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Homotopy theory, homological algebra, algebraic treatments of manifolds.
5
votes
How to get convinced that there are a lot of 3-manifolds?
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 …
3
votes
Accepted
Is the following 3-manifold irreducible?
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 …
13
votes
When can a class in $H^1(M;\mathbb{Z})$ be represented by a fiber bundle over $S^1$
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 …
3
votes
Accepted
Bases of surface groups
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 …
5
votes
Accepted
Nielsen-Thurston classification of homeomorphisms for open surfaces?
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 …
4
votes
(Short) Exact sequences with no commutative diagram between them
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 …
4
votes
Dimension of the homology group with coefficients in $\mathbb{Z}/2\mathbb{Z}$
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 …
5
votes
Uniquely geodesic and CAT(0) spaces?
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 $CAT(0)$.
9
votes
The fundamental group of a closed surface without classification of surfaces?
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 …
15
votes
Accepted
Universal covering of compact surfaces
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, …
3
votes
Classification of geometric outer automorphisms of free groups
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 …
13
votes
Compelling evidence that two basepoints are better than one
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 …
40
votes
Accepted
Connected sum of topological manifolds
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 …
8
votes
Manifolds covered by an n-dimensional torus
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 …
1
vote
Triangulation of fundamental domains for surfaces and generators
Here's an example. Start with a $4g$-gon $P$ with opposite sides glued, which yields a closed surface of genus $g$. Between any side-pair which is glued, there are $2g-1$ sides, which is the maximum p …