= Solution
Write an object of the <slice category> $\mathcal D/X$ as $a:A\to X$. Its <morphisms> are arrows in $\mathcal D$ commuting with these structure maps, and the <forgetful functor> $U_X$ removes the structure map.
Suppose a square in the slice becomes a <pullback> square in $\mathcal D$, with vertex $p:P\to X$. Given a compatible pair of slice arrows $u:Z\to A$ and $v:Z\to B$, the underlying <pullback> gives a unique $h:Z\to P$ with the required projection equations. If $r:P\to A$ is one projection, then
$$
ph=arh=au=z,
$$
where $z:Z\to X$ is the structure map. Thus $h$ is automatically a slice <morphism>, and underlying uniqueness gives uniqueness in the slice. Hence \b[$U_X$ reflects <pullbacks>].
Now let $e:(E,c)\to(A,a)$ be an <equaliser> of slice arrows $f,g:(A,a)\rightrightarrows(B,b)$. Consider any underlying arrow $u:Z\to A$ with $fu=gu$. Give $Z$ the structure map $au:Z\to X$. Then $u$ is a slice arrow into $(A,a)$, and the slice <equaliser> gives a unique $h:Z\to E$ with $eh=u$. Conversely any underlying $h$ with that equation automatically obeys $ch=aeh=au$, so there are no additional factorization possibilities outside the slice. Therefore $U_Xe$ has the ordinary <equaliser> <universal property> in $\mathcal D$. This proves \b[$U_X$ preserves <equalisers>].
Back to article page
