MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
3

1

Hi all,

Sorry if this question is not the right level for mathoverflow, but I already tried math.stackexchange and received no answers.

Suppose that $\mathcal{E}$ is a well-pointed elementary topos, that $X$ and $Y$ are objects of $\mathcal{E}$, and that $F$ is a function which maps global elements $p: 1 \to X$ to global elements $F(p): 1 \to Y$ (here $1$ is the terminal object of $\mathcal{E}$). Does there exist a (necessarily unique) arrow $f: X \to Y$ in $\mathcal{E}$ such that $fp = F(p)$ for all $p$? Equivalently, is any object in a well-pointed topos the coproduct over its global elements of $1$? It's easy to show that the answer is "yes" if the coproduct exists since the induced map $\coprod_{p \in \Gamma X} 1 \to X$ is iso. But I don't know whether the coproduct exists in general.

flag
You are asking: given a map of sets, F:E(1,X) -> E(1,Y), is there an arrow X ->Y? If from F you can cook up a natural transformation E(-,X) -> E(-,Y), apply Yoneda and you are done. – David Roberts Oct 14 2010 at 22:31
I considered trying to use Yoneda, but I think it just takes me back where I started - constructing such a natural transformation $\sigma$ would in particular involve constructing the image $\sigma_X (1_X) \in \mathcal{R}(X, Y)$, which is what I need. – Phil Wild Oct 14 2010 at 22:43
Once you mention words like topos, I can hardly imagine that you are talking kinder-level of mathunderflow (math.stackexchange), but rather for here (unless one is throwing words without understanding). – Zoran Škoda Oct 16 2010 at 15:36
(sorry for the slow reply, only just noticed your comment) Well the question arose in the course of trying to understand a book that everyone working in the field has probably already read and understood, so I didn't think it would qualify as "research level". But thanks, it's good to know I'm not wasting MO's time. – Phil Wild Oct 20 2010 at 22:58

1 Answer

4

Here's a counterexample. Take $\mathcal E$ to be the topos of sets and functions of some countable model of ZFC (or a suitable weaker set theory, if you're worried that ZFC might be inconsistent). This is a well-pointed topos. The natural-number object $N$ in $\mathcal E$ has a countable infinity of global elements. So the number of functions (in the real world, not in the countable model) from global elements of $N$ to global elements of $N$ is the cardinal of the continuum. Only countably many of these correspond to morphisms in $\mathcal E$ from $N$ to $N$, because these morphisms are elements of your countable model.

link|flag
Very nice, thank you. Are you familiar with Mac Lane and Moerdijk's book? My question arose in trying to understand Section VI.10 in which they claim to prove that a well-pointed topos with NNO which satisfies the axiom of choice gives rise to a model of restricted Zermelo set theory with the axiom of choice. They do this by identifying sets with internal trees which satisfy certain additional properties, and identify two trees if there is an internal isomorphism of trees between them. [contd.] – Phil Wild Oct 14 2010 at 22:52
In proving that the so-defined sets satisfy extensionality, the authors suppose that every point (i.e. node covered by the root) of a tree $T_1$ is isomorphic to some point of $T_2$, and then state that this gives a morphism from the subobject of points of $T_1$ to that of $T_2$. Do you know if the proof is wrong, or can the relevant morphisms be shown to exist in this case? If it's wrong, can it be fixed? – Phil Wild Oct 14 2010 at 23:01
1 
At the moment, I don't have access to Mac Lane and Moerdijk, but from your description it seems to me that their proof is either OK or easily fixable. The point is that the desired mapping could be (and should be) defined in the internal logic of the topos. That's enough to produce a morphism in the topos. The analog of this in my example would be the case of those external maps from the global elements of N to global elements of N that are not just present in the real world but definable in the countable model. – Andreas Blass Oct 15 2010 at 2:00

Your Answer

Get an OpenID
or

Not the answer you're looking for? Browse other questions tagged or ask your own question.