Every slice of a presheaf category is again a presheaf category I found the following statement in exercise 6.2.24 in Leinster's "Basic category theory" book (freely available for download here: https://arxiv.org/abs/1612.09375):

Given a small category $\mathbf{A}$ and a presheaf $X$ on $\mathbf{A}$, the slice category $\mathbf{PSh}(\mathbf{A})/X=[\mathbf{A}^{\text{op}},\mathbf{Set}]/X$ is equivalent to $\mathbf{PSh}(\mathbf{B})=[\mathbf{B}^{\text{op}}, \mathbf{Set}]$ for some small category $\mathbf{B}$.

I tried to prove it but there is at least one step with which I have some difficulty.
My attempt so far: One can at first assume that $X$ is representable, i.e. $X=H_A$
for some $A\in\mathbf{A}$ (where $H_A(B):=\text{Hom}_{\mathbf{Set}}(B,A)$. One could also write $H(A)$ if $H:\mathbf{A}\rightarrow [\mathbf{A}^{\text{op}},\mathbf{Set}]$
denotes the Yoneda embedding).
According to Proposition 3.2. of this nlab-article about over-categories,
there is an equivalence of categories
\begin{equation}
  \begin{split}
    \mathbf{PSh}(\mathbf{A}/A) \simeq \mathbf{PSh}(\mathbf{A})/H_A.
  \end{split}
\end{equation}
To obtain a correspondence between some $\mathbf{PSh}(\mathbf{B})$ and $\mathbf{PSh}(\mathbf{A})/X$, where $X$ is now any presheaf, one could use the fact that every presheaf can be written as a colimit of representables.
To this end, let $\mathbf{E}(X)$ be the category of elements of $X$ and $P:\mathbf{E}(X)\to \mathbf{A}$ its projection functor (for the definition cf. Leinster's book linked above, on p. 155, Def. 6.2.16). Then Theorem 6.2.17 on p. 155 of the book reads 
  \begin{equation}
    \begin{split}
      X \simeq \lim_{\to\mathbf{E}(X)}H_{P(-)}.
    \end{split}
  \end{equation}
As a result,
\begin{equation}
\mathbf{PSh}(\mathbf{A})/X \simeq \mathbf{PSh}(\mathbf{A})/\lim_{\to\mathbf{E}(X)}H_{P(-)}.
\end{equation}
To finish the proof, one would need to show that the equivalence shown in the nlab-article is stable under limits; in other words that
\begin{equation}
\text{(eq. (1)) }\qquad \mathbf{PSh}(\mathbf{A})/\lim_{\to\mathbf{E}(X)}H_{P(-)} \simeq \mathbf{PSh}\bigg(\mathbf{A}/\lim_{\to\mathbf{E}(X)}P(-) \bigg).
\end{equation}
If one could do that, then one could conclude $\mathbf{PSh}(\mathbf{A})/X \simeq \mathbf{PSh}(\mathbf{B})$ for $\mathbf{B}:=\mathbf{A}/\lim_{\to\mathbf{E}(X)}P(-)$.
However, I do not know how to show that (eq. (1)) holds. Many thanks.
 A: The trick is that there's no need to take a colimit. The base category can just be taken to be the category of elements. 
Here's how it goes:
Suppose I have a map of presheaves on a category $\newcommand\A{\mathcal{A}}\A$, 
$f: E\to X$.
Then if $e\in E(b)$, $f(e) \in X(b)$, and if $g:a\to b$ is a morphism in $\A$,
we know $f(e|_g)=f(e)|_g$. In other words, $f:E\to X$ is equivalent to a presheaf on the category of elements of $X$. 
Let me break down how that works. We start with an object in the slice category $f:E\to X$.
Let $(a,u)$ be an object in the category of elements of $X$, so $a\in\A$ and $u\in X(a)$.
Let $g:a\to b$ be a morphism $(a,u)\to (b,v)$ with $v|_g = u$ in the category of elements of $X$.
Then we define 
$$E_f(a,u) = f_a^{-1}(\{u\}) = \{e\in E(a) : f(e) = u\}.$$
Then the observation above says that if $g:(a,u)\to (b,v)$, and 
$e\in E_f(b,v)$ (meaning $f(e)=v$), then $f(e|_g) = f(e)|_g = v|_g=u$, so $e|_g\in E_f(a,u)$. Therefore the presheaf structure on $E$ induces a presheaf structure on $E_f$.
Similarly, a morphism $h:(f:E\to X)\to (f':E'\to X)$ in the slice category will induce a
 morphism $h : E_f \to E'_{f'}$.
Conversely, to go backwards, if $E$ is a presheaf on the category of elements of $X$, then 
define 
$$E(a) = \bigsqcup_{u\in X(a)} E(a,u),$$
and define $f$ by 
$f(u,e) = u$, for $(u,e)\in E(a)$, $u\in X(a)$, $e\in E(a,u)$.
It's not hard to check that these are inverse equivalence functors.
