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Let $\mathcal{T}$ be an algebraic theory (small category with finite products) and $\bar{\mathcal{T}}$ be its Cauchy completion.

What kind of functors (objects) yield a full subcategory of $\mathsf{Set}^{\bar{\mathcal{T}}}$ equivalent to the category of models $\operatorname{Mod} \mathcal T$ of $\mathcal T$ in $\mathsf{Set}$?

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    $\begingroup$ Yes. $Set^T$ and $Set^{\bar T}$ are canonically equivalent (by restricting along $T \to \bar T$), for any category $T$, and $Mod(T)$ is a full subcategory of $Set^T$. You probably mean to ask: how do you describe $Mod(T)$ in terms of $\bar T$. $\endgroup$ – Theo Johnson-Freyd Oct 14 '16 at 23:28
  • $\begingroup$ @TheoJohnson-Freyd Yes, you are right. I asked the wrong the question. $\endgroup$ – Stefan Perko Oct 15 '16 at 8:15
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Because $\bar{\mathcal{T}}$ is an algebraic theory itself we have:

$$\operatorname{Mod} \mathcal T \simeq \operatorname{Mod} \bar{\mathcal T}$$

because $\mathcal{T}$ and $\bar{\mathcal T}$ have the same Cauchy completion.

Taken from Adameck et al "Algebraic Theories" Chapter 15.

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