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 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?