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bio website math.cornell.edu/~hatcher
location Cornell University
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visits member for 5 years
seen 36 mins ago

Nov
13
awarded  Yearling
Nov
4
revised Reference for a fact (?) on homeomorphic knot complements
deleted 40 characters in body
Nov
3
awarded  Good Answer
Oct
30
awarded  Nice Answer
Oct
30
awarded  Nice Answer
Oct
30
answered How to compute $\pi_{3}$ of $L(p,q)\# L(p',q')$?
Oct
30
revised Reference for a fact (?) on homeomorphic knot complements
deleted 47 characters in body
Oct
30
answered Reference for a fact (?) on homeomorphic knot complements
Oct
22
comment A homological criterion for collapsibility?
The lemma as stated certainly seems false, with the counterexamples you described such as the house with two rooms, with $A$ a point. Surely K & S knew examples like this, so I wonder what they really had in mind when they stated this lemma.
Sep
21
comment Contractibility of space of embeddings of a disc
@IgorBelegradek: The formula seems to work equally well for embeddings $D^2\to {\mathbb R}^2$ and diffeomorphisms of ${\mathbb R}^2$. For embeddings the term $f(tx)$ is in effect restricting embeddings to smaller and smaller concentric disks, so one still gets embeddings, and then the $1/t$ factor re-expands to keep the derivative the identity at the origin. The limit exists, just as for diffeomorphisms, and is the identity.
Sep
21
comment Contractibility of space of embeddings of a disc
For question 2 the space is not contractible. There is a map from the space of topological embeddings to ${\mathbb R}^2 - \{0\}$ given by $f\mapsto f(1,0)$ and this map splits (i.e., has a section) by considering only embeddings which are orientation-preserving Euclidean similarities. So the space is not simply-connected. An extra condition analogous to the condition on the derivative in question 1 is needed, though it is perhaps not obvious what the best way to state this condition is.
Sep
21
comment Contractibility of space of embeddings of a disc
Please clarify what you mean by "proper". Usually a proper embedding is one which takes boundary to boundary, which in this case would force the embeddings to be diffeomorphisms or homeomorphisms, respectively, in the two cases.
Sep
18
awarded  Enlightened
Sep
9
comment Existence of $n$-connected topological groups with $m$-dimensional action extending that of $GL(m)$ on $\mathbb{R}^m$
A simple observation: $F$ must be non-orientable, since if it is orientable, then $G$, being path-connected, must act by orientation-preserving homeomorphisms, contrary to the fact that the action of $GL(m)$ on ${\mathbb R}^m$ includes orientation-reversing homeomorphisms.
Sep
6
comment “Economic” CW-structure for Eilenberg-MacLane spaces?
JUst to clarify: The usual construction of a homology decomposition, as in my algebraic topology book for example, shows that for each simply-connected space $X$ with finitely generated homology groups there is a CW complex $Y$ and a weak homotopy equivalence $Y\to X$, where $Y$ has the minimum number of cells in each dimension consistent with the structure of $H_*(X;{\mathbb Z})$, namely, 1 cell for each infinite cyclic summand of the homology and 2 cells for each finite cyclic summand. This applies in particular for $K(\pi,n)$'s with $n>1$ and $\pi$ finitely generated.
Sep
5
comment Inverse cohomological isomorphisms
@Lennart Meier: Yes, thanks. Corrected now.
Sep
5
revised Inverse cohomological isomorphisms
edited body
Sep
4
answered Inverse cohomological isomorphisms
Sep
4
comment Inverse cohomological isomorphisms
Correction: The universal covering map $S^3 \to SO(3)/I$ has degree 120, the order of the fundamental group of $SO(3)/I$.
Aug
24
awarded  Good Answer