Decide all properties for the category of Hausdorff spaces

.cspell.json

@@ -6,7 +6,6 @@
 		"notrack",
 		"sqcup",
 		"infty",
-		"chartjs",
 		"Prost",
 		"SetxSet",
 		"hilberts",
@@ -16,7 +15,8 @@
 		"varnothing",
 		"mdash",
 		"comphaus",
-		"esbuild"
+		"esbuild",
+		"coloneqq"
 	],
 	"words": [
 		"abelian",
@@ -28,6 +28,7 @@
 		"anneaux",
 		"Artin",
 		"Auslander",
+		"Axiomatising",
 		"axiomatization",
 		"bijection",
 		"bijections",
@@ -40,6 +41,7 @@
 		"catdat",
 		"Catégories",
 		"Čech",
+		"characterisation",
 		"clopen",
 		"Clowder",
 		"coaccessible",
@@ -61,6 +63,7 @@
 		"coequalizer",
 		"coequalizers",
 		"coequalizes",
+		"coexact",
 		"coexponentials",
 		"cofiltered",
 		"cofiltering",
@@ -114,6 +117,7 @@
 		"disjointness",
 		"dualizable",
 		"Dualization",
+		"Duskin",
 		"Eilenberg",
 		"endofunctors",
 		"Engelking",
@@ -139,11 +143,13 @@
 		"Haus",
 		"Heyting",
 		"homotopy",
+		"Hušek",
 		"hypercategories",
 		"hypercategory",
 		"hypercollection",
 		"hypercollections",
 		"idempotents",
+		"Ieke",
 		"infima",
 		"infimum",
 		"infinitary",
@@ -155,14 +161,18 @@
 		"Johnstone",
 		"Jónsson",
 		"Kashiwara",
+		"Kategorien",
 		"katex",
 		"Kolmogorov",
 		"Lawvere",
 		"libsql",
 		"Lindelöf",
 		"Lodha",
+		"Lokal",
+		"Luca",
 		"Makkai",
 		"Malcev",
+		"Marra",
 		"Mathoverflow",
 		"metrizable",
 		"Moerdijk",
@@ -177,6 +187,7 @@
 		"morphisms",
 		"Multialgébriques",
 		"naturality",
+		"Neves",
 		"Niefield",
 		"nilradical",
 		"nlab",
@@ -186,6 +197,7 @@
 		"Pontryagin",
 		"poset",
 		"posets",
+		"präsentierbare",
 		"preadditive",
 		"precomposed",
 		"precomposition",
@@ -207,11 +219,15 @@
 		"pushout",
 		"pushouts",
 		"quasitopos",
+		"Reggio",
+		"Renato",
 		"rng",
 		"rngs",
 		"Rosicky",
+		"Rosický",
 		"saft",
 		"Schapira",
+		"Schepler",
 		"semigroup",
 		"semigroups",
 		"semisimple",
@@ -246,12 +262,18 @@
 		"tensoring",
 		"Tietze",
 		"topoi",
+		"tripleability",
 		"Turso",
 		"Tychonoff",
+		"Ulmer",
 		"unital",
 		"unitalization",
+		"Universalis",
+		"Universitext",
 		"Urysohn",
 		"vercel",
+		"Verlag",
+		"Vincenzo",
 		"Vite",
 		"Wedderburn",
 		"well-copowered",
@@ -265,10 +287,9 @@
 		".vscode",
 		".svelte-kit",
 		".netlify",
-		"build"
+		"build",
+		"todo.txt",
+		"*.svg"
 	],
-	"ignoreRegExpList": [
-		"\\$[^$]*\\$",
-		"\\$\\$[^$]*\\$\\$"
-	]
+	"ignoreRegExpList": ["\\$[^$]*\\$", "\\$\\$[^$]*\\$\\$"]
 }

databases/catdat/data/categories/Haus.yaml

@@ -77,6 +77,20 @@ unsatisfied_properties:
   - property: accessible
     reason: In fact, it does not have any small colimit-dense subcategory by <a href="https://math.stackexchange.com/questions/4097315/" target="_blank">MSE/4097315</a>.
 
+  - property: cogenerating set
+    # cspell: disable-next-line
+    reason: 'If $S$ is a cogenerating set, then it is easy to see that also $S \setminus \{\varnothing\}$ is a cogenerating set. Between two non-empty spaces there is always a constant map. Therefore, by <a href="/content/generator_construction">this lemma</a>, the product of all spaces in $S \setminus \{\varnothing\}$ is a cogenerator $Q$. Since $Q$ is Hausdorff, $Q$ is $T_1$. By a theorem of Herrlich (<i>Wann sind alle stetigen Abbildungen in Y konstant</i>. Math. Z. 90 (1965): 152-154. <a href="http://eudml.org/doc/170472" target="_blank">EUMDL</a>), there is a regular Hausdorff space $X$ with $\geq 2$ points such that every continuous map $X \to Q$ is constant. (The author only states that $X$ is regular, but actually, $X$ is regular and $T_1$, hence Hausdorff.) But since $Q$ is a cogenerator, this implies that all maps $1 \rightrightarrows X$ are equal, i.e. that $X$ has just one point. This is a contradiction.'
+
+  - property: regular
+    reason: 'The regular epimorphisms are precisely the surjective quotient maps of Hausdorff spaces (see below). In a regular category, for every regular epimorphism $X \to Y$ and every object $Z$, the induced morphism $X \times Z \to Y \times Z$ is again a regular epimorphism. This is not the case in $\Haus$ (or $\Top$, for that matter). The standard example is the quotient map $\IR \to \IR / \IZ^+$, for which the induced map $\IR \times \IQ \to \IR/\IZ^+ \times \IQ$ is not a quotient map (<a href="https://math.stackexchange.com/questions/1907972/">MSE/1907972</a>).'
+
+  - property: coregular
+    reason: >-
+      Let $\Gamma$ be the <a href="https://en.wikipedia.org/wiki/Moore_plane" target="_blank">Moore plane</a>. Its underlying set is $\{(x,y) \in \IR^2 : y \geq 0 \}$. The open neighborhoods of points $(x,y)$ with $y > 0$ are those of $\IR^2$ (intersected with $\Gamma$), and the basic open neighborhoods of a point $(x,0)$ are open disks centered at $(x,\varepsilon)$ with radius $\varepsilon$ for some $\varepsilon > 0$. Then $\Gamma$ is Hausdorff, and the $x$-axis $A \coloneqq \{(x,0) : x \in \IR\}$ is a closed discrete subspace of $\Gamma$. In particular, by the classification of regular monomorphisms below, the inclusion map $i : A \to \Gamma$ is a regular monomorphism.
+      Consider the two subsets $A_1 \coloneqq \{(x,0) : x \in \IQ \}$ and $A_2 \coloneqq \{(x,0) : x \in \IR \setminus \IQ \}$ of $A$. They are closed in $A$ (since $A$ is closed and discrete), disjoint, but cannot be separated by disjoint open neighborhoods in $\Gamma$; this is part of the proof of the well-known fact that $\Gamma$ is not normal (<a href="https://math.stackexchange.com/questions/2528435" target="_blank">MSE/2528435</a>).
+      Let $C \coloneqq \{1,2\}$ be the discrete two-point space. The map $f : A \to C$ defined by $f(a)=1$ for $a \in A_1$ and $f(a)=2$ for $a \in A_2$ is continuous, since $A$ is discrete.
+      The pushout $C \sqcup_A \Gamma$ in $\Haus$ is the Hausdorff reflection of the pushout $Q$ in $\Top$. Notice that $Q$ is the quotient space of $\Gamma$ in which $A_1$ and $A_2$ are each collapsed to a point, denoted by $[A_1]$ and $[A_2]$. The canonical map $C \to Q$ is given by $i \mapsto [A_i]$. Now, $[A_1]$ and $[A_2]$ cannot be separated by disjoint open neighborhoods in $Q$, since such neighborhoods would pull back to disjoint open neighborhoods of $A_1$ and $A_2$ in $\Gamma$. Thus, they are identified in the Hausdorff reflection. This shows that the canonical map $C \to C \sqcup_A \Gamma$ is not injective and hence not a regular monomorphism.
+
 special_objects:
   initial object:
     description: empty space
@@ -100,3 +114,8 @@ special_morphisms:
   regular monomorphisms:
     description: embeddings with closed image
     reason: The explicit construction of equalizers shows that they are embeddings, and they have a closed image because of the well-known lemma that for a Hausdorff space the diagonal $X \to X \times X$ has closed image. For the other non-trivial direction, see <a href="https://math.stackexchange.com/questions/214045/" target="_blank">MSE/214045</a>.
+  regular epimorphisms:
+    description: surjective quotient maps
+    reason: >-
+      The coequalizer of two continuous maps $f,g : X \rightrightarrows Y$ of Hausdorff spaces is given by $Y \to Q \to Q_H$, where $Y \to Q$ is their coequalizer in $\Top$ and $Q \to Q_H$ is the Hausdorff reflection of $Q$. By the construction of coequalizers in <a href="/category/Top">$\Top$</a>, the map $Y \to Q$ is a surjective quotient map. Moreover, the construction of the Hausdorff reflection (see <a href="https://ncatlab.org/nlab/show/Hausdorff+space#HausdorffReflectionViaHomsIntoAllHausdorffSpaces" target="_blank">nLab</a>) shows that $Q \to Q_H$ is also a surjective quotient map. Hence, $Y \to Q_H$ is a surjective quotient map as well.
+      Conversely, if $p : X \to Y$ is a surjective quotient map between Hausdorff spaces, we know from the classification in $\Top$ that $p$ is the coequalizer of its kernel pair $X \times_Y X \rightrightarrows X$ in $\Top$. Since $X \times_Y X$ is Hausdorff, the same universal property remains valid in $\Haus$.