Mariano Suárez-Alvarez
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Jul
25
comment different definitions of Hopf algebras
Hatcher's Hopf algebras are graded simply because his Hopf algebras come from the homology of H-spaces.
Jul
25
comment different definitions of Hopf algebras
One can show that Hatcher's Hopf algebras have an antipode. This follows from gradedness. Classically, the Holf algebras under consideration were graded, so the antipode was not mentioned. This should be discussed in any textbook on the subject, like the one by Susan Montgomery.
Jul
25
comment Does every complex have a quasi-isomorphic projective complex?
For this you need the complex to start somewhere.
Jul
24
comment Do analytic functions on open subsets of $\mathbb{C}$ with an analytic square root form a sheaf?
You are not understanding the definition of the presheaf $\mathcal F$. If $U$ is an open set, then $\mathcal F(U)$ is the set of all homolorphic functions $h:U\to\mathbb C$ such that there exists an holomorphic $g:U\to\mathbb C$ with $h=g^2$. You are trying to glue square roots, and there is no reason to do that...
Jul
24
comment Do analytic functions on open subsets of $\mathbb{C}$ with an analytic square root form a sheaf?
If $D$ is a disc contained in $\Omega$, then $f|_D$ has a holomorphic square root, so that $f|_D$ is in $\mathcal F(D)$. This gives us a family $\{f|_D\}_{\text{$D$ a disc contained in $\Omega$}}$ of sections of the sheaf over a covering of $\Omega$. It is obvious that they glue correctly. Yet there is no section of the sheaf on the whole of $\Omega$ which restricts to them.
Jul
24
comment Do analytic functions on open subsets of $\mathbb{C}$ with an analytic square root form a sheaf?
I don't understand your question.
Jul
24
answered Do analytic functions on open subsets of $\mathbb{C}$ with an analytic square root form a sheaf?
Jul
24
comment Characterisation of the squares of the symmetric group
Since $S_n^2$ is clearly invariant under conjugation, it is a group exactly when it is equal to $A_n$, provided $n\geq5$.
Jul
23
comment Free and projective modules
If $R$ is a field, then $R^S$ is free for all $S$,simply because all $R$-modules are free in that case. You can take more generally $R$ to be any semisimple ring, like $M_n(\mathbb R)$; in this case all modules are projective.
Jul
23
comment Free and projective modules
What definition of free module do you have? Most definitions make your first question immediate!
Jul
23
comment Can we speak of derivatives of sets?
Once you fix $x$, the limit is that of a function which takes values in $\{-1,0,1\}$.
Jul
23
comment Can we speak of derivatives of sets?
Ifthat limit exists it is zero.
Jul
23
comment Prove that the 4-group V is normal subgroup of $S_4$ by using isomorphism theorem
Check that and that $vVv^{-1}=V$.
Jul
23
comment Why does the Big Oh (and similar) notations needs $n_0$?
What if $g(1)=0$?
Jul
23
answered Prove that the 4-group V is normal subgroup of $S_4$ by using isomorphism theorem
Jul
23
answered Prove that the 4-group V is normal subgroup of $S_4$ by using isomorphism theorem
Jul
23
comment Is it the case that each $\Phi^q$ of a stable cohomology operation $\{\Phi^q\}$ is a natural homomorphism?
(Most definitions I have seen make the answer to this question trivially yes for cohomology operations for the naturality part and no for the homomorphism part, and yes and yes for stable cohomology operations. See May's Concise Course, end of chapter 22, for example)
Jul
23
comment Is it the case that each $\Phi^q$ of a stable cohomology operation $\{\Phi^q\}$ is a natural homomorphism?
I think you should tell us exactly what definition of stable coh. operation you are using and, ideally, where you get stuck in trying to prove this.
Jul
23
comment Does the series $\displaystyle\sum_{j=1}^\infty -\log (1-p_j^{-3/4})$ diverge, where $\{p_j\}$ is the set of primes in increasing order?
Since the terms of the series are all positive for large $j$, the order of the terms does not really matter.
Jul
23
comment matrices in the form $AB-BA$
See this answer by Bill:math.stackexchange.com/a/99356/274