bio | website | math.cornell.edu/~hatcher |
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location | Cornell University | |
age | ||
visits | member for | 5 years, 7 months |
seen | 2 hours ago | |
stats | profile views | 2,744 |
May 11 |
answered | Multiplicative structure in the cohomological Leray-Serre spectral sequence - please elucidate a proof |
May 3 |
comment |
A Jordan Separation Theorem for Polyhedral Surfaces
The third condition in the definition of a polyhedral surface currently reads: "If an edge of a polygon in C intersects an edge of another polygon in C in a common vertex, then the two edges are also edges of a third polygon in C." This seems too restrictive since it excludes vertices of valence greater than three. Lee Mosher's answer gives the correct condition. |
Apr 5 |
awarded | gt.geometric-topology |
Apr 4 |
answered | Topological $n$-manifolds have the homotopy type of $n$-dimensional CW-complexes |
Mar 31 |
revised |
How can i change 8_19 to (3,4)-torus knot K(3,4)?
added 279 characters in body |
Mar 31 |
answered | How can i change 8_19 to (3,4)-torus knot K(3,4)? |
Feb 24 |
comment |
Maps of balls with fixed value along boundary
On the other hand, if you wish to allow only homotopies $f_t$ and $g_t$ such that $f_t=g_t$ on $S^2$ for all $t$, then such pairs $(f,g)$ are equivalent to maps $h:S^3\to X$. Homotopy classes of maps $h:S^3\to X$ without basepoint conditions are in one-to-one correspondence with orbits of the action of $\pi_1(X)$ on $\pi_3(X)$, and this set of orbits can be nontrivial. |
Feb 24 |
comment |
Maps of balls with fixed value along boundary
If you do not require $f=g$ on $S^2$ during homotopies, then the problem becomes trivial since every map $B^3\to X$ is homotopic to a constant map. Since we may assume $X$ is path-connected, this means $f$ and $g$ are homotopic. |
Feb 23 |
answered | Maps of balls with fixed value along boundary |
Feb 16 |
awarded | Nice Answer |
Feb 15 |
answered | Homology equivalence and isomorphism on $\pi_1$ not enough for homotopy equivalence? |
Feb 11 |
comment |
Inserting an open and simply-connected set between a compact set and an open set
Interesting use of language here (twice) that I don't recall seeing before: "If [X], if [Y], then [Z]." Is this equivalent to "If [X] and [Y], then [Z]"? Or maybe "If [X] such that [Y], then [Z]". |
Feb 9 |
awarded | Guru |
Feb 6 |
awarded | Good Answer |
Feb 5 |
awarded | Enlightened |
Feb 5 |
awarded | Nice Answer |
Feb 5 |
answered | Maps which induce the same homomorphism on homotopy and homology groups are homotopic |
Feb 5 |
awarded | Good Answer |
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. |