Let $\Bbbk$ be a field, $X$ affine scheme of finite type over $\Bbbk$. Let $\mathcal C\_X$ be the category of closed embeddings of $X$ into (say affine) *smooth* $Y$'s of finite type over $\Bbbk$, morphisms being closed embeddings of $Y$'s (forming commutative triangle). We have a functor $\mathcal C\_X^{\mathrm{op}}\to \text{Vect}\_\Bbbk$ by $[X\hookrightarrow Y] \mapsto\Omega^1(Y)$ You can also post-compose it with the embedding $\text{Vect}\_\Bbbk\to\text{Ch}\_\Bbbk$ to the $\infty$-category of chain complexes.


I would like to claim that *if you take the direct limit over $\mathcal C^{\mathrm{op}}\_X$ of both versions of this functor (the vector-space and the chain-complex ones), what you get is the module of Kähler differentials, resp. the Illusie cotangent complex, of $X$.* (There is a natural map in one direction and I claim it's an isomorphism of $\Bbbk$-vector spaces, resp. of objects of the derived category of $\text{Vect}\_\Bbbk$.)



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> Are such statements known? Or is there a reference for something similar? 
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(I think I have an argument at least for the first version, but I couldn't find any references.)


**Edit:** (some additional thoghts)


1. Clearly the first statment follows from the second, since the functor $H\_0$ from non-negatively (homologically) graded complexes to vector spaces preserves colimits.
2. The cotangent complex is defined as a similar colimit for some chosen "cosimplicial resolution" of $X$ by smooth affine varietis. The colimit in question can thus be thought of as "taking all such reslutions at once", but right now I can't formalize this.
3. It looks like instead of closed embeddings, one could take all maps to affines and (hope to) get the same answer.



A construction of Kähler differentials and Illusie cotangent complex as colimit over embeddings
This is the approach to the cotangent complex in the Stacks project; there one uses all maps as in your Remark 3 (yes this gives the same answer). See the Stacks project chapter on the cotangent complex. In particular for schemes over a ring in particular see [Section Tag 08V7](http://stacks.math.columbia.edu/tag/08V7).


Your remark 2 can be made more precise by saying that the homology over your category is computed by taking the usual simplicial resolution of the ring and then applying the functor. The simplest case of this is discussed in [Section Tag 08PQ](http://stacks.math.columbia.edu/tag/08PQ).


