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Denote by $HC(A,M)$ the Hochschild homological complex of an algebra $A$ with coefficients in an $A$-bimodule $M$, and let $B\rightarrow A$ be an $R$-flat extension of $R$-algebras, for some $CRing$ $R$.

On page 11 of this article: http://arxiv.org/pdf/1103.4377v2.pdf the author defines a filtration $G_n^p$ of the Hochschild complex, and makes the claim that $\underset{p \in \mathbb{N}}{\varinjlim} G_n^p$ is exactly the $n^{th}$ entry in the Hochschild complex $HC(A,M)$.

My question is short and simple.. why is this true?

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  • $\begingroup$ Can you at least tell us where in the paper is the claim made, or explain thr notation? Your question may be short but it is impossible to answer it without reading the whole paper... $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Apr 7, 2014 at 1:49
  • $\begingroup$ In any case, the claim is just claiming that the complex is the union of the layers of the filtration, which is really obvious: in degree n the layer with p=n is already the whole thing $G_n^n$ is equal to $ CH_n (A, M) $. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Apr 7, 2014 at 2:00
  • $\begingroup$ Its on page 11, I guess this is a simple one... my bad, thanks Mariano :) $\endgroup$
    – ABIM
    Commented Apr 7, 2014 at 2:23
  • $\begingroup$ What have to do Jacobi or Zariski (or the Jacobi-Zariski exact sequence) with this "excision type" exact sequence? $\endgroup$
    – Vinteuil
    Commented Apr 7, 2014 at 7:52

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As observed in that paper, $G_n^p=G_n^{p+1}$ for $p\geq n$, so that the colimit, which is simply the union, is equal to $G_n^n$. It is clear that $G_n^n=CH_n(\mathscr A,M)$, in the notation of the paper. This shows the claim you mention in each degree.

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