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Let $T$ be a small category, and $\mathrm{Mod}(T)\subset\mathrm{Fun}(T,\textbf{Set})$ the category of cartesian (finite-limit preserving) copresheaves on $T$. If $T$ is a commutative algebraic theory, then $\mathrm{Mod}(T)$ is cartesian closed (see commutative algebraic theory in nLab).

First question: Is the converse implication true? I mean: if $\mathrm{Mod}(T)$ is cartesian closed, can we prove that $T$ commutative?

Furthermore, if $T$ is the full subcategory of the simplicial category $\Delta$ with objects $[0], [1], [2], [3]$ (where $[n]$ is the order $0<1<\dotsm<n$) we have that $\mathrm{Mod}(T)= \textbf{Cat}$ is cartesian closed, in fact, such a $T$ is representable as a monoid in the cartesian-monoidal category $\mathrm{Cat}^{\mathrm{op}}\downarrow ([0]\times [0])$ (equivalently, objects are spans to $[0]$ and $[0]$ in $T$, the monoidal product is by pullbacks) and the image of a model $M$ is just a monoid in $\textbf{Set}\downarrow C_0\times C_0$ (where $C_0=M([0])$), then a small category with $C_0$ as class object. Analogous argument for functors.

Second question: Does there exist a law to recognize from the diagram structure of $T$ if $\mathrm{Mod}(T)$ is cartesian closed?

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    $\begingroup$ I believe it is closed monoidal for a not necessary cartesian monoidal structure - like abelian groups, say. $\endgroup$ – მამუკა ჯიბლაძე Feb 26 '18 at 22:48
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    $\begingroup$ Maybe I'm just missing something obvious, but in the third paragraph, don't you need to add the object $[3]$ to your $T$ to ensure associativity? (And by the way, I think you meant that $Mod(T^{op})=\mathbf{Cat}$.) $\endgroup$ – Arnaud D. Feb 27 '18 at 9:47
  • $\begingroup$ Arnaud, yes of course, right, I'll edit.. $\endgroup$ – Buschi Sergio Feb 27 '18 at 17:32

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