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Apr 4, 2013 at 16:03 history edited Zhen Lin CC BY-SA 3.0
minor clarification
Apr 4, 2013 at 15:32 comment added David Carchedi (sorry, $\mathcal{T}\left(E,\mathcal{O}\right)$ is equivalent to $\phi\left(E\right)$, of course).
Apr 4, 2013 at 15:28 vote accept David Carchedi
Apr 4, 2013 at 15:28 comment added David Carchedi I.e. the functor $$\mathcal{T} \to \mathbf{Cat}$$ sending $E$ to $$Nat\left(\mathbf{FinSet}^{op},\mathcal{T}\left(1,E\right)\right)$$ is corepresentable by an object $\mathcal{O}$ such that for each $E$ in $\mathcal{T}$, $\mathcal{T}\left(\mathcal{O},E\right)$ is equivalent to $\phi\left(E\right)$. Nice!
Apr 4, 2013 at 11:54 comment added Zhen Lin @Simon I have added a paragraph to address your comments. The answer is yes.
Apr 4, 2013 at 11:53 history edited Zhen Lin CC BY-SA 3.0
added 583 characters in body
Apr 4, 2013 at 9:48 comment added Simon Henry In Ab, the object $\mathbb{Z}$ can be characterized in purely categorical term : for example it is the only projective object $Z$ such that any other projective can be written as a coproduct of a familly of copies of $Z$. I don't know if $Set[\mathbb{O}]$ can be characterized similarly in a purely categorical way. (or equivalently, does every self-equivalence of the 2-category of toposes preserve $Set[\mathbb{O}]$ up to equivalence). This question is actually equivalent to the the initial questions.
Apr 4, 2013 at 8:09 comment added Zhen Lin Yes, but how is this any different from needing to know which object is $\mathbb{Z}$ in $\textbf{Ab}$ in order to recover the underlying set of an abelian group?
Apr 4, 2013 at 8:01 comment added Simon Henry I mean't you need to assume that you know which object of $\mathcal{T}$ is equivalent to $Set[\mathbb{O}]$. I don't think there is an actual 'universal property' defining him...
Apr 4, 2013 at 7:45 comment added Simon Henry Yes but you need to know $\phi(E)$ in order to compute $Geom(\phi(E),Set[\mathbb{O}])$, no ?
Apr 4, 2013 at 7:22 history answered Zhen Lin CC BY-SA 3.0