 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, thenwhere $z:Z\to X$ is the structure map. Thus $h$ is automatically a slice morphism, and underlying uniqueness gives uniqueness in the slice. Hence $U_X$ 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 $U_{X}e$ has the ordinary equaliser universal property in $\mathcal{D}$. This proves $U_X$ preserves equalisers.