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10h
comment The coproduct of a family of objects of a Preorder (seen as a category)
Yes, it is the least upper bound. The proof is the same, except for some fiddling with equality.
2d
comment is axiom of powers required?
Is your collection of singletons actually a set? You can only take unions over sets.
2d
comment What is the relationship between the path-loop space fibration and path induction?
Well, based path induction is more or less the statement that the space $P X$ is contractible (plus some auxiliary facts about contractible spaces). But $\Omega X$ doesn't come into it.
May
22
comment Is it possible to develop differential geometry without points?
Locales are like topological spaces, not manifolds. Besides, every Hausdorff space is sober, so the category of Hausdorff spaces embeds as a full subcategory of the category of locales.
May
21
comment Is $\coprod \subseteq \prod$ true in any (complete cocomplete) Abelian category?
Well, no: if it were true then the dual result would force it to be an isomorphism, which is absurd.
May
20
comment left adjointable functors
Well, there is also "It is a left half", which sounds less grammatical than "It is the left half". My point is that the grammaticality of "$F$ is a left adjoint" cannot be determined by general arguments.
May
20
comment left adjointable functors
Your objection is invalid. For example, "He is a father" is perfectly grammatical as a sentence.
May
20
comment Is every reduced $k$-algebra all of whose residue fields are $k$ finitely generated?
Sorry, I misremembered. Your hypothesis implies that $A$ is zero-dimensional.
May
20
comment left adjointable functors
For me, "$F$ is left adjointable" sounds more like "$F$ has a left adjoint", which is the opposite of "$F$ is a left adjoint".
May
20
comment Relation between Noether's one-sided ideals and Polish notation?
There is no relation with Polish notation. In fact, Polish notation makes the relationship more obscure: the difference in question is between $r \, a \, b \, {+} \, {\times}$ and $a \, b \, {+} \, r \, {\times}$.
May
20
comment Intuitionistically, are these inequivalent? $P \rightarrow Q,\; \neg Q \rightarrow \neg P,\; P \wedge \neg Q \rightarrow \bot,\; \neg P \vee Q$
$\lnot Q \to \lnot P$ and $P \land \lnot Q \to \bot$ are equivalent. The rest are different.
May
19
comment Geometrically, why do line bundles have inverses with respect to the tensor product?
For me the problem comes even earlier: what is the geometric significance of the tensor product of line bundles?
May
19
comment Regarding Espace Etale of a presheaf, is $\bar s$ an open map?
Yes, sections of espaces étalés are always open embeddings.
May
19
comment An injective morphism between varieties that is not an immersion
Sure. Take the projective closures of $X$ and $Y$.
May
18
comment show that the theory of fields cannot be axiomatized by horn sentences
I am sure such a result appears in any textbook that discusses universal Horn theories.
May
18
comment show that the theory of fields cannot be axiomatized by horn sentences
Why not use a theorem about theories that are axiomatisable by Horn sentences? Say, the one that has something to do with products of models?
May
18
comment Abstract interpretation of isomorphism between tensor product with dual and hom
No, you can't do it in an abelian category. You need a symmetric monoidal closed category – and in fact that's all you need to define the morphism. Showing that it is (sometimes) an isomorphism is not abstract nonsense, however.
May
18
comment Abstract interpretation of isomorphism between tensor product with dual and hom
Yes, it does. Isn't that what I said? You might like to think about how to do it in, say, the category of modules over a commutative ring.
May
18
comment Abstract interpretation of isomorphism between tensor product with dual and hom
As stated, $\mathscr{F}^\vee \otimes \mathscr{G} \to \mathscr{H}om (\mathscr{F}, \mathscr{G})$ is canonical. You can describe it abstractly if you like. What does require work is showing that it is sometimes an isomorphism.
May
16
comment Derived functors - homotopical vs homological approach
You can avoid using the model structure – it is enough to have acyclic resolutions. The proof is easy – as I said, you just calculate explicitly using acyclic resolutions instead of using their respective universal properties.