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As Mark Bennet said, it means every two regions have a nontrivial piece of boundary in common: nontrivial means containing a topological arc, i.e., not being a point. This is a standard detail in Map coloring problems, and Thurston's exercise is in the context of map coloring.


The answer depends on what you mean by a solid genus-2 handlebody, and the trouble is that this is ambiguous and requires some interpretation. Had you asked about "the complement of the solid genus-2 handlebody then I would assume you meant this and the answer would be yes, as in the answer of @QuangHoang. But since you asked about "the complement of a ...


Yes, that's correct. To see why, one can think of $S^3$ as $\Bbb R^3$ with an infinity point whose a neighborhood is the complement of $B^3$ in $\Bbb R^3$. It is left to see that $B^3-T\sharp T$ is equivalent to the punctured $T\sharp T$. Note 1: it may help to work with the solid torus first. Note 2: It is somewhat obvious in the view of Morse Theory, but ...


To give one possible elaboration of studiosus's counterexample, let $M = M' = S^1 \times D^2$, let $f,g\colon \partial M \to \partial M'$ be given by $(x,y) \mapsto (x,y)$ and $(-x,y) \mapsto (x,-y)$. These maps are not isotopic (think what they do to the fundamental group of $\partial M$), but both $M \cup_f M'$ and $M \cup_g M'$ are $S^1 \times S^2$.


Actually, you do not need the Chech cohomology, just a little bit of analysis. Namely, start with the characteristic function $\chi: C\to \{0,1\}$ of a proper clopen subset of $C$. Then, by the Tietze-Urysohn extension theorem, the function $\chi$ will extend continuously to a map $S^3\to [0,1]$. Then you can modify the extension to make it smooth outside of ...


The number $N_s$ must be independent of $S$; however we should restrict $S$ to $S\ge s$. Allowing arbitrary positive $S$ would make all nontrivial geodesic spaces fail the condition. E.g., in $\mathbb R$ we need at least $(S+s)/S$ balls of radius $S$ to cover a ball of radius $S+s$; and $(S+s)/S\to\infty$ as $S\to 0$. The answer appears to be too short, so ...


In general there is no such curve $\gamma$. One way to find the conjugating map is to use the classification of surfaces. (That is, use the fact that $S - \alpha$ is homeomorphic to $S - \beta$.) Another approach is given in Lickorish's famous paper "A representation of orientable combinatorial 3-manifolds". He uses the fact that if $\alpha$ and $\beta$ ...

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