CompHaus does not have exact cofiltered limits

Author: dschepler

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

@@ -130,4 +130,12 @@ VALUES
 	'filtered-colimit-stable monomorphisms',
 	FALSE,
 	'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.'
+),
+(
+	'CompHaus',
+	'exact cofiltered limits',
+	FALSE,
+	'Consider the $\IN$-codirected systems $X_n := [0, 1] \times [0, 1/n]$ with the maps $X_{n+1} \to X_n$ being inclusion maps, and $Y_n := [0, 1+1/n]$ with the maps $Y_{n+1} \to Y_n$ also being inclusion maps. We define $f_n : X_n \to Y_n$, $(x, y) \mapsto x$ and $g_n : X_n \to Y_n$, $(x, y) \mapsto x+y$. It is straightforward to check these give morphisms of $\IN$-codirected systems in $\CompHaus$.<br>
+	Now for each $n$, the coequalizer of $f_n$ and $g_n$ is a single-point space. On the other hand, $\lim X_n \simeq [0, 1] \times \{ 0 \}$; $\lim Y_n \simeq [0, 1]$; and $\lim f_n = \lim g_n$, $(x, 0) \mapsto x$. Thus, the coequalizer of $\lim f_n$ and $\lim g_n$ is $[0, 1]$, showing that this coequalizer is not preserved under limits.'
 );
+