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

Co-authored-by: Script Raccoon <scriptraccoon@gmail.com>

Author: dschepler

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

@@ -59,7 +59,7 @@ VALUES
 	'CompHaus',
 	'epi-regular',
 	TRUE,
-	'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>
+	'First, any epimorphism $f : X\to Y$ is surjective: if not, its image would be a proper subset of $Y$, which is compact and hence closed. Then by Urysohn''s lemma, there would be a non-zero continuous function $g : Y \to [0, 1]$ which is $0$ on the image; but then $g \circ f = 0 \circ f$, giving a contradiction.<br>
 	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 $h$.'
 ),
 (