bio  website  math.berkeley.edu/~ianagol 

location  Berkeley  
age  45  
visits  member for  5 years, 7 months 
seen  3 hours ago  
stats  profile views  12,393 
I'm a professor at UC Berkeley working predominantly in the field of lowdimensional topology.
1d

comment 
Injectivity of the DehnNielsenBaer map?
I guess they address the issue of homotopy/homotopy equivalence vs. diffeomorphism/isotopy in Chapter 1. 
2d

comment 
Penrose tiling substitution is bijective
This paper and it's references might shed some light on this question: ams.org/journals/tran/199634811/S0002994796016406 
2d

comment 
Injectivity of the DehnNielsenBaer map?
The mapping class group is usually defined as diffeos. modulo isotopies. To show equivalence with $Out(\pi_1(M))$, first one may show equivalence with selfhomotopy equivalences up to homotopy, and then show that homotopy equivalences up to homotopy are equivalent to diffeomorphisms up to isotopy. 
2d

comment 
Injectivity of the DehnNielsenBaer map?
As for homotopy equivalence implies homeomorphism, one may prove this by induction on a hierarchy, a la Waldhausen. Similarly for homotopy equivalent homeos. are isotopic. 
2d

comment 
Is there a non rightorderable torsionfree factor of the Braid group on 3 strands?
I don't understand your logic  why can't there be finitely many relators? 
2d

comment 
Is there a non rightorderable torsionfree factor of the Braid group on 3 strands?
Yes, those two relations (I didn't compute m and l, but this should be easy). 
May 26 
comment 
Trigonal loci in Teichmueller spaces
I think the result is essentially due to Hilden  look at the top of p. 8 of the paper, where the reference is given. 
May 26 
comment 
Is there a non rightorderable torsionfree factor of the Braid group on 3 strands?
Yes, it is easy to obtain such representations. One adds a relation of the form m^a*l^b, where r=a/b and m is the longitude, l the meridian. 
May 23 
answered  How to show whether a given knot and its mirror image are the same or not? 
May 23 
revised 
Degrees of selfmaps of aspherical manifolds
deleted 6 characters in body 
May 20 
answered  hyperbolic structure on Figure–8 knot complement 
May 18 
comment 
When can a class in $H^1(M;\mathbb{Z})$ be represented by a fiber bundle over $S^1$
In dimension 4, it might suffice that $ker(\pi_1(M)\to \mathbb{Z})$ is a PD(3) group; if it were, conjecturally this group would be an aspherical 3manifold group, hence there ought to be a homotopy equivalent 4manifold with the same fundamental group. However, the Borel conjecture is still open for 4manifolds. I'm not sure if there is a natural conjecture in the case that $\pi_2(M)\neq 0$. 
May 15 
awarded  Nice Answer 
May 6 
comment 
Kleinian groups containing an isomorphic copy of itself
Yes, Scott (and Shalen) actually prove that the manifold is homotopy equivalent to a compact 3manifold. In fact, even with torsion, it is homeomorphic to the interior of a compact orbifold by tameness  of negative euler characteristic if it is nonelementary. In higher dimensions, this strategy at least should apply to geometrically finite Kleinian groups with nonzero Euler characteristic. 
May 6 
comment 
Kleinian groups containing an isomorphic copy of itself
In 3dims. (discrete groups in $PSL_2(\mathbb{C})$), either a finitely generated group has finite covolume, in which case @YCor's comment applies, or it is infinite covolume, in which case if it is nonelementary, the Euler characteristic is $<0$ (by the Scott core theorem, finitely generated groups are finitely presented, so Euler characteristic makes sense), and therefore it cannot be isomorphic to a finiteindex subgroup of itself. So Rivin's answer is overkill at least in 3D (coHopfian is much stronger than what you're asking for). 
May 5 
comment 
Do unit quaternions at vertices of a regular 4simplex, one being 1, generate a free group?
Ok, excellent, that shows it is a lattice! 
May 5 
comment 
Do unit quaternions at vertices of a regular 4simplex, one being 1, generate a free group?
Nice description. One thing I still don't quite get is why $\Gamma_0$ is generated by the 4 given elements? I don't see how you can deduce this without a bit more information (a free group on 4 elements has many 4 generator subgroups). 
May 4 
comment 
Can every $\mathbb{Z}^2$ disk be pinballreached?
It seems that you're searching for something like "Sinai billiards". 
May 4 
comment 
Avoiding meancurvature flow dumbbell neckpinch by inflating a surface
@JosephO'Rourke: Did you mean to discuss Ricci flow? I was confused by the title of your question, since you actually seem to be looking for something related to mean curvature flow. 
May 3 
comment 
Do unit quaternions at vertices of a regular 4simplex, one being 1, generate a free group?
@QiaochuYuan: Ok, I was using the term "BassSerre tree" incorrectly. I really meant the tree associated to $SL_2$ of a local field, which is a special case of a BruhatTits building. Serre's book Chapter II section 1 is the usual reference. Shalen's notes (linked in my other comment) Chapter 3 also has a nice discussion. 