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revised Fundamental group of Antoine's necklace
added 1 character in body
Aug
15
comment Show the wedge product of two tori is not homotopy equivalent to a 2-manifold.
You can find a proof of the fact that the fundamental group of non-compact surface is free here: chiasme.wordpress.com/2014/08/27/on-subgroups-of-surface-groups (Property 2). For the fact that the fundamental group of a surface with boundary is free as well, just take a fundamental polygon $P$ defining your surface, and remove $n \geq 1$ points. Then, notice that the punctured surface deformation retracts to a graph, whose fundamental group is free.
Aug
15
comment Show the wedge product of two tori is not homotopy equivalent to a 2-manifold.
Erratum: of course, in my previous comment, "manifold" should be replaced with "surface".
Aug
15
comment Show the wedge product of two tori is not homotopy equivalent to a 2-manifold.
The fundamental group of a non-compact manifold or of a compact manifold with boundary is necessarily free. On the other hand, $\pi_1(\mathbb{T} \wedge \mathbb{T}) \simeq \mathbb{Z}^2 \ast \mathbb{Z}^2$ is not free.
Aug
14
comment No continuous function switches $\mathbb{Q}$ and the irrationals
I think so. But now, your solution does not seem to be really simpler than the original.
Aug
12
comment No continuous function switches $\mathbb{Q}$ and the irrationals
If $f([0,1])=[-1,1]$, then you are suggesting to consider $g(x)=f(x)$ in order to find a function $[0,1] \to [0,1]$, so it doesn't seem to work.
Jul
31
answered Open math problems which high school students can understand
Jul
20
revised Isometric group of hyperbolic 3-dim manifolds
added 8 characters in body; edited tags
Jul
20
answered Isometric group of hyperbolic 3-dim manifolds
Jul
18
comment $G=\langle x,y\ |\ x^{-1}y^2x=y^{-2}, y^{-1}x^2y=x^{-2} \rangle$ is torsion-free.
For me, it seems very natural, because a motivation behind small cancellation is to determine whether or not an element belongs to the normal closure of some subgroup. Afther that, I agree that my answer is probably not the simplest solution to the problem. But I found interesting to give a "standard" approach to this kind of question.
Jul
18
answered $G=\langle x,y\ |\ x^{-1}y^2x=y^{-2}, y^{-1}x^2y=x^{-2} \rangle$ is torsion-free.
Jul
1
awarded  Yearling
Jun
20
accepted When is a right-angled Coxeter group one-ended?
Jun
19
answered When is a right-angled Coxeter group one-ended?
Jun
19
comment When is a right-angled Coxeter group one-ended?
Six hundred pages on the geometry of Coxeter groups, what a huge work! Thank you for the reference, I didn't know it. I will add an answer applying these results to right-angled Coxeter groups. +1
Jun
19
asked When is a right-angled Coxeter group one-ended?
Jun
7
revised Monomorphisms into direct products
added 43 characters in body
Jun
7
asked Monomorphisms into direct products
Jun
4
revised Presentation of the additive group of the rational numbers
added 14 characters in body
Jun
4
answered Presentation of the additive group of the rational numbers