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Nov
1
comment How is exponentiation defined in Peano arithmetic?
Adding function symbols is not allowed – that is tantamount to changing the logical language.
Nov
1
revised How to prove exactness implies complex?
edited tags
Nov
1
comment Some questions on abelian category
What definition of abelian category are you using?
Nov
1
comment Are there such things as non-extensional set theories?
Set theory with atoms is technically not extensional, but that's just a problem with the definition.
Oct
31
comment on the extension module of a pair (some module, a finite free module)
Well, $\mathrm{Ext}^i(M, -)$ is an additive functor, and additive functors preserve biproducts (i.e. finite direct sums/products).
Oct
31
comment Is $R$ initial in the category of $R$-algebras?
You are not working in the right category. There is only one $\mathbb{C}$-algebra homomorphism $\mathbb{C} \to \mathbb{C}$.
Oct
31
answered Is this “set quotient” known?
Oct
31
comment Is this “set quotient” known?
It has many names: Heyting implication, relative pseudocomplement...
Oct
31
answered Cancellation of products in an arbitrary category does not hold; does it hold with this extra condition?
Oct
30
comment A definition check.
That's just terminology. I suppose some people think "cocone" sounds silly.
Oct
30
answered Reflector for isomorphisms
Oct
30
reviewed Reject What do $x\in[0,1]^n$ and $x\in\left\{ 0,1\right\}^n$ mean?
Oct
30
reviewed Approve Hilbert's Nullstellensatz and maximal ideals
Oct
30
comment Why is $\mathbf{Rel} \cong \mathbf{Rel}^{\mathbb{op}}$?
I do not recommend this practice as it can be rather confusing in a category that has an involution, such as $\mathbf{Poset}$.
Oct
30
comment Why is $\mathbf{Rel} \cong \mathbf{Rel}^{\mathbb{op}}$?
$F$ is supposed to be a contravariant functor. So it sends a relation $C \to D$ to a relation $D \to C$.
Oct
29
revised Free Metric Space?
deleted 54 characters in body
Oct
29
answered Free Metric Space?
Oct
29
comment ENS is an abbreviation of?…
Incidentally, it also stands for École Normale Supérieure...
Oct
29
comment If $H_1, H_2\leq G$ are such that $H_1\cong H_2$ then $G/H_1\cong G/H_2$?
@PtF It holds if there is an automorphism of $G$ that carries $H_1$ to $H_2$.
Oct
28
answered Visualizing a homotopy pull back