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seen Jul 9 at 0:23

Jul
2
awarded  Curious
May
21
comment showing exact functors preserve exact sequences (abelian categories, additive functors, and kernels)
Let $\phi$ denote the monomorphism. Then exactness at $FA$ tells us $\ker Fg = \operatorname{im} \phi$. And then maybe it helps if we identify $\operatorname{im} \phi$ with $F(\ker f)$? (Maybe part of my confusion is due to the fact that I'm still used to thinking of kernels and images as objects?)
May
21
comment showing exact functors preserve exact sequences (abelian categories, additive functors, and kernels)
Hmm, exactness at $F(\ker f)$ means that $F(\ker f) \to FA$ is a monomorphism. Our goal is to show that $\ker Fg = \operatorname{im} Ff$, so we need to extract some information about $\ker Fg$ from the monomorphism $F(\ker f) \to FA$, somehow?
May
21
asked showing exact functors preserve exact sequences (abelian categories, additive functors, and kernels)
Jan
16
awarded  Yearling
Aug
14
awarded  Popular Question
Jun
26
comment How to show that this function (related to the zeta function) is even?
Sorry, what do you mean by "harmonic sum"? A sum that arises when dealing with Fourier series?
Jun
25
comment How to show that this function (related to the zeta function) is even?
Wow, thanks! Could you explain how you were able to come up with all of that?
Jun
13
asked How to show that this function (related to the zeta function) is even?
May
21
asked Does $f, f' \in L^1([0, \infty))$ imply that $\lim_{x \to \infty} xf(x) = 0$?
May
6
awarded  Caucus
Apr
14
accepted Transpose of a linear operator on functions
Apr
11
comment Transpose of a linear operator on functions
Ah, thanks! I was thinking of using the divergence theorem, but I wasn't sure if it was applicable (partly because I didn't know what was the exact space of functions $L$ was acting on).
Apr
11
asked Transpose of a linear operator on functions
Feb
17
awarded  Popular Question
Jan
30
awarded  Enthusiast
Jan
23
revised Alternative to Axler's “Linear Algebra Done Right”
edited tags
Jan
22
awarded  Nice Answer
Jan
22
comment Why is $y = \sqrt{x-4}$ a function? and $y = \sqrt{4 - x^2}$ should be a circle
@vincentbelkin, I edited my answer.
Jan
22
comment Why is $y = \sqrt{x-4}$ a function? and $y = \sqrt{4 - x^2}$ should be a circle
@AsafKaragila, thanks!