37,209 reputation
241108
bio website
location
age
visits member for 3 years, 7 months
seen 32 mins ago

Nov
26
comment Exact sequences and Hom
A ring is a group under multiplication if and only if it is the zero ring. You will have to be much more precise about what you're doing here.
Nov
25
comment Who first called the Grothendieck's schéma scheme?
Sometimes people refer to axiom schemas as "axiom schemes", so either way...
Nov
25
comment Topology-Composition of homeomorphism
Are you asking whether the composite of two homeomorphisms is a homeomorphism? In that case the answer is yes, but the nature of the proof depends on the definition you choose.
Nov
25
comment Coproducts in $\text{Ab}$
(You're right that the theorem is false as stated. The correct statement is given in lemma 6.2.5 here.)
Nov
25
comment Coproducts in $\text{Ab}$
Of course, the coproduct and product functors are literally equal provided we use the canonical coproduct structure, so the natural isomorphism from $(A \times B, \iota_1, \iota_2)$ to the canonical coproduct structure is also a natural isomorphism from $(A \times B, \iota_1, \iota_2)$ to $(A \times B, \pi_1, \pi_2)$. Obviously this reasoning means we can't apply Lack's theorem here, but nonetheless it suggests Clive's definition is at worst unwieldy, rather than wrong.
Nov
25
comment Coproducts in $\text{Ab}$
There are some subtle points here. There is a natural isomorphism, but it isn't the identity transformation. Once you have a semiadditive structure, you can show that $\langle \textrm{id}_A, 0 \rangle : A \to A \times B$ and $\langle 0, \textrm{id}_B \rangle : B \to A \times B$ make $A \times B$ into a coproduct. Let's call it the canonical coproduct structure. Now, if $A \times B$ has any other coproduct structure $(A \times B, \iota_1, \iota_2)$ , we get a unique natural isomorphism from $(A \times B, \iota_1, \iota_2)$ to the canonical coproduct structure on $A \times B$.
Nov
25
comment Coproducts in $\text{Ab}$
@QiaochuYuan I don't think so. There is a not-so-well-known theorem of Lack saying that in any pointed category, the existence of any natural isomorphism $X + Y \cong X \times Y$ is enough to ensure the canonical map $X + Y \to X \times Y$ is an isomorphism.
Nov
25
awarded  Generalist
Nov
25
awarded  Enlightened
Nov
24
comment Can every proof by contradiction also be shown without contradiction?
Of course. And it depends on what things you take as primitive. (Is $=$ more primitive than $\ne$? Sometimes constructivists take an "apartness" relation $\mathrel{\#}$ to be primitive, in which case $=$ becomes the negation of $\mathrel{\#}$!)
Nov
24
comment Can every proof by contradiction also be shown without contradiction?
Well, that just shifts the contradiction elsewhere, no? Consider how the rest of the proof would go: we then assume $\sqrt{2} = n / m$, in which case $0 \ge 1 / (3m^2) > 0$, a contradiction.
Nov
24
answered Can every proof by contradiction also be shown without contradiction?
Nov
24
comment Can every proof by contradiction also be shown without contradiction?
There isn't any "must" about it. If $\phi$ is a compound formula such as $\psi \lor \theta$ you can just push the $\lnot$ inward and get $(\lnot \psi) \land (\lnot \theta)$. But that just causes the number of negative statements to multiply...
Nov
24
comment Compactness is closed-hereditary - significance of closed property?
A space where every subspace is compact is a called a noetherian topological space.
Nov
24
comment Can every proof by contradiction also be shown without contradiction?
The definition of "irrational" is "not rational", so the statement "$\sqrt{2}$ is irrational" is inherently a negative one. As such there is no "direct" proof (unless one somehow comes up with a definition of "irrational" that does not invoke "rational").
Nov
24
comment Can every proof by contradiction also be shown without contradiction?
Formally, a negative statement such as $\lnot \phi$ is an abbreviation for $\phi \to \bot$, i.e. it is the statement that $\phi$ leads to an absurdity; accordingly, to prove $\lnot \phi$, one must show that assuming $\phi$ leads to contradiction. This is completely orthodox logic.
Nov
24
comment Sufficient condition for a locally ringed space to be an affine scheme
Proposition (2) is probably false. If I'm not mistaken, a punctured affine plane has a cover by open affine subsets of the form you ask for, but is not itself affine.
Nov
24
comment Bijection between $\mathbb{R}$ and $\mathbb{R}/\mathbb{Q}$
@PatrickDaSilva That has nothing to do with category theory: it's like saying "haven't done enough English" when someone uses a word you don't know. What Thomas is saying is actually something very simple and you surely know examples of this phenomenon: there are surjective group homomorphisms with no left inverse, there are surjective continuous maps with no left inverse, etc. etc.
Nov
24
comment Does the ring of global sections functor on the category of locally ringed spaces have an adjoint functor?
It's Lemma 01I1 in Stacks.
Nov
23
comment Is there a $\mathbb{C}$ vector space structure over $\mathbb{R}$ with scalar multiplication as defined?
If you don't require any compatibility on the pre-existing $\mathbb{R}$-vector space structure on $\mathbb{R}$, then there is a $\mathbb{C}$-vector space structure on the set $\mathbb{R}$ simply because $\mathbb{R}$ and $\mathbb{C}$ have the same cardinality.