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\begin{document}
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\parbox[b]{3.25in}{
  {\LARGE\bf Cheat Sheet 3D Topology}
  \newline \footnotesize Follows Hatcher's notes [Ha] and Hempel's book [He].
}
\hfill\parbox[b]{4.75in}{\small
  \null\hfill\sheeturl
  \newline\null\hfill initiated 31/8/13;
    continues \href{http://drorbn.net/AcademicPensieve/2013-08/}{2013-08};
    continued \href{http://drorbn.net/AcademicPensieve/2013-10/}{2013-10};
    modified \today
}

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{\bf Definition.} $M$ prime: $M=P\#Q\Rightarrow(P=S^3)\vee(Q=S^3)$. M Irreducible: an embedded 2-sphere in $M$ bounds a 3-ball. (Irreducible $\Rightarrow$ Prime).

{\bf Theorem} (Alexander, 1920s). $S^3$ is irreducible.

{\bf Theorem.} Orientable, prime, not irreducible $\Rightarrow\ S^2\times S^1$. Nonorientable? Also $S^2\widetilde{\times}S^1$ (Klein 3D).

{\bf Theorem.} Compact connected orientable 3-manifolds have unique decomposition into primes.

{\bf Proof. }
$\bullet$ Given a system of splitting spheres (sss) and a $\theta$-partition of one member, at least one part will make an sss.
$\bullet$ An sss can be simplified relative to a fixed triangulation $\tau$: circle and single-edge-arc intersections with faces of $\tau$ can be eliminated. $\bullet$ The size of an sss is bounded by $4|\tau|+\operatorname{rank}H_1(M;\bbZ/2)$ and hence prime-decompositions exist. $\bullet$ Uniqueness. \qed

Nonorientable $M$? Same but $M\#(S^2\times S^1)=M\#(S^2\widetilde{\times}S^1)$.

{\bf Theorem.} If a covering is irreducible, so is the base. ([Ha] proof is fishy).

{\bf Examples.} Lens spaces, surface bundles $F\to M\to S^1$ with $F\neq S^2,\RP^2$. Yet $S^1\times S^2/(x,y)\sim(\bar{x},-y)=\RP^3\#\RP^3$, a prime covers a sum.

{\bf Definition.} $S\subset M^3$ a 2-sided surface, $S\neq S^2$, $S\neq D^2$. {\em Compressing disk for $S$} is a disk $D\subset M$ with $D\cap S=\partial D$. If for every compressing $D$ there's a disk $D'\subset S$ with $\partial D'=\partial D$, $S$ is {\em incompressible}.

{\bf Claims.}
$\bullet$ $\pi_1(S)\hookrightarrow\pi_1(M)$ $\Rightarrow$ $S$ incompressible.
$\bullet$ No incompressibles in $\bbR^3$/$S^3$.
$\bullet$ In irreducible $M^3$, $T^2$ is 2-sided incompressible iff $T$ bounds a $D^2\times S^1$ or $T$ is contained in a $B^3$.
$\bullet$ A $T^2$ in $S^3$ bounds a $D^2\times S^1$ on at least one side.
$\bullet$ $S\subset M$ incompressible $\Rightarrow$ ($M$ irreducible iff $M|S$ irreducible).
$\bullet$ $S$ a collection of disjoint incompressibles or disks or spheres in $M$, $T\subset M|S$. Then $T$ is incompressible in $M$ iff in $M|S$.

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{\bf Dehn's Lemma} (Dehn 1910 (wrong), Papakyriakopoulos 1950s). $M$ a 3-manifold, $f\colon B^2\to M$ s.t.\ for some neighborhood $A$ of $\partial B^2$ in $B^2$ the restriction $F\mid_A$ is an embedding and $f^{-1}(f(A))=A$. Then $f\mid_{\partial B^2}$ extends to an embedding $g\colon B^2\to M$.

{\bf The Loop Theorem} (Stallings 1960, implies Dehn's lemma). $M$ a 3-manifold, $F$ a connected 2-manifold in $\partial M$, $\ker(\pi_1(F)\to\pi_1(M)\not\subset N\triangleleft\pi_1(F)$. Then there is a proper embedding $g\colon(B^2,\partial B^2)\to(M,F)$ s.t.\ $[g\mid_{\partial B^2}]\not\in N$.

{\bf The Sphere Theorem.} $M$ orientable 3-manifold, $N$ a $\pi_1(M)$-invariant proper subgroup of $\pi_2(M)$. Then there is an embedding $g\colon S^2\to M$ s.t.\ $[g]\not\in N$.

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{\bf To do.}
$\bullet$ JSJ.

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