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Given an algebraic/equational theory $T$, let $\mathcal{B}(T)$ be the classifying topos of $T$, regarded as the category of presheaves on the category of finitely presentable models of $T$ in Sets.

Given a cocomplete topos $\mathcal{E}$ and a $T$-model $M$ in $\mathcal{E}$, we get a corresponding uniquely determined (up to isomorphism) geometric morphism $p_M : \mathcal{E} \to \mathcal{B}(T)$ such that $p_M^*(U_T) \cong M$, where $U_T$ is the universal model of $T$ in $\mathcal{B}(T)$.

My question is, has anyone ever computed an explicit description of this geometric morphism $p_M : \mathcal{E} \to \mathcal{B}(T)$ corresponding to the model $M$, by chasing through all the equivalences in the proof that $\mathcal{B}(T)$ (as defined above) is the classifying topos of $T$? I.e. given a presheaf $F$ on the category of finitely presented models of $T$ in Sets, do we know what the object $p_M^*(F)$ is in $\mathcal{E}$?

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  • $\begingroup$ Such a presheaf $F$ is the colimit of representable presheaves (by density of the Yoneda embedding). A representable presheaf, that is a functor of the form $\mathrm{Hom}_{T\mathrm{-Mod}(\mathrm{Set})}(A, \cdot)$ for a finitely presented model $A$ of $T$ in $\mathrm{Set}$, is in turn a finite limit of the universal model $U_T$. Since $p_M^*$ preserves arbitrary colimits and finite limits, the value $p_M^*(F)$ is therefore determined by $p_M^*(U)_T$. I guess this is in Moerdijk/Mac Lane. $\endgroup$
    – Ingo Blechschmidt
    Commented Sep 23, 2016 at 22:14

1 Answer 1

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I believe it's basically straightforward, although I haven't carefully verified the following.

Let $C$ be the category of the category of finitely presented algebras. Then $p_M$ corresponds to a left exact functor $L : C^\circ \to \mathcal{E}$.

Letting $F_n$ be the free algebra on $n$ elements, the whole thing is basically set by $L(F_1) = M$.

Since $F_n$ is the coproduct of $n$ copies of $F_1$, we have $L(F_n) = M^n$.

Recall from universal algebra that an element of $\hom(F_1, F_n)$ is the same thing as an element of $F_n$, which is the same thing as an $n$-ary operation. Thus, for $f : F_1 \to F_n$, $L(f) : M^n \to M$ is the corresponding $n$-ary operation.

More generally, $\hom(F_m, F_n) = \hom(F_1, F_n)^m$, so for $f : F_m \to F_n$, $L(f) : M^n \to M^m$ is just an $m$-tuple of $n$-ary operations.

The remaining finitely presented algebras are given by coequalizers corresponding to relations asserting various pairs of operations are the same. $L$ applied to such an algebra gives the equalizer defining the subobject of tuples that have the same vale.

Finally, objects of $\mathcal{B}(T)$ are colimits of things in $C^\circ$, and $p_M^*$ maps colimits to colimits.

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