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Aug
14
comment Is any splitting field algebraic?
Actually, $K[X] / (f)$ may not be a splitting field for $f$.
Aug
14
comment Why is $\mathsf{Type} : \mathsf{Type}$ a contradiction?
@user40276 No. This is Girard's paradox.
Aug
13
comment Is the parametrization $(t^3,t^6)$, a reparametrization of $(t,t^2)$?
Well, then you don't have a reparametrisation, because the inverse is not smooth.
Aug
13
comment Is the parametrization $(t^3,t^6)$, a reparametrization of $(t,t^2)$?
Obviously that theorem depends on the definition of "reparametrisation".
Aug
13
answered Is this assignment of the topos of sheaves functorial?
Aug
12
comment degree of an etale cover of the affine line
If $X \to \mathbb{A}^1_k$ is a finite étale cover and $X$ is connected then it is an isomorphism. No? For $k = \mathbb{C}$ this corresponds to the fact that $\mathbb{C}$ is simply connected.
Aug
11
comment Choosing projective replacement to be functorial
1. Classical derived functors are functors – between the derived categories. 2. Look up Cartan–Eilenberg resolutions. 3. A functorial projective replacement should be enough. 4. There is only one possible meaning.
Aug
11
comment Can you integrate on a scheme?
Integrate what?
Aug
11
comment Derived functors definition
That is the assumption, yes. Please read carefully.
Aug
10
comment Sifted colimits of models of a Lawvere theory.
I'm not so certain that filtered colimits + reflexive coequalisers suffice for all sifted colimits, but at any rate the conclusion about finitary monads on $\mathbf{Set}$ is correct. It's less clear to me what happens for monads of higher accessibility rank, though.
Aug
10
comment Examples of preadditive categories
The total category of modules is not preadditive.
Aug
10
awarded  Enlightened
Aug
10
awarded  Nice Answer
Aug
9
comment Definition of exact sequence of functors.
That is correct. In fact, the category of functors (with values in a given abelian category) is itself an abelian category, so this does not need a separate definition.
Aug
7
comment Show that $A \lor B ⊢ B \lor A$
There's more than one way of stating the elimination rule as well. For instance, we could get rid of $\to$ and instead have the premises $\Gamma, p \vdash r$ and $\Gamma, q \vdash r$.
Aug
6
answered This is just the Eilenberg-Moore category, right?
Aug
6
answered Construction of a ring from a category
Aug
6
comment This is just the Eilenberg-Moore category, right?
I guess you mean to ask about the category of functors that preserve small products. The answer is yes, the proof is in the comments at the second linked page.
Aug
6
comment Show that $A \lor B ⊢ B \lor A$
I did not downvote. And, as you say, the OP has not specified the rules. It is entirely possible that a constructive proof is required.
Aug
6
comment Show that $A \lor B ⊢ B \lor A$
This is terrible – unnecessary proof by contradiction!