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10:16 AM
Hi all, Im looking at Lurie's SAG, 25.3.1.5 page 1712. https://www.math.ias.edu/~lurie/papers/SAG-rootfile.pdf

The problem:We want to show the space of derivations is corepresentabe. i.e:
1. We have a functor Mod_A^cn -> S , from connective A-modules to spaces, given by
M \mapsto Der(A,M)
The latter object is space of derivations of A into M given by
Map_{CAlg/A}(A, A\oplus M) = "space of sections of projection of sqzero extension"

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Confusion: To show that this functor is corepresentable. The text claims it is accessible and limit preserving.
 
 
5 hours later…
3:15 PM
@MikeMiller it might be worth checking this out arxiv.org/abs/math/0406270
 
@JonathanBeardsley This is good evidence of your initial claim that the answer would be ugly! But it's also probably exactly what I need if I want to pursue this. Thanks, this saves me a lot of effort making attempts to spell things out, especially since this looks perhaps cleaner than the more naive approach I started fiddling with!
 
Bbb
4:11 PM
@BryanShih, I think in the simplicial setting in that chapter, you can see that it preserves small limits explicitly: he defines A \otimes M (25.3.1.1) as image of the classical construction “A \oplus M” under the equivalence of infinity categories 25.2.1.13, and you can check by hand that the classical construction commutes with small limits.
Conceptually, this might be clearer in the spectral setting, (similar statement might be true in the simplicial case too, not sure), where we know that A\oplus M is the image of the right adjoint in a stabilization adjunction to algebras over A.
 
 
1 hour later…
5:37 PM
@MikeMiller Perhaps. I mean, it's also evidence that my claim that it's impossible is inaccurate.
However, it does seem like one would need to extremely motivated to engage with this stuff.
 
6:00 PM
@Bbb Maybe it's also good to point out the intuition for this, e.g. if you look at a product in $C_{/A}$, you need to take the pullback in $C$ of the obvious span, in particular for $A\oplus M\to A \leftarrow A\oplus N$, this will clearly give you $A\oplus (M\oplus N)$; and it works similarly for other limits
 

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