The question I found myself wondering about was: is the functor $$\mathscr{F} \mapsto \mathscr{F}_\infty := \varinjlim_{R\to \infty} \Gamma((R, \infty), \mathscr{F})$$ a point of the topos $\mathbb{R}$?
I've found some comments that since $\mathbb{R}$ should presumably be a "tame" topological space, every topos point of $\mathbb{R}$ would be the functor of stalks at some element of $\mathbb{R}$, implying that my functor is not a point. What I was looking for, though, was a more direct proof that the given functor is not a point.
It does seem that this functor preserves finite limits and small coproducts; so the failure must be a failure to preserve coequalizers.
What I came up with was: since a point must preserve small colimits, and since any epimorphism in a topos is a coequalizer of a parallel pair, that means that a point must preserve epimorphisms. On the other hand, if we consider the epimorphism $$\coprod_{N \in \mathbb{Z}} (N-1, N+1) \to 1$$ then the $\infty$ stalk of the left hand side is $\emptyset$, whereas the $\infty$ stalk of the right hand side is 1, so the $\infty$ stalk functor does not preserve epimorphisms.
(The original counterexample I had, which I simplified to the above, was: consider the morphism $C^1(\mathbb{R}) \to C(\mathbb{R})$ defined by $f \mapsto f' - f^2$. I think this should be an epimorphism essentially by Picard's theorem on solutions to ODEs. On the other hand, any inverse image of $1 \in C(\mathbb{R})$ is locally of the form $f(x) = \tan(x - a)$ for some $a$, so the connected components of the domain are at most of length $\pi$. Thus, no inverse image of 1 has a stalk at $\infty$, again showing that the $\infty$ stalk functor does not preserve epimorphisms. I just wanted to include this in case it might seem like a more "naturally arising" counterexample.)
So, the question is: is this a valid proof, or is there something silly that I'm missing?