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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