69,167 reputation
486171
bio website
location Buenos Aires, Argentina
age 40
visits member for 4 years
seen 3 hours ago

Jul
20
comment What does it mean to demonstrate “por doble inclusión” in Spanish?
If you want to be fancy, you can also say «probar por inclusión mutua»...
Jul
20
answered What does it mean to demonstrate “por doble inclusión” in Spanish?
Jul
19
comment find matrix A given its exponential
Exactly. (A more gentlemanly way of putting it is to observe that $A=\tfrac{d}{dt}(e^{At})(e^{At})^{-1}$)
Jul
19
comment manifold structure on a finite dimensional real vector space
@Berci, your comment is akin to telling «don't worry» to someone who is worrying: useless.
Jul
19
comment Fibonacci series, which is most pure mathematically?
Thre is no such thing as being «most mathematically pure»...
Jul
17
comment Kähler differentials of the cuspidal cubic
Rank is additive in short exact sequences; $A_{(X,Y)}\operatorname{d}X \oplus A_{(X,Y)}\operatorname{d}Y $ has rank $2$. What is the rank of the denominator?
Jul
16
answered an ideal of matrix ring which is projective
Jul
16
comment Does $f(dx)$ have any meaning?
Urgh. ${}{}{}{}$
Jul
15
comment Proper terminology for a pair of magmas
A «pair of magmas»? Why do you expect it to have a name?
Jul
14
comment Complex (topological) K-theory of $S \Sigma$ for a surface $\Sigma$.
I don't recall how involved it is to see that $\pi_2U=0$, honestly, but up to that (and the clutching construction, of course) it's very elementary.
Jul
14
comment Complex (topological) K-theory of $S \Sigma$ for a surface $\Sigma$.
Write the surface as the mapping cone of a map $S^1\to(S^1)^{\vee g}$. The map $[f]\in[\Sigma,U]\to\pi_1(f)\in\hom(\pi_1\Sigma,\pi_1U)$ is surjective because of the usual property of mapping cones, and it is injevtive because $\pi_2U=0$, I guess.
Jul
13
comment Complex (topological) K-theory of $S \Sigma$ for a surface $\Sigma$.
Another approach is to use the clutching construction, which leaves you with describing maps out of the abelianization of the fundamental group of the surface, which is very easy (and shows that most of the technology you mention is not really needed :-))
Jul
10
comment Homeomorphism are equivalence relations, so what are the equivalence classes?
You are mixing a few things, so that what you wrote does not make a lot of sense.
Jul
10
comment A question on function notation
What I do is $f:x\in X\mapsto x^2\in X$, because I really dislike the two step thing :-)
Jul
10
comment Is $H^0(S^0;G)\simeq G\oplus G$ or $G$?
if you consider the reduced version, then the statement becones false for $S^n$ with $n>0$ :-)
Jul
10
comment A question on function notation
Your second option is not very often seen in papers.
Jul
10
answered Is $H^0(S^0;G)\simeq G\oplus G$ or $G$?
Jul
8
comment Null homotopic by simplicial approximation
You mixed up $m$ and $n$, which in this question is problematic. In any case, you cannot take a point out of the domain of functions and expect anything sensible with respect to homotopy. In this case, you have little else but to show that you can take a point out of the codomain, and that is what the simplicial approximation lemma allows you to do.
Jul
8
comment How do Homology Groups work
Well, that is precisely what the equivalence relation of two cycles being homologous means! (for an appropriate interpretation of what «a cobordism within a space» means)
Jul
8
comment If $A,B$ are equinumerous, then so are their complements
If this were true, everything would be extremely boring. I don't think this is reeally unfortunate :-)