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An amateur general topology researcher.


Nov
11
revised An implication involving filters
added 6 characters in body
Nov
11
comment An implication involving filters
@BrianM.Scott: I've corrected my question
Nov
11
revised An implication involving filters
added 97 characters in body
Nov
11
asked An implication involving filters
Nov
1
revised A proposition about filters on cartesian product of two sets
added 56 characters in body
Nov
1
answered A proposition about filters on cartesian product of two sets
Oct
31
asked A proposition about filters on cartesian product of two sets
Oct
26
comment Coequalizer in $\mathsf{Sets}$
What is $c$? You have not defined it. And if it is a quotient map then $h$ would be also forced to be equal on objects of the same equivalence class. I misunderstand something
Oct
26
comment Check a theorem about the category Set
@TobiasKildetoft: So I suppose my statement is an erroneous formulation of the UMP for coequalizers in Set. What is the correct formulation?
Oct
26
asked Check a theorem about the category Set
Oct
26
comment Right adjoint of forgetful functor from Top
Can it be generalized for any concrete category with an initial and a terminal object?
Oct
26
accepted Right adjoint of forgetful functor from Top
Oct
26
comment How to construct co-equalizers in $\mathbf{Top}$?
Adjointness warrants only that the limit in Top IF IT EXISTS coincides with the limit in Set, but not its existence. Right?
Oct
26
asked Right adjoint of forgetful functor from Top
Oct
26
comment How to construct co-equalizers in $\mathbf{Top}$?
Also: How to prove that it has a right adjoint, namely the functor Set→Top which equips a set with the indiscrete topology?
Oct
26
comment How to construct co-equalizers in $\mathbf{Top}$?
But how to prove that it is really a coequalizer in Top? Can it be proved elementarily (without adjoints)?
Oct
26
accepted How to construct co-equalizers in $\mathbf{Top}$?
Oct
25
asked How to construct co-equalizers in $\mathbf{Top}$?
Oct
25
accepted “Moving” a filter from one set to an other set, two equivalent formulations
Oct
25
comment “Moving” a filter from one set to an other set, two equivalent formulations
@ArthurFischer: No, to be useful this thing should work for every possible pair of sets $\mathfrak{A}$, $\mathfrak{B}$.