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A feature of derived algebraic geometry is that we have internal homs. Furthermore, we can think of $B\mathbb{Z}$ as the derived algebraic geometric analogue of $S^1$. Thus we have an analogue of the loop space, the derived loop space $$LX:=\mathbb{R}\operatorname{Map}(B\mathbb{Z},X)$$ using the notation of "Chern character, loop spaces and derived algebraic geometry" (https://hal.archives-ouvertes.fr/hal-00772859/document). My question is how does this interact with the étale homotopy type? Specifically I would like to know if the following is true $$\operatorname{Et}(LX)=\Omega \operatorname{Et}(X)$$ where the right hand side denotes the loop space of the étale type of $X$. A naive hope for a proof would be $$\operatorname{Et}(\mathbb{R}\operatorname{Map}(B\mathbb{Z},X))\cong \operatorname{Map}(\operatorname{Et}(B\mathbb{Z}),\operatorname{Et}(X)) \cong \operatorname{Map}(S^1,Et(X))$$ but I don't know if this makes sense.

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    $\begingroup$ The proof should be a bit simpler than that, as you can write $LX = X \times^h_{X \times X} X$. (You can always express maps from a simplicial set as a homotopy limit.) The etale homotopy type won't see any of the derived structure, though. $\endgroup$
    – Jon Pridham
    Commented Aug 3, 2020 at 17:19
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    $\begingroup$ By @Jon Pridham's comment you can see that the etale homotopy type of $LX$ is $Et(X)$, when $X$ is a scheme. This disagrees with the free loop space of $Et(X)$. $\endgroup$
    – Phil Tosteson
    Commented Aug 3, 2020 at 18:00
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    $\begingroup$ Could we at least hope that there exists some (derived) scheme $Y$ such that $\Omega Et(X)\cong Et(Y)$? $\endgroup$
    – curious math guy
    Commented Aug 3, 2020 at 18:05
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    $\begingroup$ It seems unlikely you'd have a natural way of constructing $Y$. Reverting to the original question, Friedlander found a class of smooth morphisms which become fibrations in etale homotopy (after fairly mild completion), so it would be enough to take $X$ to be a higher stack whose diagonal takes lies in that class. In practice, that would mean $BG$ and its higher analogues, for $G$ smooth (no derived structure). $\endgroup$
    – Jon Pridham
    Commented Aug 3, 2020 at 19:31
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    $\begingroup$ You could try to cheat and construct a scheme $Y(Z)$ with the homotopy type of an arbitrary simplicial set $Z$, by gluing algebraic simplices together. I'm not sure if this works /what finiteness conditions you need to impose on $Z$ to get a scheme versus an Ind scheme. $\endgroup$
    – Phil Tosteson
    Commented Aug 3, 2020 at 19:47

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