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If you are willing to quote the Schoenflies theorem and the classification of surfaces, a quick proof of this result is a standard exercise. First, using algebraic topology (as in the proof of the Jor …

Read Chapter 4 of Thurston's notes http://library.msri.org/books/gt3m/. He produces infinitely many closed hyperbolic 3-manifold of different volumes, and hence non-homeomorphic, just by doing Dehn su …

Yes, $Y$ is still irreducible, and this holds by a simple connectivity argument. Suppose that $\Sigma \subset Y \subset X$ is a smoothly embedded 2-sphere. Since $X$ is irreducible, $\Sigma$ bounds …

If $M$ is a compact and irreducible 3-manifold, one answer is provided by a theorem of Stallings, in his 1962 paper "On fibering certain 3-manifolds": $\alpha$ is represented by a fibration $f : M \to …

There is indeed a surface basis for $H$ containing $x_1,\ldots,x_k$. I'll give a topological proof, basically the same as the proof suggested in the comment of @HJRW. First, I'll give a topological re …

In the topological category the proof that connected sum is well-defined depends on the Annulus Theorem, first proved by Kirby; the necessity of the Annulus Theorem is seen from Bruno Martelli's answe …

The theories for an open surface $S$ and for a compact surface with boundary $\overline S$ whose interior is identified with $S$ are the same. The inclusion of $S$ into $\overline S$ defines an isomor …

Here is an answer involving finitely generated groups in the quotient, and free groups in the kernel and the total group. Suppose that $G$ is an infinite, finitely generated group. Let $F_n = \langl …

One way this formula is proved is using intersection number of oriented curves. The proof works just fine on any orientable surface using coefficients in any field, both for defining Betti numbers and …

A good counterexample is the Teichmuller space of a closed oriented surface $S$. It is uniquely geodesic by Teichmuller's theorem, but it is not $CAT(0)$.

I will answer the question of whether this also gives the classification cheaply. No. It gives the classification at the expense of proving that every surface group has a free cocompact action on th …

I think the short answer is "No", you cannot deduce from this result a classification of all geometric outer automorphisms. I think it might eventually be possible to obtain a classification of geom …

You can build a certain covering space of the surface $S$ rather explicitly as a nested union of closed discs $D_1 \subset D_2 \subset D_3 \subset \cdots$, each contained in the interior of the next, …

In my proof that mapping class groups are automatic, Ann. of Math. (2) 142 (1995), no. 2, 303–384, I used a theorem from ECHLPT "Word Processing in Groups" which says that if a groupoid is automatic …

Assuming that you DO mean that $T^n$ is a finite sheeted covering space, at the very least one can say that $\pi_1(M)$ is a torsion free $n$-dimensional crystallographic group. This follows from the B …