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Copy file name to clipboardExpand all lines: database/data/003_properties/003_limits-colimits-behavior.sql
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'infinitary codistributive',
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TRUE
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),
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(
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'countably distributive',
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'is',
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'A category is <i>countably distributive</i> if it has finite products, countable coproducts, and for every object $A$ the functor $A \times -$ preserves countable coproducts. Concretely, for every countable family of objects $(B_i)$ the canonical morphism $\coprod_i (A \times B_i) \to A \times \coprod_i B_i$ must be an isomorphism.',
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NULL,
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'countably codistributive',
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TRUE
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),
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(
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'countably codistributive',
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'is',
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'A category is <i>countably codistributive</i> if it has finite coproducts, countable products, and for every object $A$ the functor $A \sqcup -$ preserves countable products. Concretely, for every countable family of objects $(B_i)$ the canonical morphism $A \sqcup \prod_i B_i \to \prod_i (A \sqcup B_i)$ must be an isomorphism.',
Copy file name to clipboardExpand all lines: database/data/004_property-assignments/CAlg(R).sql
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),
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(
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'CAlg(R)',
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'infinitary codistributive',
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'countably codistributive',
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FALSE,
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'The canonical homomorphism $A \otimes_R R^{\mathbb{N}} \to A^{\mathbb{N}}$ is given by $a \otimes (r_n)_n \mapsto (r_n a)_n$ and does not have to be surjective: Since $R \neq 0$, there is a commutative $R$-algebra $K$ which is a field. Now take $A := K[X]$ and consider the sequence $(X^n)_{n} \in A^{\mathbb{N}}$.'
Copy file name to clipboardExpand all lines: database/data/004_property-assignments/CRing.sql
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),
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(
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'CRing',
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-
'infinitary codistributive',
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'countably codistributive',
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FALSE,
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'The canonical homomorphism $\mathbb{Q} \otimes \mathbb{Z}^{\mathbb{N}} \to (\mathbb{Q} \otimes \mathbb{Z})^{\mathbb{N}} = \mathbb{Q}^{\mathbb{N}}$ is not an isomorphism: its image consists of those sequences of rational numbers whose denominators can be bounded.'
Copy file name to clipboardExpand all lines: database/data/004_property-assignments/Man.sql
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'[Sketch] Since $\mathbf{Top}$ is infinitary extensive, a continuous map $f : M \to \coprod_i N_i$ corresponds to a decomposition $M = \coprod_i M_i$ (as topological spaces) with continuous maps $f_i : M_i \to N_i$. Endow the open subset $M_i \subseteq M$ with the smooth structure inherited from $M$. Now remark that $f$ is smooth iff each $f_i$ is smooth.'
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),
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(
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-- TODO: maybe add "countably extensive" to make this more conceptual
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'Man',
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'countable coproducts',
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'countably distributive',
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TRUE,
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-
'Take the usual disjoint union, which is clearly locally Euclidean and Hausdorff, and it is second countable since we are using only countable many spaces. (Without that condition, all coproducts would exist.)'
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'To construct countable coproducts, take the usual disjoint union of spaces, which is clearly locally Euclidean and Hausdorff, and it is second countable since we are using only countable many spaces. (Without that condition, all coproducts would exist.) Now we need to check that the canonical smooth map $\coprod_i X \times Y_i \to X \times \coprod_i Y_i$ is a diffeomorphism (for countable families). It is a homeomorphism since $\mathbf{Top}$ is infinitary distributive. The inverse $X \times \coprod_i Y_i \to \coprod_i X \times Y_i$ is smooth since the domain is covered by the open subsets $X \times Y_i$ on which the map is clearly smooth.'
Copy file name to clipboardExpand all lines: database/data/005_implications/008_topos-theory-implications.sql
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),
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(
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'nno_criterion',
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'["infinitary distributive"]',
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'["countably distributive"]',
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'["natural numbers object"]',
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'Consider the copower $N := \coprod_{n \in \mathbb{N}} 1$ with inclusions $i_n : 1 \to N$ for $n \in \mathbb{N}$. We define $z := i_1 : 1 \to N$ and $s : N \to N$ by $s \circ i_n = i_{n+1}$. Since the category is infinitary distributive, we have $A \times N \cong \coprod_{n \in \mathbb{N}} A$ for every object $A$. Given morphisms $f : A \to X$, $g : X \to X$, a morphism $\Phi : A \times N \to X$ therefore corresponds to a family of morphisms $\phi_n : A \to X$ for $n \in \mathbb{N}$. The condition $\Phi(a,z)=f(a)$ becomes $\phi_0 = f$. The condition $\Phi(a,s(n)) = g(\Phi(a,n))$ becomes $\phi_{n+1} = g \circ \phi_n$. This recursively defines the morphisms $\phi_n$. (We are basically using that $\mathbb{N}$ is a natural numbers object in $\mathbf{Set}$.) Concretely, $\phi_n = g^n \circ f$.',
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'Consider the copower $N := \coprod_{n \in \mathbb{N}} 1$ with inclusions $i_n : 1 \to N$ for $n \in \mathbb{N}$. We define $z := i_1 : 1 \to N$ and $s : N \to N$ by $s \circ i_n = i_{n+1}$. Since the category is countably distributive, we have $A \times N \cong \coprod_{n \in \mathbb{N}} A$ for every object $A$. Given morphisms $f : A \to X$, $g : X \to X$, a morphism $\Phi : A \times N \to X$ therefore corresponds to a family of morphisms $\phi_n : A \to X$ for $n \in \mathbb{N}$. The condition $\Phi(a,z)=f(a)$ becomes $\phi_0 = f$. The condition $\Phi(a,s(n)) = g(\Phi(a,n))$ becomes $\phi_{n+1} = g \circ \phi_n$. This recursively defines the morphisms $\phi_n$. (We are basically using that $\mathbb{N}$ is a natural numbers object in $\mathbf{Set}$.) Concretely, $\phi_n = g^n \circ f$.',
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