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 Jan 4 answered The Hairy ball theorem and Möbius transformations Jan 4 comment irrep of a non unit element in the finite group I like this argument but not this particular hint. Regular representation has $\rho(g) \neq E$ by the very definition, so there is nothing to conclude. The real work is in reducing into irreducibles. Jan 4 answered irrep of a non unit element in the finite group Jan 4 comment Are any of those quotient rings isomorphic to other well known rings? Yes, (1) seems to be something like $C(\beta {\bf R} \setminus {\bf R})$. Essentially, any function with non-trivial or non-existent limit corresponds to a new point in the compactification. Jan 4 revised “Proof” that $g(t) = (t,t,t,…)$ is not continuous with uniform topology Removed the wrong remark about the box topology. Jan 4 comment “Proof” that $g(t) = (t,t,t,…)$ is not continuous with uniform topology @Stefan: 1) you are correct, it seems your $B(x,r)$ generate a topology that lies strictly between uniform and box topology. I'll remove the note. 2) any open set containing $z$ must also contain $B(z, \epsilon)$ for some $\epsilon$ (that's the definition of basis). Jan 4 comment “Proof” that $g(t) = (t,t,t,…)$ is not continuous with uniform topology @Daniel: done. And thanks for the discussion. I stared at this question for quite some time until I realized where the problem is; ..it's so natural to take openness for granted :) Jan 4 revised “Proof” that $g(t) = (t,t,t,…)$ is not continuous with uniform topology Added explanation of unopenness Jan 4 comment “Proof” that $g(t) = (t,t,t,…)$ is not continuous with uniform topology @Daniel: ah, I see what you mean by the radius, fine. Still, consider the OP's $B(x, 1)$ with $x = (1, 1/2, \dots)$. $(1, 1, \dots) \in B(x, 1)$ but there is no open subset of $B(x, 1)$ containing it. Jan 4 comment “Proof” that $g(t) = (t,t,t,…)$ is not continuous with uniform topology @Daniel: are you saying that uniform topology is finer than the box topology? And also that $g$ is not continuous? Jan 3 answered “Proof” that $g(t) = (t,t,t,…)$ is not continuous with uniform topology Jan 3 awarded Good Question Jan 3 comment $H^1$ of a constant sheaf @Georges: indeed, thank you. But OP didn't mention irreducibility in their reasoning at all, therefore my comment. Jan 3 comment $H^1$ of a constant sheaf The argument you offer doesn't really use properties of either $H^1$ or $k(X)$. Are you sure you haven't just proved that $H^1(X, A)$ vanishes for any constant sheaf $A$ whatsoever (an absurdity to be sure)? Jan 3 comment $H^1$ of a constant sheaf The nerve certainly need not be a simplex, are you sure Serre said that? Jan 1 comment How prove this limit $\lim_{n\to\infty}a_{n}=0$ Let $f(x) = x$, $a_n = 1$, $b_n = 1$, $c_n = 0$. Then the conditions on the sequences are satisfied but $\lim_{n\to\infty} a_n = 1$. Jan 1 comment When is the converse of Casorati–Weierstrass false? I'd say the converse is the trivial direction. If the function has a removable singularity at a point, it's obviously bounded near it, so the image can't be dense. Similarly, if the function has a pole then its modulus must be bounded away from zero and the image will now miss the nbhd of 0. I suspect this is why the theorem is not stated as an equivalence. Jan 1 comment Why does $\operatorname{rank}(\mathbf{X})$ equal $\operatorname{rank}(\mathbf{X^TX})$? What is $\dim(\mathbf{X^TX})$? OP mentions all of this in the question, why repeat it? On the other hand, I have no idea what the real question is supposed to be. Jan 1 comment Computing complex integrals Are you familiar with the residue theorem? It lets you relate the integral over the real line to the residues of $f$ provided you can show the integral over the half circle vanishes in the limit of $r \to \infty$. Dec 24 comment Are these functions isomorphisms of the first binary structure with the second? - Fraleigh p.34 3.13, 3.15 It was not magic :) If you work on your own and try to prove e.g. injectivity, you are quickly lead to consider values at $x=0$. After that, any function that is not equal to $0$ at $x=0$ works. Thankfully, B.S. posted full details, check his answer.