % % \begin{figure}[h!tbp] \epsfysize=0.7in \centerline{\epsfbox{papernearrestrict.eps}} \end{figure} % \par\noindent % Likewise for points of $(h(C) \sm C)$. In other words, pictures (1)(iii), (1)(v), (2)(iii) and (2)(v) in the definition of pseudo-transverse do not occur. \clcl{Claim 3} Every circle of $\partial K$ has non-zero degree in the cylinder of $\partial_v L$ that contains it. Here $\dv L$ refers to the vertical boundary of $L$, namely ${\partial L}\cap (A \cup A' \cup (\de \F \times I))$. \clcl{Claim 4} $\pi_1(K) \rightarrow \pi_1(L)$ is injective. \clcl{Claim 5} Either $\pi_1(K) \rightarrow \pi_1(L)$ is surjective, or else $K$ is an annulus that is parallel to $\dv L$. \clcl{Claim 6} If $K$ is an annulus, then the two loops of $\partial K$ go to two distinct cylinders of $\dv L$. \smallskip\noindent \bigskip \clcl{Proof of Claim 1} Let $\mu$ be an arc contained in $\X \cap (A \sm A')$ and suppose both endpoints of $\mu$ lie in one vertical line of $A \cap A'$. Suppose that $\mu$ does not wrap all the way around an annulus of $A$, and let $\alpha$ be the segment of $A \cap A'$ that connects the endpoints of $\mu$. Then $\alpha \cup \mu$ bounds a half-disk $E$ in $A$. Notice that $E \cap A' = \alpha$. The set $E \cap \X$ cannot contain any closed loops of $\X \cap A$, since any such loop would bound a move 1 subdisk of E, which should have been removed in Step 2. But $E \cap \X$ may contain other arcs besides $\mu$ with endpoints on $\alpha$. (See Figure~\ref{fig-arcs}.) By replacing $\mu$ and $E$ with an arc and subdisk closer to $\alpha$ if necessary, I can assume that $E \cap \X = \mu$. Nudge $E$ relative $\alpha$ off of $A$ to get a new disk $E'$ bounded by the arcs $\alpha$ and $\mu'$, where $\mu' \subset \X$ and $\interior(\mu') \subset \interior(\X)$. Since $E \cap \X = \mu$ and $E \cap A' = \alpha$, I can assume that $E' \cap \X = \mu'$ and $E' \cap A' = \alpha$. Also, $E' \cap A = \alpha$. So $E'$ is a move 2 disk, in violation of Step 2. Thus, the endpoints of $\mu$ must lie in distinct components of $A \cap A'$ after all. The same argument applies to arcs contained in $\X \cap (A' \sm A)$. \medskip \begin{figure}[htbp] \epsfxsize=3.0in \centerline{\epsfbox{arcs.eps}} \caption{$E \cap \X$ may contain additional arcs.} \label{fig-arcs} \end{figure} \clcl{Proof of Claim 2} Suppose picture (1)(iii) or (1)(v) does occur. (The argument is similar if picture (2)(iii) or (2)(v) occurs.) Let $\alpha$ be the segment of ${{p^{-1}}(U)} \cap \X \cap A$ drawn vertically in these pictures, extended in $A$ until it first hits $\de \X$ or $A \cap A'$. See Figure~\ref{fig-ignore}. Label the first endpoint of $\alpha$ as $\partial_1\alpha$ and the second endpoint as $\partial_2 \alpha$. So $\partial_1 \alpha$ lies on $C \times 1$. Suppose first that $\partial_2 \alpha$ lies on $\partial \X$ but not on $A \cap A'$. Recall that $\de \X = (C \times 1) \cup (C' \times 0) \cup B$; therefore $(\de \X \cap A) \sm (A \cap A') \subset ((C \times 1) \cap A) \cup (B \cap A)$. Since $p|_B$ is an embedding, it follows that $B \cap A \subset C \times 1$. So $\partial_2 \alpha$ lies in $\de \X \cap (C \times 1))$. Thus, $\alpha$ cuts off an arc $\beta$ of $C \times 1$ such that $\alpha \cup \beta$ bounds a move 1 disk. This disk should already have been removed in Step 2. Next, suppose that $\partial_2\alpha$ lies on $A \cap A'$, and let $\beta$ be the arc of $\partial \X$ such that $\partial_1\beta = \partial_1\alpha$ and $\partial_2\beta$ lies on the same vertical line of $A \cap A'$ as $\partial_2\alpha$, and so that $\alpha \cup \beta$ does not wrap all the way around an annulus of $A$. Then $\alpha \cup \beta$ is an arc contained in $\X \cap A$ with both endpoints in the same vertical line which should not exist by Claim 1. \medskip \begin{figure}[htbp] \epsfxsize=5.0in \centerline{\epsfbox{ignore.eps}} \caption{The arc $\alpha$ in the proof of Claim 2.} \label{fig-ignore} \end{figure} \clcl{Proof of Claim 3} Let $K$ be a component of $\X \sm (A \cup A')$ and let $L$ be the component of $(\F \times I) \sm (A \cup A')$ that contains it. Let $\dv L$ denote the vertical boundary of $L$, that is, $\dv L = \de L \cap (A \cup A' \cup (\de \F \times I))$. Suppose that a circle $\gamma$ of $\partial K$ gets sent to a cylinder $G$ of $\dv L$ by degree 0. $G$ may be an annulus of $A$ or an annulus of $A'$. Or $G$ may consist of rectangles of $A \sm A'$ and $A' \sm A$, joined together along vertical lines of $A \cap A'$ and possibly along vertical strips of $\de \F \times I$. Suppose first that $\gamma$ is contained in a single component of $A \sm A'$ or $A' \sm A$. (This happens, in particular, if $G$ is an annulus of $A$ or $A'$.) Since $\gamma$ has degree 0 in $G$, it must bound a disk in $G$, which I can assume has interior disjoint from $\X$ by replacing $\gamma$ with an innermost loop if necessary. But this disk is a move 1 disk, so it should have been removed already in Step 2. See Figure~\ref{fig-degree0}. \begin{figure}[htbp] \epsfysize=1.7in \centerline{\epsfbox{degree0.eps}} \caption{Suppose a circle $\gamma$ of $\partial K$ gets sent to a cylinder of $\dv L$ with degree 0.} \label{fig-degree0} \end{figure} Now suppose that $\gamma$ is not contained in a single component of $A \sm A'$ or $A' \sm A$. Since $p|_B$ is an embedding, $\gamma$ must be disjoint from $\de \F \times I$. Again, $\gamma$ bounds a disk in $G$. The vertical lines of $A \cap A'$ cut the disk into subdisks. Consider an outermost subdisk and the arc $\sigma \subset \gamma$ that forms half its boundary. This arc $\sigma$ is contained in $\X \cap (A \sm A')$ (or in $\X \cap (A' \sm A)$) and both its endpoints lie in the same vertical line. Furthermore, $\sigma$ cannot wrap all the way around an annulus of $A$ (or $A'$). Claim 1 says that such arcs do not exist. \medskip \clcl{Proof of Claim 4} The following diagram commutes, and the map $\pi_1(\X) \rightarrow \pi_1(\F \times I)$ is injective. So it will suffice to show that $\pi_1(K) \rightarrow \pi_1(\X)$ is injective. \begin{picture}(150, 80)(-30, 0) \thinlines \put(20, 60){$\pi_1(K)$} \put(80, 60){$\pi_1(L)$} \put(140, 60){$\pi_1(\F \times I)$} \put(80, 20){$\pi_1(\X)$} \put(38, 43){\vector(3,-2){23}} \put(117, 27){\vector(3,2){25}} \put(50, 62){\vector(1,0){17}} \put(110, 62){\vector(1,0){17}} \end{picture} If $\pi_1(K) \rightarrow \pi_1(\X)$ does not inject, then some loop in $\partial K$ must bound a disk $D$ in $\X \sm K$. I claim that $D$ contains at least one loop $\omega$ of $\X \cap A$ or $\X \cap A'$. Since $\de D$ bounds $D$ on one side and $K$ on the other, $\de D$ must be disjoint from $\de X$. So $\de D$ must intersect $A$ or $A'$. Assume without loss of generality that $X$ intersects $A$. Possibly $\partial D$ itself is contained in $\X \cap A$. In this case set $\omega = \partial D$. Otherwise, take an arc of $\partial D \cap (\X \cap A)$ and let $\omega$ be the component of $\X \cap A$ that contains it. Notice that $\omega$ must lie in $D$, since it cannot intersect $\interior(K)$. Since $D \cap \de \X = \emptyset$, $\omega$ is a closed loop rather than an arc. Since $\omega \subset D$, $\omega$ is null-homotopic in $\F \times I$. Since each annulus in $A$ $\pi_1$-injects into $\F \times I$, $\omega$ must bound a disk $E$ in $A$. If $E \cap A' = \emptyset$, then $E$ is a move 1 disk, which should already have been removed. If $E \cap A' \neq \emptyset$, then $\de E \cap (A \sm A')$ will contain an arc whose endpoints lie in the same vertical line of $A \cap A'$, as in the proof of Claim 3. But Claim 1 says that such arcs do not exist. \medskip \clcl{Proof of Claim 5} Let $R$ be the surface $p(L)$. From Claim 4, the map $\pi_1(K) \rightarrow \pi_1(L)$ is injective. Therefore, it is possible to write $\pi_1(L)$ either as an amalgamated product or as an HNN extension over $\pi_1(K)$, depending on whether or not $K$ separates $L$. %See \cite{HNN} for definitions and facts about amalgamated products and % HNN extensions. In fact, $K$ separates $L$, because the inclusion of $R \times 1$ into $L$ induces an isomorphism of fundamental groups, which could not happen if $\pi_1(L)$ were an HNN extension. Let $M_1$ be the component of $L \sm K$ that contains $R \times 1$, and let $M_2$ be the other component. Then the composition of maps \[\pi_1(R \times 1) \rightarrow \pi_1(M_1) \rightarrow\pi_1(L)\] is an isomorphism. So $\pi_1(M_1) \rightarrow \pi_1(L)$ is surjective. But \[\pi_1(L) = \pi_1(M_1)\ast_{\pi_1(K)}\pi_1(M_2)\] so the injection $\pi_1(K) \rightarrow \pi_1(M_2)$ must be an isomorphism. By the h-cobordism theorem \cite[Theorem 10.2]{Hempel}, $(M_2, K) \cong (K \times I, K \times 1)$. Now there are two possibilities: either $R \times 0 \subset M_1$ or $R \times 0 \subset M_2$. If $R \times 0 \subset M_1$, then $\partial M_2 \sm K$ is a subset of $\dv L$. So $K$ is parallel to a subsurface of $\dv L$, and therefore must be a disk or an annulus. By Claim 3, $K$ must be a boundary parallel annulus. If $(R \times 0) \subset M_2$, then it follows as above that the composition \[\pi_1(R \times 0) \rightarrow \pi_1(M_2) \rightarrow \pi_1(L) \] is an isomorphism. Therefore, $\pi_1(M_2) \rightarrow \pi_1(L)$ is surjective. Also, $\pi_1(M_2) \rightarrow \pi_1(L)$ is injective since \[\pi_1(L) = \pi_1(M_1)\ast_{\pi_1(K)}\pi_1(M_2)\] So $\pi_1(M_2) \rightarrow \pi_1(L)$ is an isomorphism. In addition, $\pi_1(K) \rightarrow \pi_1(M_2)$ is an isomorphism, from above. Therefore, $\pi_1(K) \rightarrow \pi_1(L)$ is an isomorphism, and the claim is proved. \medskip \clcl{Proof of Claim 6} Let $K$ be any annulus component of $\X \sm (A \cup A')$, let $L$ be the corresponding component of $(\F \times I) \sm (A \cup A')$, and suppose that both circles of $\partial K$ go to the same cylinder $G$ of $\partial L$. Since both circles have degree $\pm 1$ in $\partial_v L$, they bound an annulus $W$ in $\dv L$. Notice that $W$ is disjoint from $\de \F \times I$, since by assumption, $p|_B: B \rightarrow \de \F$ is an embedding. If $W$ is entirely contained in $A$ or $A'$, then $W$ is a move 3 annulus, which should have been removed in Step 2. So $W$ must consist of alternating rectangles of $A$ and $A'$. The union $W \cup K$ forms a torus, which is embedded in $L$ since $\interior(K) \cap \partial_v L = \emptyset$. The torus $W \cup K$ compresses in $L$, because $L$ is homotopic to a surface with boundary and therefore $\pi_1(L)$ cannot contain a $\Z \times \Z$ subgroup. Since $L$ is irreducible and $W$ lies on the boundary of $L$, $W \cup K$ bounds a solid torus $T$ in $L$. I will construct a move 2 disk as follows. Start with a vertical arc $\alpha$ of $W \cap (A \cap A')$. Connect its endpoints with an embedded arc $\sigma$ of $K$ so that $\alpha \cup \sigma$ is null homotopic in $T$. This is possible because each component of $\partial K$ generates $\pi_1 (T)$, so I can replace a poor choice of $\sigma$ by one that wraps around $\partial K$ an additional number of times and get a good choice of $\sigma$. Now $\alpha \cup \sigma$ bounds an embedded disk $E$ in $T$, which I can assume has interior disjoint from $\X$ by replacing it with a subdisk if necessary. In addition, $E \cap A = E \cap A' = \alpha$, so $E$ is a move 2 disk. But Step 2 already eliminated all disks of this form. \bigskip \stst{Step 3} Recall that $p: {\F \times I} \rightarrow \F$ is the projection map. I will isotope $\X$ relative to $\de \X$ so that for any arc or loop $\gamma$ of $\X \cap (A \sm A')$ or $\X \cap (A' \sm A)$ $p|_\gamma$ is a local homeomorphism onto its image. I will consider arcs first and loops next. Take any arc $\mu$ of $\X \cap (A \sm A')$. If $\interior(\mu) \cap \de \X \neq \emptyset$, then $\mu \subset \de \X$ by Claim 2. The map $p$ is already injective on arcs of $\de \X \cap A$ and $\de \X \cap A'$, so I can leave $\mu$ alone. If $\interior(\mu) \cap \de \X = \emptyset$, then isotope $\mu$ relative to $\partial \mu$ to an embedded arc $\mu'$ in $A \sm A'$ such that $p|_{\mu'}$ is a homeomorphism onto its image. This is possible because by Claim 1, either $\mu$ wraps all the way around an annulus of $A$ or else the endpoints of $\mu$ lie in distinct vertical lines of $A \cap A'$. The isotopy can be done in such a way that $\mu'$ does not intersect any other arcs of $\X \cap (A \sm A')$ that may lie in the same vertical rectangle. Perturb $\X$ in a neighborhood of $\mu$ to extend the isotopy on $\mu$. Pick another arc of $\X \cap (A \sm A')$ and repeat the procedure. When all the arcs of $\X \cap (A \sm A')$ have been pulled taut, continue with arcs of $\X \cap (A' \sm A)$. Next, consider any loop $\lambda$ of $\X \cap A$ that does not intersect $A'$. By Claim 3, the loop $\lambda$ has degree 1 in $A$, so it can be isotoped to a loop $\lambda'$ such that $p|_{\lambda}$ an homeomorphism onto its image. As before, the isotopy can be done in such a way that $\lambda'$ does not intersect any other loops of $\X \cap A$, and the isotopy can be extended to a neighborhood of $\lambda$ in $\X$. Loops of $\X \cap A'$ that are disjoint from $A$ can be isotoped similarly. I claim that at this stage, for any component $K$ of $X \sm (A \cup A')$, $p|_{\de K}$ is a local homeomorphism except along vertical lines of $K \cap (A \cap A')$. Every point of $\de K$ is either a point on the interior of an arc of $\de \X \cap (\de \F \times I)$, a point of $\X \cap A \cap (\de \F \times I)$ or $\X \cap (A' \cap (\de \F \times I))$, a point of $A \cap A'$, or a point on the interior of an arc or loop of $\X \cap (A \sm A')$ or $\X \cap (A' \sm A)$. The map $p|_{\de K}$ is a local homeomorphism near the first type of point by assumption. It is a local homeomorphism near the second type of point by Claim 2. It is a local homeomorphism near the third type of point (away from vertical twist lines) because $X$ is pseudo-transverse. Finally, $p|_{\de K}$ is a local homeomorphism near the fourth type of point by the work done in Step 3. \smallskip \stst{Step 4} I will finish isotoping $\X$ so that $p$ is locally injective except at twist lines. I will do this by using a fact about maps between surfaces. \stst{Fact} {\it Let $f: (G, \de G) \rightarrow (H, \de H)$ be a map such that $f|\de G$ is a local homeomorphism and $f_\ast: \pi_1(G) \rightarrow \pi_1(H)$ is injective. Then there is a homotopy $f_\tau:(G, \de G) \rightarrow (H, \de H)$, with $\tau \in I$, $f_0 = f$, and $f_\tau|_{\de G} = f_0|_{\de G}$ for all $\tau$, such that either (1) or (2) holds: \begin{enumerate} \item $G$ is an annulus or Mobius band and $f_1(G) \subset \de H$, or \item $f_1: G \rightarrow H$ is a covering map \end{enumerate} } The case when $G$ is a disk is easy to verify; all other cases are covered by \cite[Theorem 13.1]{Hempel}. Pick a component $K$ of $\X \sm (A \cup A')$, and let $L$ be the corresponding component of $(\F \times I) \sm (A \cup A')$. Assume first that $\de K$ does not contain any vertical arcs of $A \cap A'$. By Claim 4, the map $\pi_1(K) \rightarrow \pi_1(L)$ is injective. Since $p_\ast: \pi_1(L) \rightarrow \pi_1(p(L))$ is an isomorphism, the composition $(p|_K)_\ast: \pi_1(K) \rightarrow \pi_1(p(L))$ is injective. Furthermore, by the discussion following Step 3, $p|{\de K}: \de K \rightarrow \de p(L)$ is a local homeomorphism on $\de K$. Therefore, there is a homotopy $f_\tau: K \rightarrow p(L)$ with $f_0 = p|_K$ and $f_\tau|_{\de K} = p|_{\de K}$ such that either $K$ is an annulus or Mobius band and $f_1(K) \subset \de p(L)$, or $f_1$ is a covering map. If $K$ is a Mobius band and $f_1(K) \subset \de p(L)$, then $f_0(\de K)$ is a degree 2 loop in $\de p(L)$ which is impossible since $\de K \subset \dv L$ is embedded. By Claim 6, it not possible for $K$ to be an annulus and $f_1(K)$ a subset of $\de p(L)$. Therefore, $f_1$ must be a covering map. By Claims 5 and 6, the map $\pi_1(K) \rightarrow \pi_1(L)$ is surjective. So the map $(p_K)_\ast: \pi_1(K) \rightarrow \pi_1(p(L))$ is surjective. Therefore, $f_1$ must be a homeomorphism. If $\de K$ contains vertical arcs of $A \cap A'$, then $p|_{\de K}$ is still very close to a local homeomorphism -- in fact, if $\hat{K}$ is the surface obtained by collapsing each vertical arc of $\de K \cap (A \cap A')$ to a point, then $p|_{K}$ factors through a map $\hat {p}: \hat{K} \rightarrow p(L)$ such that $\hat{p}|_{\de \hat{K}}$ is a local homeomorphism. So it is still possible to homotope $p|_K$ relative to $\de K$ to a map $f_1$ such that ${f_1}|_{\interior(K)}$ is a homeomorphism. The homotopy $f_\tau$ of $K$ relative to $\de K$ in $p(L)$ induces a homotopy of $K$ relative to $\de K$ in $L$ which keeps the vertical coordinate of each point of $K$ constant and changes its horizontal coordinate according to $f_\tau$. At the end of the homotopy, the new surface $K'$ itself will be embedded in $\F \times I$, since $p|_{K'}$ is a homeomorphism. Homotope $\X$ as described above for every other component of $\X \sm (A \cup A')$. At this stage, each component of $\X \sm (A \cup A')$ is embedded in $\F \times I$. But two components $K'_1$ and $K'_2$ in the same piece $L$ of $(\F \times I) \sm (A \cup A')$ might intersect each other. If that happens, an additional homotopy of $K$ can be tacked on to clear up the problem, as follows. Let $K_1 \subset L$ and $K_2 \subset L$ be two components of $\X \sm (A \cup A')$ before the homotopy of Step 4 and let $K'_1$ and $K'_2$ be these components after the homotopy. Suppose that $K'_1$ intersects the component $K'_2$. Since $K_1$ and $K_2$ are disjoint, the component of $L \sm K_1$ that meets $\F \times 0$ either contains all of $K_2$ or else contains no part of $K_2$. Therefore, either all loops of $K_1$ lie above the corresponding loops of $K_2$ or else they all lie below the corresponding loops. Since the homotopies of $K_1$ and $K_2$ did not move $\de K_1$ and $\de K_2$, the same statement holds for loops of $\de K'_1$ and $\de K'_2$. Therefore, it is possible to alter the vertical coordinates of $K'_1$ and $K'_2$ to make the two surfaces parallel, so that $K'_1$ lies entirely above $K'_2$ or entirely below $K'_2$. Therefore all components of $\X \sm (A \cup A')$ can be assumed disjoint after the homotopy of Step 4. In its final position, $\X$ is embedded in $\F \times I$. A theorem of Waldhausen \cite[Corollary 5.5]{W} states that if $G$ and $H$ are incompressible surfaces embedded in an irreducible 3-manifold, and there is a homotopy from $G$ to $H$ that fixes $\de G$, then there is an isotopy from $G$ to $H$ that fixes $\de G$. Therefore, the above sequence of homotopies can be replaced by an isotopy. \section{Remarks} It is not true that every surface in near-horizontal position is incompressible. For example, the surface in Figure~\ref{fig-example} compresses in $\R^2 \times I$. I hope to find simple conditions for when a surface in near-horizontal position is incompressible. A special case is considered in \cite{thesis}. \bibliography{bib} \bibliographystyle{plain} \end{document}