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Mar
8
comment Poincare Duality Reference
The fact that the star complexes constitutes a cellular filtration only depends on the homology of the manifold, just as Serfert & Threlfall shows. I cannot see how to determine the topology, not just homology of that, and how to take advantage of the tubular neighborhood you've mentioned. In addition, sorry for my ignorance, I don't know the tubular neighborhood theorem for triangulated manifolds, but not for differentiable manifolds.
Mar
8
comment Poincare Duality Reference
Sorry, I cannot follow your idea. I want to go back to the language introduced in Seifert and Threlfall. The dual cell decomposition is constructed as follows: there's a point, the barycenter, associated to each $n$ dim simplex, then a $1$ dim star complex associated to each $n-1$ dim simplex, whose center is the barycenter of the complex and whose outer boundary is the totality of star complexes associated to each incident $n$ dim simplex, and so forth. Maybe it's easy to prove that such a star complex is a cellular decomposition, i.e. $H_m(X_{n+1},X_n)=0$ for $m\neq n$.
Mar
8
revised The hyper-derived functors $\mathbb L_\bullet F$ are just derived functors of $H_0F$?
added 2 characters in body
Mar
7
comment Four Color Theorems: Graphs vs. Maps
In fact, it's just a matter of duality, therefore the conditions of duality theorems should be satisfied. For more details, see Alexander duality, or Poincaré duality, etc. I'm not familiar with these stuff.
Mar
7
asked The hyper-derived functors $\mathbb L_\bullet F$ are just derived functors of $H_0F$?
Mar
7
comment Four Color Theorems: Graphs vs. Maps
Since the graph is embedded as a subset of $\mathbb R^2$, therefore the complement is just the set-theoretic complement. For example, if the graph is $S^1$ embedded in $\mathbb R^2$, then the complement is of two connected components (Jordan curve theorem).
Mar
7
comment Four Color Theorems: Graphs vs. Maps
It seems to me that a (finite) planar graph is a (finite) graph embedded in $\mathbb R^2$. The regions are the connected components of the complement of the embedded graph.
Mar
7
revised A quasi-isomorphism between the total complex of a Cartan-Eilenberg resolution and the complex per se.
added 32 characters in body
Mar
7
comment A quasi-isomorphism between the total complex of a Cartan-Eilenberg resolution and the complex per se.
@ZhenLin Yes, since for fixed $n$, in order to compute $H_n$ etc, we only need a finite portion.
Mar
7
asked A quasi-isomorphism between the total complex of a Cartan-Eilenberg resolution and the complex per se.
Mar
7
comment Homology and (co)Limits
@Exterior In general, chain homology functor commutes with exact functors, a consequence of the FHHF theorem (cf. Ravi Vakil's notes, FOAGjan2915, 1.6.H). If, instead of exact functors, consider arbitrary right exact functor, then maybe a spectral sequence intervenes.
Mar
7
comment Poincare Duality Reference
@AndrewMarshall Well, with these abstract nonsense (say five lemma or even spectral sequence argument), the combinatorial meaning is hidden. I think that original approach is still valuable.
Mar
7
comment Poincare Duality Reference
Is there any reference for the proof that the dual decomposition is really a CW-decomposition? Apparently, it depends on the fact that the original simplicial complex is a manifold. I don't know how to take advantage of this homogeneity explicitly.
Mar
7
comment Poincare Duality Reference
In which section of Schubert's book? I didn't find that.
Feb
24
revised Convergence in measure and convergence of norm implies convergence in L^p
added 30 characters in body
Feb
24
answered Convergence in measure and convergence of norm implies convergence in L^p
Feb
24
comment Commutative ring where $r$, $s$ are associates but $r \neq us$ for any $u$ unit.
Consider the ring $A=\mathbb Z[X,Y,Z]/(XYZ-X)=\mathbb Z[x,y,z]$, where $x,y,z$ are residues of $X,Y,Z$ in $A$.
Feb
24
asked Long exact sequence for a triple follows from long exact sequence for a pair?
Feb
23
awarded  Nice Question
Feb
23
comment If $G$ is an uncountable group and $H$ is a subgroup then $G$ \ $H$ is uncountable ?
But different $gH$'s are disjoint, therefore $G\setminus H$ is uncountable.