bio | website | math.cornell.edu/~hatcher |
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location | Cornell University | |
age | ||
visits | member for | 5 years, 2 months |
seen | 2 days ago | |
stats | profile views | 2,214 |
Dec 17 |
comment |
Cap product à la Poincaré
Also, $T^*$ is a cellulation (a decomposition into cells), not a triangulation as stated in the question. |
Dec 17 |
comment |
Cap product à la Poincaré
You are talking about a product that involves just homology and not cohomology, so this is not the cap product. Instead it is usually called the intersection product. A classical reference for the intersection product is the textbook by Seifert and Threlfall. A more recent textbook source is Bredon's "Topology and Geometry". Bredon attributes the intersection product to Lefschetz. The intersection product is a perfectly well-behaved product, with the right hypotheses, so it's not clear why you use the words "evil", "fails", and "ill-defined" in reference to it. |
Dec 9 |
answered | Lens spaces and generalized Petersen graphs |
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 |