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Aug
26
comment Full subcategory of abelian category is abelian
@RobertCardona I misread the definition of $g$, I thought it was defined as the cokernel. But in any case, you now have an answer below, so it doesn't matter anymore, I guess
Aug
26
comment Full subcategory of abelian category is abelian
I think it would follow from the universal property as follows: since $\mathrm{coker} f$ and $g$ are both cokernels for $f$, you can show that there exists an invertible morphism $\theta$ such that $\theta g = \mathrm{coker} f$ (existence follows from $g$ being a cokernel; invertibility follows from applying the universal property to both). Therefore $$f = \mathrm{ker} \theta g = \mathrm{ker} g.$$ But I don't see how you can conclude that $g$ is a morphism in $\mathcal{S}$, unless $\mathcal{S}$ is strictly full (the codomains of $g$ and $\mathrm{coker} f$ are isomorphic, via $\theta$).
Aug
24
revised Meaning Behind Mapping from a Compact Subset to Another Set
formatting
Aug
24
comment Average of sum and Sum of average of 2 random non zero distributions of numbers
I think this could be a quite difficult problem to study... For example, if you take mean zero and variance 1, for both $A$ and $B$, then the ratio $A_i/B_i$ is a standard Cauchy (so the average is also a standard Cauchy) and the ratio of sums is also Cauchy (I think). So even studying their correlation does not make sense...
Aug
17
comment Analytic continuation of Gamma function
Did you have a look at this other MSE question: math.stackexchange.com/questions/159333/…?
Jun
30
comment Boundedness in a topological space?
@mam I don't agree. We may disagree on the definition, but I would say Concept B is an extension of Concept A if there exists a universe where they are both defined and equivalent, or I would even say that, whenever they are both defined, they should then be equivalent. "Compactness" and "boundedness" are almost never equivalent (not even on the real line), and on metric spaces you can only prove that compact sets are bounded... I don't think it helps anyone to think that either is an extension of the other, it just adds to the confusion...
Jun
30
comment Boundedness in a topological space?
@mma yes, open balls form a basis for the topology generated by the metric. But all your comments are aimed at metric spaces... there are no open balls in a general topological space... and no notion of boundedness
Jun
29
comment Boundedness in a topological space?
@mma In a discrete metric space, all sets are closed and bounded, and a set is compact if and only if it is finite. Therefore, all infinite sets are bounded with non-compact closure.
Jun
28
comment Boundedness in a topological space?
@mma You're absolutely right, I can't believe this typo has been there all this time...
Jun
28
revised Boundedness in a topological space?
typo
Apr
28
awarded  Popular Question
Apr
17
revised Instructive proofs in functional analysis
clarified what looked like a confusing sentence
Mar
30
comment Constructive proof of the existence of an algebraic closure
@Rubertos You can show that there exists a set that contains all algebraic extensions up to isomorphism
Jan
11
comment What properties do you lose when you extend your number set?
@fvel If they don't fully answer your questions, you should explain why. Then maybe someone can answer whatever is left that you want to know.
Jan
9
revised What properties do you lose when you extend your number set?
edited tags
Jan
9
answered What properties do you lose when you extend your number set?
Dec
23
revised function such that $f(x\cdot t)=f(x)g(t)$
the title was confusing, and different from the statement of the question
Dec
23
comment Proof If $AB-I$ Invertible then $BA-I$ invertible.
@JaVaPG Yes, exactly. But as MPW (and Kaladin) mentioned, the two problems are basically equivalent (since $AB-I=-(I-BA)$).
Dec
23
comment Proof If $AB-I$ Invertible then $BA-I$ invertible.
@JaVaPG Correct.
Dec
23
revised Proof If $AB-I$ Invertible then $BA-I$ invertible.
improved formatting