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In his paper "Numerology in Topoi" available here:

Peter Freyd defines an object $A$ in a topos $\mathcal{E}$ to be geometrically finite if whenever $f:\mathcal{B}\rightarrow\mathcal{E}$ is a geometric morphism with $\mathcal{B}$ Boolean, $f^{*}(A)$ is K-finite. He claims without proof that the following object $P$ in the topos $Sh(\mathbb{R})$ has this property: $P$ is the sheaf of germs of functions $f:\mathbb{R}\rightarrow\mathbb{N}$ such that $f(x)=0$ for all but finitely many rational $x$, and $f(p/q)\leq q$. The fact that $P$ has this property is not at all obvious to me. Attempting to prove this, here's what I have so far:

Geometric morphisms as above correspond via Diaconsecu's theorem to continuous flat functors $F:\mathcal{O}(\mathbb{R})\rightarrow\mathcal{B}$. Now $\mathcal{O}(\mathbb{R})$ has all finite limits, so $F$ is flat iff it is left exact, and in this case continuous is equivalent to preserving arbitrary joins, so such an $F$ is essentially a frame homomorphism from $\mathcal{O}(\mathbb{R})$ to $Sub_{\mathcal{B}}(1)$. The inverse image of the geometric morphism corresponding to $F$ is given by tensoring with $F$, so $f^{*}(P)$ is $P\otimes F$ which can be computed as the colimit of the composite $\int P\rightarrow \mathcal{O}(\mathbb{R})\rightarrow\mathcal{B}$ where $\int P$ is the category of elements of $P$, the first functor is the forgetful functor, and the second is $F$.

This colimit can be computed as the coequalizer of $\tau,\theta:\coprod_{U'\subseteq U}P(U)\times F(U')\longrightarrow\coprod_{U}P(U)\times F(U)$ where $P(U)$ means the $P(U)$-indexed coproduct of copies of $1$ in $\mathcal{B}$ (which is necessarily cocomplete), $\tau|_{P(U)\times F(U')}: P(U)\times F(U')\rightarrow P(U')\times F(U')$ is given by restriction and $\theta|_{P(U)\times F(U')}:P(U)\times F(U')\rightarrow P(U)\times F(U)$ is given by the inclusion $F(U')\subseteq F(U)$.

In nice situations where $f$ is, say, the inclusion of a point, this colimit is easy to compute, and K-finite as claimed, but I don't see any way to show that it must be K-finite in general Boolean $\mathcal{B}$. Any thoughts would be appreciated.

Edit: Some further progress: If $f$ is dense, it factors through $Sh_{\neg\neg}(\mathbb{R})$. A calculation I won't post here shows that the restriction of $P$ to $Sh_{\neg\neg}(\mathbb{R})$ is 1, so everything works fine. If $f$ is not dense we can let $U=\bigcup\{V\in\mathcal{O}(\mathbb{R})|F(V)=0\}$ and if $C=\mathbb{R}-U$, we have that $f$ will factor through $Sh(C)$, and this factorization will be dense, so in fact $f$ factors through $Sh_{\neg\neg}(C)$. Thus it suffices to show that the restriction of $P$ to double negation sheaves on a closed subtopos of $Sh(\mathbb{R})$ is K-finite. Hopefully nothing too pathological happens here.

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No, continuous does not mean preserving arbitrary joins. It would be more precise to say that it preserves "jointly epimorphic families", in the sense that if $\mathfrak{U} = \{ U_i \to V \}$ is an open cover of $V$, then the image of $\mathfrak{U}$ under $F$ is jointly epimorphic. However it is true that a left exact functor $F : \mathcal{O}(\mathbb{R}) \to \mathcal{B}$ must have image in the subterminals of $\mathcal{B}$, so in this case it is just a matter of preserving joins. – Zhen Lin Feb 19 '13 at 22:40
Yeah, my language was a bit clumsy there. Edited for clarity. – Shawn Henry Feb 19 '13 at 22:48

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