nicolas
Reputation
277
Top tag
Next privilege 500 Rep.
Access review queues
 Jan 14 comment initial algebra and free monoid awesome...... thank you ! Jan 11 comment codiagonal functor and faithfullness I see. this "notation" a+b $\in C+C$ is bad enough for it to be called wrong Jan 11 comment codiagonal functor and faithfullness I dont understand how your construction of C+C is different from the one I present, as a universal arrow constructed from a pair of arrow coming from C.. Jan 10 comment codiagonal functor and faithfullness yes, we call them sum types in programming. I guess the terminology differs a bit. Awodey writes it C + C though Jan 10 comment codiagonal functor and faithfullness left and right are actually some standard injection to sum types in Haskell or Agda, but point well taken about the notation. Jan 10 comment codiagonal functor and faithfullness well left a or (0,a) .. ;) Jan 10 comment codiagonal functor and faithfullness Agree about the sum category. I think it's the category whose objects are either left of some object of C or right of some object of C. Arrows are either from a left or from a right, into a object of C+C Jan 10 comment codiagonal functor and faithfullness It is at the beginning of chapter 7 définition 7.1 about faithfullness Nov 15 comment right inverse and supplement of kernel in a banach oh yes, sorry. I'll fix that Nov 15 comment right inverse and supplement of kernel in a banach it sounds like a more elegant formulation. Nov 13 comment If $f$ is continuous at $a$, is it continuous in some open interval around $a$? @Kaz what do you mean ? Nov 12 comment How to show that $(\operatorname{Im}{A})^⊥ = \ker(A^⊤)$, $\operatorname{Im}{(A^⊤)}= (\ker A)^⊥$? in infinite dimension, one only has $closure(Im A^T) \subset {ker A}^\perp$ Nov 11 comment show: $\overline{\overline X} = \overline X$ this is the way to prove it. dealing with epsilon is overkill for this Nov 11 comment How to show that $(\operatorname{Im}{A})^⊥ = \ker(A^⊤)$, $\operatorname{Im}{(A^⊤)}= (\ker A)^⊥$? @Timbuc he refers to another question I think Nov 11 comment How to show that $(\operatorname{Im}{A})^⊥ = \ker(A^⊤)$, $\operatorname{Im}{(A^⊤)}= (\ker A)^⊥$? @qexi this is another question. and one for which you might be surprised by the answer :) Nov 11 comment How to show that $(\operatorname{Im}{A})^⊥ = \ker(A^⊤)$, $\operatorname{Im}{(A^⊤)}= (\ker A)^⊥$? this is not always true in infinite dimension I think btw Nov 10 comment Can't argue with success? Looking for “bad math” that “gets away with it” That was a test by the student to see if teacher reads the proof Nov 10 comment Can't argue with success? Looking for “bad math” that “gets away with it” @joriki if only mathematics was using "expressions", binders, higher order functions, and other rigorous CS notations (de brujin), the work would be a better place.. how many bad conceptual math have been produced by wrong notations... Nov 10 comment Can't argue with success? Looking for “bad math” that “gets away with it” @Squirtle I just did Nov 10 comment Can't argue with success? Looking for “bad math” that “gets away with it” @GrumpyParsnip that counts as luck in my book