Let $\mathcal{E}$ be a topos, $X$ and $Y$ objects and $i$ a closed inclusion $Y \to X$. Let $c_{(-)}$ be a universal closure operation.
The function $i^*: Sub(X) \to Sub(Y)$ is a well defined bijection when restricted to dense subobjects.
I think the inverse is its left adjoint $\Sigma_i$ defined by $\Sigma_i (m) = i \circ m$.
The function $i^*$ is well defined because $c_Y(i^*(m)) = i^*(c_X(m)) = i^*(id_X) = id_Y$. I struggle proving $\Sigma_i$ is well defined. Assuming the statement
For monomorphisms $A \xrightarrow{m} B \xrightarrow{n} C$, we have that $c_B(g \circ f) = c_B(g \circ c_A(f))$
it follows because $c_X(i \circ m) = c_X(i \circ c_Y(m)) = c_X(i \circ id_Y) = c_X(i) = id_X$ (as $i$ is a closed inclusion), but I have not been able to prove that statement and am not sure whether it is even true.
The composite $i^* \circ \Sigma_i$ is the identity on the subobjects of $Y$ because $i$ is mono, so in particular it is the identity on the dense subobjects of $Y$.
We have a defining pullback
$\require{AMScd}$ \begin{CD} i^*(n) @>{f}>> N\\ @V{}VV @V{n}VV\\ Y @>{i}>> X \end{CD}
so $i^*(n)$ is a subobject of $N$ and $\Sigma_i (i^*(n)) = i^*(n) \xrightarrow{f} N \xrightarrow{n} X$. Now if $f$ is an isomorphism, this must mean that $\Sigma_i (i^*(n))$ is $n$. But how do we prove this?
I am also getting more and more convinced that perhaps $\Sigma_i$ is not the inverse. As $\mathcal{E}$ is regular, we can send $m$ to the image of $i \circ m$, which I feel is another good candidate, but I am getting the same problems as with $\Sigma_i$.
A fact that may be useful: as $f$ is actually $n^*(i)$, it is a closed inclusion.
Proof: $c_N(f) = c_N(n^*(i)) = n^*(c_X(i)) = n^*(id_X) = id_N$.