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Let $T$ be a sheaf topos and $I$ a small category. Then the functor category $[I,T]$ is also a sheaf topos. Now let $E$ be an elementary topos (cartesian closed category with finite limits + subobject classifier).

Are there additional conditions that we can impose on $E$ to guarantee that $[I,E]$ is an elementary topos?

I think we need completeness at least (since this ensures that $[I,E]$ is cartesian closed). The limits are also computed pointwise, so I guess the only issue is how to guarantee that $[I,E]$ has a subobject classifier...

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If $E$ is small complete, then $[I, E]$ is an elementary topos, by the following argument.

If $I_0$ is the discrete category of objects of $I$, then $[I_0, E]$ is just an $I_0$-indexed product of elementary toposes, and such is always an elementary topos.

Then $[I, E]$ is the category of coalgebras for a left exact comonad on a topos $[I_0, E]$, and such is always a topos. In detail, the comonad is a composite

$$[I_0, E] \stackrel{[d_1, E]}{\to} [I_1, E] \stackrel{\Pi_{d_0}}{\to} [I_0, E]$$

that is right adjoint to a monad

$$[I_0, E] \stackrel{[d_0, E]}{\to} [I_1, E] \stackrel{\Sigma_{d_1}}{\to} [I_0, E]$$

whose algebras are functors $I \to E$. Here $d_0, d_1: I_1 \rightrightarrows I_0$ are the domain and codomain functions on the set $I_1$ of morphisms of $I$, and $\Pi_{d_0}$ takes a $I_1$-tuple of objects $\langle c_f \rangle_{f \in I_1}$ to the $I_0$-tuple $\langle \prod_{f: d_0(f) = i} c_f \rangle_{i \in I_0}$ (which is where we use completeness). The category of coalgebras of the comonad is equivalent to the category of algebras of its left adjoint monad, by an old result which can be found in the famous 1965 paper of Eilenberg and Moore that introduced the category of algebras construction for decomposing a monad into a left adjoint followed by its right adjoint.

The results I'm mentioning are thoroughly discussed in the Mac Lane-Moerdijk book on topos theory. While I'm at it, let me mention that small completeness of a topos $E$ is equivalent to small cocompleteness of $E$: that completeness implies cocompleteness follows by the theorem of Paré that the power object functor $P: E^{op} \to E$ is monadic so that $E^{op}$ is small-complete if $E$ is, and the converse direction is not difficult.

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