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Let $\mathcal{C}$ be a (possibly enriched) category with all finite product, and $\mathbf{M}$ a class of morphisms.
Can one construct completion of $\mathcal{C}$ w.r.t. all pullbacks along morphisms in $\mathbf{M}$, which is product-preserving? if so, how to do it?

I am not familiar with this kind of constructions, so I'd be glad to see detailed answers or references. for example, what is the term "closure" in this answer mean and how to form it?

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  • $\begingroup$ About pullback completion see "SIMPLY CONNECTED LIMITS" R. Parè, Can. J. Math., Vol. XLH, No. 4, 1990, pp. 731-746. $\endgroup$ May 27, 2015 at 22:18

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In general, "the closure of X in Y under Z" means the smallest sub-thing of Y that contains X and is closed under Z; this can generally be obtained by intersecting all the sub-things with these properties. Generally Z is an operation, and "closed under Z" means that if the inputs to the operation lie in the sub-thing in question, so does the output. If X and Y are categories and Z is a class of limits or colimits, it means that if a Z-diagram lies in the subcategory then so does its limit or colimit.

In the case of your particular question, one possibility would be to consider the closure of (the image under the Yoneda embedding of) $C$ in its presheaf category $\mathrm{Set}^{C^{\mathrm{op}}}$ under pullbacks of $M$-morphisms. More precisely, you could define consider the smallest full subcategory $C'$ of $\mathrm{Set}^{C^{\mathrm{op}}}$ that contains (the image under the Yoneda embedding of) $C$ and has the property that if $X \to Y \leftarrow Z$ lies in $C'$ and $Z\to Y$ is in (the image under the Yoneda embedding of) $M$, then the pullback also lies in $C'$. The embedding of $C$ in $C'$ will preserve all limits that exist in $C$ (and hence in particular finite products), since the Yoneda embedding does.

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  • $\begingroup$ is there any difference (regarding completions of categories) between using the yoneda embedding $C\to [C^{op},\mathbf{Set}]$ and the opposite co-yoneda embedding $C\to [C,\mathbf{Set}]^{op}$? $\endgroup$
    – Yitzhak Z
    May 28, 2015 at 17:44
  • $\begingroup$ @YitzhakZ yes, there is. The yoneda embedding preserves all limits that exist in C, whereas it is the free cocompletion, meaning that it preserves almost no colimits that exist in C. The opposite Yoneda embedding has the opposite properties. $\endgroup$ May 29, 2015 at 20:07

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