Bruno Stonek

less info
reputation
11244
bio website bruno.stonek.com location Montevideo, Uruguay age 24 member for 3 years, 5 months seen 4 hours ago profile views 1,414

Math student in Université Paris 13.

781 Actions

 Jun18 comment if $s \implies \lnot w$, does $\lnot s \implies w$? If it is sunny, it doesn't rain. But if it doesn't rain, it doesn't mean it has to be sunny. --Edit: ah, I hadn't seen Didier's answer below ;P Jun15 comment Why is an empty function considered a function? ...so, the initial element in Set (--see en.wikipedia.org/wiki/Initial_and_terminal_objects) Jun15 comment Equivalent conditions for a preabelian category to be abelian @Theo: Sorry, I hadn't had time to finish checking the argument. I did now, it's perfect. Thank you once again, you've been of huge help! Jun15 accepted Equivalent conditions for a preabelian category to be abelian Jun14 comment Equivalent conditions for a preabelian category to be abelian @Theo: If I understand correctly, since $e=coker(f)$ and $y$ is another map such that $yf=0$, then there exists a unique $y'$, by definition of cokernel, no? I don't see how it is needed to use again that $e$ is epi. Jun14 comment Equivalent conditions for a preabelian category to be abelian @Theo: Thanks! I'm digesting your answer. I don't understand this part, though: "Since e is epi, this factorization is unique and hence e is a cokernel of k." Jun14 comment Equivalent conditions for a preabelian category to be abelian @Theo: again, thank you very much. I'm absorbing your previous comment right now. I don't quite get your "on the other hand", it seems you're using that $g$ is a retraction, but that isn't necessarily so. In any case, dually you get a right inverse, and existence of right and left inverse implies iso. Also, you're using that if f is epi then it's the cokernel of its kernel: I didn't know that (I think that's what Freyd proves on 2.11... "Ker and Cok are inverse functions", come on...) Jun14 comment Equivalent conditions for a preabelian category to be abelian @Theo: I still can't do it. The answer seems to be hidden in Freyd's book, which I find quite unreadable. I put it aside and tried to prove that $\overline{f}$ is mono-epi, so this would prove that it is iso. I couldn't do it, and in fact, I'm also unable to prove that in an abelian category mono-epi implies iso. What a failure! I wonder whether it is true, in fact, that mono implies section and epi implies retraction? Jun12 comment Equivalent conditions for a preabelian category to be abelian @Theo: I see. Combining Freyd's theorems 2.12, 2.18 and 2.18* with Mac Lane's CWM proposition 1 in page 195 (the analogue of Freyd's "unique factorization"), I believe I can write a tidy proof. I shan't do so today, though. Thank you for your assistance. Jun12 comment Equivalent conditions for a preabelian category to be abelian @Theo: Thank you, I'm aware that product and coproduct coincide with biproduct. I shall take a look at Freyd; I already had, but I didn't recognize it under that name and (to my taste) awkward wording. Jun12 comment Equivalent conditions for a preabelian category to be abelian @Theo: would you consider posting an answer? Thank you very much. Jun12 comment Equivalent conditions for a preabelian category to be abelian @Theo: I'm having difficulties proving that $\overline{f}$ is mono, anyway. I'm trying to prove that $Ker(\overline{f})=0$ by using the universal property of ker, but I'm not seeing how every other map $X\to Coim(f)$ factors through zero... Jun12 comment Equivalent conditions for a preabelian category to be abelian @Theo: I'm trying. But is it true in an abelian category that mono-epi is iso? Jun12 asked Equivalent conditions for a preabelian category to be abelian Jun12 comment Additive category that is not abelian The problem with free modules over a ring is that submodules of free modules need not be free? If that's the case, the category of all free modules over a PID is abelian... please correct me if I'm wrong. Jun8 comment How to prove uniform continuity? Observe that what you're actually proving is that a differentiable function with bounded derivative is Lipschitz-continuous (and that Lipschitz-continuity implies uniform continuity -- see en.wikipedia.org/wiki/Lipschitz_continuity ) Jun8 accepted For an integrable function $f$, do continuity conditions on its integral affect continuity of $f$? Jun7 comment For an integrable function $f$, do continuity conditions on its integral affect continuity of $f$? @user9176: edited to clarify. @Theo: thank you for a very detailed and informative answer! Jun7 revised For an integrable function $f$, do continuity conditions on its integral affect continuity of $f$? added 78 characters in body Jun7 revised For an integrable function $f$, do continuity conditions on its integral affect continuity of $f$? deleted 259 characters in body; edited title