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

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?

-
Could you define "geometric" morphism? – magma Feb 29 '12 at 3:10
a couple of adjoint functors $f^*\dashv f_*$, such that $f^*\colon \cal F\to E$ commutes with finite limits. – Fosco Loregian Feb 29 '12 at 15:57
thank you, I just found it in Wikipedia too – magma Feb 29 '12 at 16:34
you're welcome; does the question is so hard nobody has a clue? – Fosco Loregian Feb 29 '12 at 17:32

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 ${\cal G}$.
Sorry, I did not notice your comment earlier. The category ${\cal G}$ is just the presheaf category ${{\cal S}et^{\mathfrak{a}^op}}$ where $\mathfrak{a} = \{D\rightrightarrows A\}$ is the full subcategory of $\cal G$ given by $D=\bullet$ and $A=\bullet\to\bullet$. As to the loops of $\Omega$, note that the maps in ${\cal G}(1,\Omega)$ must correspond o the (three different) subobjects of $1$ (which are $\emptyset$, $D$ and $1$ itself). – Marc Olschok Mar 12 '12 at 20:04