73,396 reputation
493185
bio website
location Buenos Aires, Argentina
age 40
visits member for 4 years, 6 months
seen 3 mins ago

28m
comment how to prove that two sets have the same homotopy type
In any case, what you are talking about in the third paragraph of your question has very little to do with homotopy equivalences... In general, it is rather non trivial to show that two sets have the same homotopy type. Without being more specific about what sets you have in mind, there is pretty much nothing more useful we can say that «Look for a homotopy equivalence».
42m
comment How to “find” this Lie algebra: proof that $\mathfrak{sl}$ is trace zero matrices
Keep in mind thatmthe exponential map is not a homomorphism except in extremely special circumstances.
43m
answered How to “find” this Lie algebra: proof that $\mathfrak{sl}$ is trace zero matrices
1h
comment Reducibility of Representations over Finite fields
Compute its endomorphism ring, for example.
1h
comment Citation: earliest incidence of the Borel localization theorem
Hsm.se? ${}{}{}{}$
1h
comment Is every $S^1$ knot orientable?
But before you read up on orientations you must decide whether $\mathbb Z$ is isomorphic to one of its subgroups or not!
2h
comment The quotient map and isomorphism of cohomology groups
@LewisZhang, as we observed in the comments above yours, that fact is in fact probably more difficult than the one you want to prove.
5h
comment Proving that two summations are equivalent
You do not want to prove those two expresions are equivalent: you want to prove they are equal.
6h
comment Torsion elements of a module not a subgroup
Any textbook which covers modules.
6h
comment Torsion elements of a module not a subgroup
If $A$ is a ring, then there is a canonical way of considering $M=A$ as an $A$-module. You should certainly review this...
6h
comment On finite generation of certain $\operatorname{Ext}$'s
As for your question regarding the finite generation of Hochschild cohomology: for example, if $A\otimes A$ is noetherian, then those Ext's are finitely generated (if $A$ has a resolution by finitely generated bimodules, this works also, and sometimes you have this and not noetherianity of $A\otimes A$). Notice that the noetherianity of $A$ does not imply in general that of $A\otimes A$ (sadly!)
7h
comment Homology group $H_1(G;\mathbb{R})$ is a vector space?
Before doing anything, ypu should consider the question of whether those maps do or do not form a vector space. It is quite a basic matter of good ol' linear algebra, which you should have a firm grasp before even thinking about the homology of anything!
7h
comment Homology group $H_1(G;\mathbb{R})$ is a vector space?
Homology with coefficients in R is not the abelianization, actually.
8h
comment Mipoint between points in projective space
Well, that depends on what properties exactly you want that 'distance' to have. If you want a distance which is projectively invariant, then no.
8h
comment Mipoint between points in projective space
I mean precisely what my second sentence says :-) All the points in the line joining your two points are indistinguisable from the point of view of projective geeometry.
8h
comment Questions of Hyperspace of Compact Sets
You are taking limits in $K(X)$ but never told us what topology $K(X)$ has.
8h
comment Mipoint between points in projective space
No, not if you want it to have any projective characterization. Indeed, the projective group acts transitively on the set of points of the line joining your two points (with them excluded)
10h
comment The quotient map and isomorphism of cohomology groups
I did not say that the 1st one depends only on excision: what you mention is the little work I referred to, but the basic fact is that you can use excision to see that the relative groups involved do not depend much on what's inside of the X-B. As I daid, the 2nd one you can get by shoing that you can extend orientations to the whole of B; see the treatment of orientability on manifolds in the book by Greenberg and Harper.
20h
answered I need to identify $G/H$ upto isomorphism.
21h
answered Galois comodules