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I am trying to get some sense of the proof of Theorem 10 in this link.

https://www.math.ias.edu/%7Elurie/278xnotes/Lecture6-Completeness.pdf

Immediately after it, Exercise 11 says a model of coherent category with a chosen point is equivalent to a model of the slice category.

May I please ask how precisely should I build this correspondence? Even just a quick construction without proof details reply would be appreciated. I would like to check it.

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We have a coherent functor $\mathcal{C}\to\mathcal{C}_{/X}, Y\mapsto (\mathrm{pr}_2\colon Y\times X\to X)$, so if we have a model $M'\colon\mathcal{C}_{/X}\to\mathrm{Set}$ we get a model $M\colon \mathcal{C}\to\mathrm{Set}$ via compositon. The chosen element of $M[X]$ is induced by the diagonal map $X\to X\times X$ in the following sense: the diagonal $X\to X\times X$ in $\mathcal{C}$ is also a map $$(\mathrm{id}_X\colon X\to X)\to (\mathrm{pr}_2\colon X\times X\to X)$$ in $\mathcal{C}_{/X}$, and applying $M'$ to it gives us a map $*\cong M'[\mathrm{id}_X]\to M'[\mathrm{pr}_2\colon X\times X\to X]=:M[X]$. This selects the chosen element of $M[X]$.

Conversely, given a model $M\colon\mathcal{C}\to\mathrm{Set}$ with a chosen element $m\colon *\to M[X]$, we can define $M'[p\colon Y\to X]$ via a pullback square $$ \require{AMScd} \begin{CD} M'[p\colon Y\to X]@>>> M[Y]\\ @VVV @VV{M[p]}V\\ * @>{m}>> M[X] \end{CD} $$ in $\mathrm{Set}$. You can check that $M'\colon\mathcal{C}_{/X}\to\mathrm{Set}$ is indeed a functor and coherent, and that these assignments define a bijection between the appropriate sets Lurie talks about. If you want details about this, let me know and I will edit them in.

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  • $\begingroup$ Thanks a lot for this instant answer! May I please ask what is meant by the element of $MX$ is induces by the diagonal map? I would appreciate for more details because after seeing the construction, it is still not such clear to me that it is the obvious thing to do. $\endgroup$
    – Y.X.
    Commented May 16 at 20:08
  • $\begingroup$ I edited in the definition of the chosen element of $M[X]$. When you ask for more details, do you mean you also would like more details about why in the second paragraph $M'$ as defined there is coherent and why we get this bijection between the sets Lurie talks about? Or was the definition of the chosen element of $M[X]$ the only thing? (It's fine in either case, but I can read your comment either way, which is why I ask for this clarification.) $\endgroup$ Commented May 16 at 20:52
  • $\begingroup$ Thanks so much! Let me try to unfold it and see if there is any outstanding confusion! If there remains unclear detail, I would comment here. $\endgroup$
    – Y.X.
    Commented May 16 at 23:03

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