6
$\begingroup$

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 $\mathcal{E}$?

$\endgroup$
  • $\begingroup$ 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 Sep 23 '16 at 22:14
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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