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12h
comment Homology Whitehead theorem for non simply connected spaces
@studiosus Oh, looks like you're right
14h
reviewed Close Problem with a sequence with multiple integrals
14h
reviewed Close how to solve the current problem of evaluating limits approching zero
14h
comment Is there an injective continuous map $\mathbb{R}^2 \rightarrow \mathbb{R}$?
See en.wikipedia.org/wiki/Invariance_of_domain
15h
revised Topological invariance of chern classes
edited tags
15h
answered Topological invariance of chern classes
16h
comment Homology Whitehead theorem for non simply connected spaces
@Olivier The first part (the inclusion of the meridian is a homology iso) follows from Alexander duality. The reference for the second part (codimension 2 sphere is unknotted iff its complement is homotopy equivalent to $S^1$) is, I believe, J. Levine, Unknotting spheres in codimension 2, Topology 4, 9–16 (1965)
17h
answered Homology Whitehead theorem for non simply connected spaces
2d
reviewed Close Coefficient question in generating functions
2d
reviewed Close permutation with four fixed numbers
2d
reviewed Close Gentle introduction to discrete vector field
2d
reviewed Close Closed form questions
2d
comment Homology Whitehead theorem for non simply connected spaces
(tangentially) related: Two CW complexes with isomorphic homotopy groups and homology, yet not homotopy equivalent
Apr
13
comment Complex number with 3 dimensions
@BenMillwood The link answers the OP's question «Why is there no complex number in 3 dimensions?» [and what it means exactly].
Apr
12
awarded  Enlightened
Apr
12
awarded  Nice Answer
Apr
12
comment Complex number with 3 dimensions
related: Why are the only division algebras over the real numbers the real numbers, the complex numbers, and the quaternions?
Mar
18
comment closed form for $\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+…+\binom{n}{n}$
Related: How do I count the subsets of a set whose number of elements is divisible by 3? 4?
Mar
4
comment Show that $\displaystyle\sum_{k=0}^n\binom{2n}{2k}^{\!2}-\sum_{k=0}^{n-1}\binom{2n}{2k+1}^{\!2}=(-1)^n\binom{2n}{n}$
Related: math.stackexchange.com/q/497470
Mar
3
reviewed Reject suggested edit on Prove that the intersection of convex sets is convex using the following three points…