Does a geometric morphism $f\colon \cal E\to F$ preserves and reflects the subobject classifier?

Question

I'm stuck in the apparently easy exercise in the title; I tried to prove it twice but both arguments were flawed (one of the two: one can easily obtain a natural map $Sub_\mathcal E(A)\to Sub_\mathcal E(f_*A)$, but this is rarely an equivalence). A friend of mine proposed me a counterexample but I can't retrieve his proof... any clue?

Answer 1

A good candidate for a counterexamples is a "global section" functor $\Gamma={\cal E}(1,-):{\cal E}\to{\cal S}et$ from ${\cal E}$ to Sets. If ${\cal E}$ has all (small) coproducts (e.g. is a presheaf topos) then $\Gamma$ is the direct part of a geometric morphism.

Now consider the topos ${\cal G}$ of directed Graphs (with loops and multiple edges allowed). Then ${\cal G}(1,X)$ are just the loops of the graph $X$. Since the subobject classifier $\Omega$ has 3 different loops, the set $\Gamma(\Omega)$ has $3$ elements and cannot be equal to $2$. Also $\Gamma(2)=2$ but $2\not\cong\Omega$ in ${\cal G}$.