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Dec
29
comment Can we construct $\Bbb C$ without first identifying $\Bbb R$?
...in particular, the answer is yes: first construct $\bar{\mathbb Q}$, then take trans. extension of size continuum.
Dec
29
comment Closed form for series $\sum_{m=1}^{N}m^n\binom{N}{m}$
@Mhenni Be it as it may, next time you learn a question you've answered is a duplicate please mark it as such.
Dec
28
comment Closed form for series $\sum_{m=1}^{N}m^n\binom{N}{m}$
@Mhenni AFAICS you're linking to an exact duplicate of this question — but instead of voting to close you're reposting a part of your old answer. That's borderline gaming the system for rep, IMO
Dec
27
comment How prove this $\sum_{k=0}^n \binom{n}{k} \binom{(p-1)n}{k} \binom{pn+k}{k} = \binom{pn}{n}^2 $
This is a special case of math.stackexchange.com/questions/280481
Dec
27
comment Why are higher homology groups not the abelianizations of higher homotopy groups?
Tangentially: in a sense, higher homology groups are higher homotopy groups of the abelianization of the space (i.e. by Dold-Thom/Kan $H_i(X)\cong\pi_i(\mathbb Z[X])$).
Dec
26
comment Concerning Adam's theorem/binary continous operation on Spheres
For $S^0$ / $S^1$ / $S^2$ / $S^3$ / $S^7$ binary operation is the product of unit norm real numbers / complex numbers / quaternions / octonions respectively
Dec
26
comment Prove that $\sum_{k=0}^{m}\binom{m}{k}\binom{n+k}{m}=\sum_{k=0}^{m}\binom{n}{k}\binom{m}{k}2^k$
cf. math.stackexchange.com/q/461762
Dec
26
comment Law of sines: uniform proof of Euclidean, spherical & hyperbolic cases
Could you please add at least a summary of Bär's uniform proof?
Dec
25
comment Manifolds with vanishing Stiefel-Whitney classes but are not stably parallelizable
Related: math.stackexchange.com/questions/46297/…
Dec
25
comment Steenrod Algebra as automorphisms of additive group
Related: mathoverflow.net/q/83096
Dec
24
comment Number of Sets of Partitions
it's clear from the example that question is not about [formula for] the partition function
Dec
23
comment Reference to finite coverings causing injections on deRham cohomology
related: math.stackexchange.com/questions/184529/…
Dec
22
comment Why is ${n-1 \choose -1} = 0$ when (-1)! is undefined?
You also can't choose k things from $-5$ or $3/2$ object and yet binomial coefficients $\binom{-5}k$ and $\binom{3/2}k$ are [usually defined to be] non-zero.
Dec
22
comment How to prove that the sum of squared binomials equals $\binom{2n}{n}$
or of math.stackexchange.com/questions/373122/… or of math.stackexchange.com/questions/320348/… or of math.stackexchange.com/questions/404715/… ...
Dec
22
comment Understanding the Definition of the Tensor Product of Chain Complexes
...details can be found in most algebraic topology textbook (e.g. in Hatcher's). And IMHO it's very useful in general to study (some) algebraic topology in parallel with homological algebra.
Dec
22
comment Understanding the Definition of the Tensor Product of Chain Complexes
@Exterior For cellular homology this is [not some complicated quasiisomorphism but] an almost tautological isomorphism: on the level of elements that's just $\mathbb R^n\times\mathbb R^m=\mathbb R^{n+m}$ — and for the differential that's exactly what motivates the definition of the differential in $C\otimes D$. As for the Kunneth formula, aforementioned lemma allows one apply Kunneth formuma from homological algebra (computing $H(C\otimes D)$ in terms of $H(C)$ and $H(D)$) to compute $H(X\times Y)$.
Dec
22
comment Understanding the Definition of the Tensor Product of Chain Complexes
@Exterior Details depend on the version of (co)homology you're using — but if, say, $X$ and $Y$ are CW-spaces then $C(X\times Y)\cong C(X)\otimes C(Y)$ where $C(-)$ is the cellular chain complex.
Dec
22
comment Understanding the Definition of the Tensor Product of Chain Complexes
If you're computing (co)homology of topological spaces, tensor product of complexes corresponds to the usual product of spaces.
Dec
22
comment Counting subsets with r mod 5 elements
Dear Adriano, that edit a) was quite unnecessary; b) title now looks quite ugly. I'm rolling back.
Dec
21
comment An example where $\frac{(2m)!(2n)!}{m!n!(m + n)!}$ is the number of ways of counting something?
(For me the interpretation is combinatorial enough, but the proof is not. It would be nice, of course, to have a combinatorial proof of the fact that $S(m,n)$ indeed counts these paths.)