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Minä puhun varias linguas mais toutes very poorly: seien Sie nett! Teşekkür ederim و ¡hasta luego!


Feb
26
awarded  Pundit
Feb
14
awarded  Yearling
Nov
19
revised Is the quotient map a homotopy equivalence?
added 116 characters in body
Nov
19
comment Is the quotient map a homotopy equivalence?
Oh, drat. You're right, of course. I wanted to avoid technicalities, but I can't. I will modify the question. Thank you very much.
Nov
18
awarded  Promoter
Nov
15
comment connected sum of torus with projective plane
No worries. I was genuinely curious.
Nov
15
comment connected sum of torus with projective plane
Nice. Where do these pictures come from?
Nov
15
comment Is the quotient map a homotopy equivalence?
Yes: for me, contractibility implies connectedness.
Nov
15
asked Is the quotient map a homotopy equivalence?
Nov
15
comment fundamental group of a graph is free
It's proposition 0.17 in Hatcher's book, but you still need some kind of homotopy extension.
May
1
comment Braid Group of a Weyl Group
Do you know the usual braid group? It corresponds to the case where the Weyl group is simply the symmetric group. en.wikipedia.org/wiki/Braid_theory The omission of the relations T^2 = 1 really changes the structure of the group...
Apr
26
comment Proving Two Complexes are Not Quasi-Isomorphic
Short story shorter: $K$ is abstractly isomorphic to $\Bbb C[x,y]$, but no ambient morphism $\Bbb C[x,y] \oplus \Bbb C[x,y] \to \Bbb C[x,y]$ restricts to such an isomorphism.
Apr
26
answered Proving Two Complexes are Not Quasi-Isomorphic
Apr
26
comment If $ G $ has no non-trivial automorphism, then $ G $ is abelian and $ g^2 = e $ for all $ g \in G $ .
If G is infinite, you have to believe in Choice! math.stackexchange.com/questions/28145/…
Apr
25
comment Compactness of covering space
Yes. mathoverflow.net/questions/32111/…
Apr
25
comment Van Kampen with complicated attaching map
en.wikipedia.org/wiki/Lens_space
Apr
25
comment Van Kampen with complicated attaching map
I don't understand your example: in that case, you know that $U \cap V$ looks like the common boundary of $U$ and $V$, so a torus $S^1 \times S^1$, don't you?
Apr
25
comment If $ G $ has no non-trivial automorphism, then $ G $ is abelian and $ g^2 = e $ for all $ g \in G $ .
The first step is to show that the group is Abelian, using the fact that the inner automorphisms are trivial. Then, you have won something: if (and only if) a group is Abelian, the inversion $x \mapsto x^{-1}$ is an automorphism. Using the hypothesis, you then have $\forall x \in G, x^{-1} = x$, which is equivalent to all elements of the group being involutions.
Apr
25
comment Prove that the only operator on $\mathbb{C}$ for which his inner product is zero is zero
An orthogonal projection on $\Bbb R$ isn't a counterexample (look at vectors in the image). A $90^\circ$ rotation in the plane is.
Apr
24
comment Binary tetrahedral group and $\rm{SL}_2(\mathbb F_3)$
@QiaochuYuan: I doubt it: finite subgroups of $\rm{GL}_2(\mathbb Z)$ preserve a scalar product, so they are cyclic or dihedral.