Reputation
2,810
Top tag
Next privilege 3,000 Rep.
Cast close & reopen votes
Badges
1 8 21
Newest
 Yearling
Impact
~30k people reached

Sep
26
suggested approved edit on $A$ retract of $X$ and $X$ contractible implies $A$ contractible.
Sep
20
comment Why is every category not isomorphic to its opposite?
Right, needless to say that the only morphism going into the empty set is its own identity, which is indeed an iso.
Sep
18
comment Does the comma category functor have an adjoint?
Sorry but your proof does not hold. You make 2 mistakes: 1) $[A, C] = [B, C]= C^1$ is not $\mathrm{Set}(\mathrm{ob} \ C)$ , it is exactly the category $C$ itself. 2) $s\downarrow t= Hom(s, t)$ is the discrete category whose objects are the morphisms from s to t and only has the identity morphisms. When you then rewrite the adjoint iso, you see that it actually holds if we set $F(X) = s$. The category $\mathrm{Arr}(s, t))$ that you mention is actually $C\downarrow C$ also known as $C^2$.
Sep
15
comment Equalizers and Basic limit theorem in Category theory
@Did I know, and I did it! But as I said it was rejected. I am glad it is fixed now.
Sep
14
awarded  Citizen Patrol
Sep
14
comment Equalizers and Basic limit theorem in Category theory
....@Did and has nothing to do with category theory. So the current title is misleading. I tried to edit it yesterday, but apparently some (half asleep?) moderator did not accept it. Besides, as you point out, the OP himself calls it BASIC limit theorem in the body. So please some other moderator should consider reaccepting my edit or at least fix the title appropriately.
Sep
14
comment Equalizers and Basic limit theorem in Category theory
@Did Exactly!!! The central limit theorem is a fundamental result in statistics en.wikipedia.org/wiki/Central_limit_theorem
Sep
13
suggested rejected edit on Equalizers and Basic limit theorem in Category theory
Sep
10
comment Identity in a category
Omar's demonstration is typical in these kind of proofs: identities (or units or neutral elements) are uniquely determined by their defining equations, if they are already known to exist
Sep
6
comment Where can I learn more about order-reflecting functions? And is the following result well-known?
Nice remark that the proof doesn't use the properties of ≤. I am a bit confused by your definition though. Perhaps there are some typos. The op-functor is $F$ or is $F^{-1}$ ? Also, the op-functor should go in the opposite direction to $F$, correct? "op" stands for (or reminds to) "opposite"?
Sep
1
answered What can we conclude about the natural projection maps?
Aug
31
comment Direct products in a partially ordered category
I am not sure I understand your question, but if you are trying to characterize a product in a poset category, then you should reverse the arrows of $\pi_0$ and $\pi_1$ and $\phi$ should be the largest lower bound to the family of objects
Aug
25
revised Understanding adjoint functors
Edited isomorphism formula
Aug
25
suggested approved edit on Understanding adjoint functors
Aug
17
revised Complete abelian categories with projectieve generators are fully abelian.
edited small typos
Aug
17
comment Complete abelian categories with projectieve generators are fully abelian.
Welcome to Math.SE :-) I suggest you choose a nickname (your real name or a fantasy name), so that other users may more easily recognize you later.
Aug
17
suggested approved edit on Complete abelian categories with projectieve generators are fully abelian.
Aug
9
revised Functionally structured spaces and manifolds
edited small typos
Aug
9
suggested approved edit on Functionally structured spaces and manifolds
Aug
6
comment Concrete balanced category
Would the downvoter mind to explain his/her downvote? The OP wrote: if every bimorphism is an iso we have a balanced category, if we have a balanced category , is it true that every bimorphism is an iso? I simply said : yes it is true. What's wrong with it?