Here's the idea.
The monomorphisms in the presheaf category are pointwise monomorphisms, so we can identify subobjects of a presheaf $X\in[\mathbf{A}^\text{op},\newcommand\Set{\mathbf{Set}}\Set]$ with subpresheaves of $X$, in the sense of presheaves $F$ such that $F(a)\subseteq X(a)$ for all objects $a$, and for $f:a\to a'$, $X(f)$ sends elements of $F(a')$ to $F(a)$.
Now suppose we have a subpresheaf $F$ of some presheaf $X$. We want to construct a natural transformation $X\to \newcommand\Sub{\operatorname{Sub}}\Sub(H_{-})$.
Thus for $a\in \newcommand\A{\mathbf{A}}\A$, $\alpha\in X(a)$, we need to construct a subpresheaf of $H_a$. By Yoneda, $\alpha$ corresponds to a natural transformation
$H_a\to X$, so we can just take the preimage of $F$ in $H_a$. In other words, define $G_\alpha\newcommand\into\hookrightarrow\into H_a$ by
$$G_\alpha(a') = \{ f : a'\to a \text{ such that } f^*\alpha \in F(a')\subseteq X(a')\}.$$
Then we define $\eta : X\to \Sub(H_-)$ by $\eta \alpha = G_\alpha$.
Conversely, from a natural transformation, $\eta : X\to \Sub(H_-)$, we can recover the subobject $F$ by
$$F(a) =\{ \alpha\in X(a) \text{ such that } 1_a \in (\eta_a\alpha)(a)\subseteq H_a(a)\}.$$
Side note: a subfunctor is a subobject of a functor, and a sieve is a subobject of a representable functor, but we don't really need to use these words to prove the claim.
Edit:
To see that $\eta$ is natural, let $f:a\to a'$, let $\alpha\in X(a')$.
We need to show that $\eta f^*\alpha = f^*\eta\alpha$.
Now
$$
(\eta f^*\alpha)(a'')
=
\{
g: a''\to a \text{ such that } g^*f^*\alpha \in F(a'')
\},
$$
and
$$
(f^*\eta\alpha)(a'')
=
(f_*)^{-1}((\eta\alpha)(a''))
=
\{
g: a''\to a \text{ such that } (f\circ g)^* \alpha \in F(a'')
\}.
$$
Thus, since $(f\circ g)^* = g^*f^*$, we have naturality.
Edit 2
I've been asked how we show that if we start with a natural transformation $\eta : X\to \Sub(H_-)$ and construct the associated subobject $F$ of $X$ how we show that the natural transformation $\overline{F}$ associated to $F$ is in fact $\eta$.
Let $a,a'\in \mathbf{A}$. Recall that
$$F(a) = \{ \alpha \in X(a) \text{ such that } 1_a \in \eta_a(\alpha)(a) \}.$$
We also know that if $\alpha \in X(a)$, then
$$
\overline{F}_a(\alpha)(a')
=
\{
g:a'\to a
\text{ such that }
g^*\alpha \in F(a)
\}.
$$
Putting these together we can compute
$$
\begin{aligned}
\overline{F}_a(\alpha)(a')
&=
\{
g:a'\to a
\text{ such that }
1_{a'}\in \eta_{a'}(g^*\alpha)(a')
\}
\\
&=
\{
g:a'\to a
\text{ such that }
1_{a'}\in g^*(\eta_{a}\alpha)(a')
\}
\\
&=
\{
g:a'\to a
\text{ such that }
1_{a'}\in (g_*)^{-1}(\eta_{a}\alpha)(a')
\}
\\
&=
\{
g:a'\to a
\text{ such that }
g\circ 1_{a'}\in (\eta_{a}\alpha)(a')
\}
\\
&=
\{
g:a'\to a
\text{ such that }
g\in (\eta_{a}\alpha)(a')
\}
\\
&=(\eta_a\alpha)(a').
\end{aligned}
$$
Thus as subobjects of $H_a$, we have that
$\eta_a\alpha = \overline{F}_a\alpha$, as desired.
The construction you want definitely works in any elementary topos (with an NNO), even those which are not Boolean, and in fact it will also work in any arithmetic universe ("predicative toposes, toposes without power objects").
I'm in a hurry, there are surely more elegant ways to construct $Q$, but here is one:
- Construct $List(P)$, the object of (finite) lists of values from $P$.
- Construct the map $N \to List(N)$ thought as mapping $n$ to $[0,1,\ldots,n-1]$.
- Intersect $List(P)$ with the image of that map (as subobjects of $List(N)$).
- The image of the resulting object under the map $length : List(N) \to N$ is the desired object $Q$.
Best Answer
The 2 objects of $\mathcal{E}$ and of $\mathcal{E}_{\lnot\lnot}$ are not necessarily the same object. (In fact, I think what your argument shows is that they are the same object if and only if $\mathcal{E}$ is a De Morgan topos).
As an example, consider the case $\mathcal{E} := \operatorname{Sh}(\mathbb{R})$. Then $2_{\mathcal{E}}$ can be described as the sheaf where $2_{\mathcal{E}}(U)$ is the locally constant functions $U \to 2$. Thus, in the case $U := \mathbb{R} \setminus \{ 0 \}$, $2_{\mathcal{E}}(U)$ has a section which is 0 on $(-\infty, 0)$ and 1 on $(0, \infty)$. We also see that the given section of $2_{\mathcal{E}}(U)$ does not extend to a global section of $2_{\mathcal{E}}(\mathbb{R})$. On the other hand, $h_U$ is $\lnot\lnot$-dense in $h_{\mathbb{R}} \simeq 1_{\mathcal{E}}$. This shows that $2_{\mathcal{E}}$ is not a $\lnot\lnot$-sheaf since $\operatorname{Hom}(h_{\mathbb{R}}, 2_{\mathcal{E}}) \to \operatorname{Hom}(h_U, 2_{\mathcal{E}})$ is not surjective.