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| 1 | +INSERT INTO category_property_assignments ( |
| 2 | + category_id, |
| 3 | + property_id, |
| 4 | + is_satisfied, |
| 5 | + reason |
| 6 | +) |
| 7 | +VALUES |
| 8 | +( |
| 9 | + 'CompHaus', |
| 10 | + 'locally small', |
| 11 | + TRUE, |
| 12 | + 'This is trivial.' |
| 13 | +), |
| 14 | +( |
| 15 | + 'CompHaus', |
| 16 | + 'generator', |
| 17 | + TRUE, |
| 18 | + 'The one-point space is a generator because it represents the forgetful functor to $\Set$, which is faithful.' |
| 19 | +), |
| 20 | +( |
| 21 | + 'CompHaus', |
| 22 | + 'complete', |
| 23 | + TRUE, |
| 24 | + 'It is monadic over $\Set$.' |
| 25 | +), |
| 26 | +( |
| 27 | + 'CompHaus', |
| 28 | + 'cocomplete', |
| 29 | + TRUE, |
| 30 | + 'It is monadic over $\Set$.' |
| 31 | +), |
| 32 | +( |
| 33 | +-- TODO: rework this when Barr-exact property is added |
| 34 | + 'CompHaus', |
| 35 | + 'regular', |
| 36 | + TRUE, |
| 37 | + 'It is monadic over $\Set$ and therefore Barr-exact.' |
| 38 | +), |
| 39 | +( |
| 40 | +-- TODO: rework this when Barr-exact property is added |
| 41 | + 'CompHaus', |
| 42 | + 'effective congruences', |
| 43 | + TRUE, |
| 44 | + 'It is monadic over $\Set$ and therefore Barr-exact.' |
| 45 | +), |
| 46 | +( |
| 47 | + 'CompHaus', |
| 48 | + 'cogenerator', |
| 49 | + TRUE, |
| 50 | + 'The unit interval $[0, 1]$ is a cogenerator: Suppose we have $f, g : X \to Y$ with $f \ne g$. Choose $x\in X$ such that $f(x) \ne g(x)$. Then by Urysohn''s lemma, there is a function $h : Y \to [0, 1]$ such that $h(f(x)) = 0$ and $h(g(x)) = 1$. Therefore, $h\circ f \ne h\circ g$.' |
| 51 | +), |
| 52 | +( |
| 53 | + 'CompHaus', |
| 54 | + 'extensive', |
| 55 | + TRUE, |
| 56 | + 'This follows as for $\Top$ or $\Haus$ since finite coproducts in $\CompHaus$ are formed as disjoint union spaces with the disjoint union topology.' |
| 57 | +), |
| 58 | +( |
| 59 | + 'CompHaus', |
| 60 | + 'epi-regular', |
| 61 | + TRUE, |
| 62 | + 'First, any epimorphism $f : X\to Y$ is surjective: if not, its image would be a closed subset of $Y$. Then by Urysohn''s lemma, there would be a function $g : Y \to [0, 1]$ which is 0 on the image but is not the zero function; but then $g \circ f = 0 \circ f$, giving a contradiction.<br> |
| 63 | + Therefore, we see that if $g, h : E \rightrightarrows X$ is the kernel pair of $f$, and $U : \CompHaus \to \Set$ is the forgetful functor, then $U(f)$ is the coequalizer of $U(g)$ and $U(h)$. Since $U$ is fully faithful, that implies $f$ is the coequalizer of $g$ and $h$.' |
| 64 | +), |
| 65 | +( |
| 66 | + 'CompHaus', |
| 67 | + 'well-powered', |
| 68 | + TRUE, |
| 69 | + 'This is clear from the description of monomorphisms as closed embeddings.' |
| 70 | +), |
| 71 | +( |
| 72 | + 'CompHaus', |
| 73 | + 'semi-strongly connected', |
| 74 | + TRUE, |
| 75 | + 'Every non-empty compact Hausdorff space is weakly terminal (by using constant maps).' |
| 76 | +), |
| 77 | +( |
| 78 | + 'CompHaus', |
| 79 | + 'Malcev', |
| 80 | + FALSE, |
| 81 | + 'This is clear since $\FinSet$ is not Malcev and can be interpreted as the subcategory of discrete spaces.' |
| 82 | +), |
| 83 | +( |
| 84 | + 'CompHaus', |
| 85 | + 'skeletal', |
| 86 | + FALSE, |
| 87 | + 'This is trivial.' |
| 88 | +), |
| 89 | +( |
| 90 | + 'CompHaus', |
| 91 | + 'cartesian closed', |
| 92 | + FALSE, |
| 93 | + 'Consider the unit interval $I := [0, 1]$. If there were an exponential $I^I$ in $\CompHaus$, by using $1$ as a test object we can see its underlying set would have to be the set of continuous functions $I \to I$, and the evaluation morphism $I^I \times I \to I$ would have to be $(f, x) \mapsto f(x)$. Now consider the sequence $f_n : I \to I$, $x \mapsto x^n$. By assumption, this would have a convergent subnet $f_{n_j} \to g$ for some directed set $J$ and cofinal map $J \to \IN$. But then, since the evaluation morphism $I^I \times I \to I$ is continuous, this would mean $f_{n_j}(x) \to g(x)$ for each $x \in I$. For $x \in [0, 1)$, this implies that $g(x) = 0$, and for $x = 1$, this implies that $g(x) = 1$, contradicting the fact that $g$ must be continuous.' |
| 94 | +), |
| 95 | +( |
| 96 | + 'CompHaus', |
| 97 | + 'regular subobject classifier', |
| 98 | + FALSE, |
| 99 | + 'The proof is almost identical to the one for <a href="/category/Haus">$\Haus$</a>.' |
| 100 | +), |
| 101 | +( |
| 102 | + 'CompHaus', |
| 103 | + 'infinitary extensive', |
| 104 | + FALSE, |
| 105 | + 'Consider the coproduct of $\aleph_0$ copies of the one-point space, which is homeomorphic to $\beta \IN$. Take a non-principal ultrafilter of $\IN$, and let $f : 1 \to \beta \IN$ be the corresponding morphism. Then the pullback of $f$ and the coproduct inclusion $i_n : 1 \to \beta \IN$ is empty for each $n$. Therefore, this coproduct is not stable under pullbacks.' |
| 106 | +), |
| 107 | +( |
| 108 | + 'CompHaus', |
| 109 | + 'filtered-colimit-stable monomorphisms', |
| 110 | + FALSE, |
| 111 | + 'The proof is similar to <a href="/category/Haus">$\Haus$</a>. For $n \geq 1$ let $X_n$ be the pushout of $[1/n, 1] \hookrightarrow [0, 1]$ with itself. That is, $X_n$ is the union of two unit intervals $[0, 1] \times \{ 1 \}$ and $[0, 1] \times \{ 2 \}$ where we identify $(x,1) \equiv (x,2)$ when $x \geq 1/n$. As in the construction for $\Haus$, we see that the colimit in $\Haus$ is $[0, 1]$ where all corresponding points of both unit intervals are identified. Since this is compact Hausdorff, it also provides the colimit in $\CompHaus$. Again, the injective continuous maps $\{1,2\} \to X_n$, $i \mapsto (0,i)$ (where $\{1,2\}$ is discrete) become the constant map $0 : \{1,2\} \to [0,1]$ in the colimit, which is not a monomorphism.' |
| 112 | +), |
| 113 | +( |
| 114 | + 'CompHaus', |
| 115 | + 'accessible', |
| 116 | + FALSE, |
| 117 | + 'For any small regular cardinal $\kappa$, consider the $\kappa$-directed system $\kappa \to \CompHaus$, where for ordinal $\alpha < \kappa$ we have $\alpha \mapsto [0, \alpha]$, and for $\alpha \le \beta < \kappa$ the morphism $\alpha \to \beta$ maps to the inclusion map $[0, \alpha] \hookrightarrow [0, \beta]$. Then the direct colimit of the underlying sets is $\kappa$, so the direct colimit in $\CompHaus$ is the Stone-Čech compatification $\tilde\kappa$ of $\kappa$.<br> |
| 118 | + We now claim that any nonempty object $K$ of $\CompHaus$ is not $\kappa$-presentable for any small regular cardinal $\kappa$. If so, then choose a point $x \in \tilde \kappa \setminus \kappa$, and consider the constant map $K \to \tilde \kappa$ with value $x$. By the assumption, this would have to factor through $[0, \alpha]$ for some ordinal $\alpha < \kappa$. But since the topological space $\kappa$ is Tychonoff, the map $\kappa \to \tilde \kappa$ is injective; thus, the map $[0, \alpha] \to \tilde\kappa$ is also injective. That means that the factor must be a constant map with value in $[0, \alpha]$, giving a contradiction.<br> |
| 119 | + Now any colimit in $\CompHaus$ of empty spaces is empty, showing that $\CompHaus$ is not accessible.'); |
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