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Add more property assignments
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.vscode/settings.json

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"surjectivity",
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"Tarski",
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"tensoring",
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"Tietze",
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"Turso",
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"Tychonoff",
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"unital",
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"unitalization",
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"Urysohn",
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"vercel",
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"Vite",
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"Wedderburn",

databases/catdat/data/003_category-property-assignments/CompHaus.sql

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TRUE,
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'Every non-empty compact Hausdorff space is weakly terminal (by using constant maps).'
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),
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(
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'CompHaus',
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'coregular',
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TRUE,
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'Since $\CompHaus$ is mono-regular and epi-regular, the regular-epi-mono factorization system from the regularity of $\CompHaus$ also serves as a regular-epi-mono factorization system for $\CompHaus^\op$. Therefore, it suffices to show that pushouts preserve (regular) monomorphisms in $\CompHaus$. Thus, suppose we have a pushout square
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$$\begin{CD}
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A @> i >> B \\
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@V f VV @VV g V \\
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C @>> j > D,
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\end{CD}$$
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with $i : A \hookrightarrow B$ a monomorphism. Then for any pair of distinct elements $c, c'' \in C$, by Urysohn''s lemma there exists $\gamma : C \to [0, 1]$ with $\gamma(c) = 0$ and $\gamma(c'') = 1$. Also, by Tietze''s extension theorem, there exists $\beta : B \to [0, 1]$ such that $\beta \circ i = \gamma \circ f$. By the pushout property, there is a unique $\delta : D \to [0, 1]$ such that $\delta \circ g = \beta$ and $\delta \circ j = \gamma$. Since $\delta(j(c)) \ne \delta(j(c''))$, we conclude that $j(c) \ne j(c'')$. This shows that $j$ is injective, so it is a regular monomorphism.'
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),
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(
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'CompHaus',
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'Malcev',
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),
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(
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'CompHaus',
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'infinitary extensive',
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'natural numbers object',
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FALSE,
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'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.'
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'Let $I := [0, 1]$. If a natural numbers object $N$ existed, then we could iterate the initial conditions $I\to I\times I$, $x \mapsto (x, x)$ and $I\times I \to I \times I$, $(x, y) \mapsto (x, xy)$ to get a continuous function $N \times I \to I \times I$ such that $(n, x) \mapsto (x, x^n)$ for $x\in I$, $n \in \IN$. The sequence $(n) \in N$ has a convergent subnet $(n_\lambda)_{\lambda \in \Lambda}$, say with limit $y$. Thus, for any $x\in I$ and $\lambda \in \Lambda$, we have $(n_\lambda, x) \mapsto (x, x^{n_\lambda})$. Taking limits, we see $(y, x) \mapsto (x, 0)$ if $x \ne 1$ or $(y, x) \mapsto (x, 1)$ if $x = 1$. In other words, $(y, x) \mapsto (x, \delta_{x, 1})$ for all $x\in I$. However, that contradicts the fact that the composition $I \overset{y \times \id}\longrightarrow N\times I \to I\times I \overset{p_2}\longrightarrow I$, $x \mapsto (y, x) \mapsto (x, \delta_{x,1}) \mapsto \delta_{x,1}$, would have to be continuous.'
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),
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(
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'CompHaus',

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