How to construct birelative Hochschild homology?definition of (semi)group (co)homologyHochschild homology...
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How to construct birelative Hochschild homology?
definition of (semi)group (co)homologyHochschild homology with trivial coefficients: how to make $K$ an $M_n(K)$-moduleHochschild (co)homology and derived functorsTwo-sided bar construction for algebras: $B_*(M,R^e,R)$ is quasi-isomorphic to $Motimes_{R^e}B(R,R,R)$Hochschild homology of dgas with nontrivial differentialMaps inducing identity in Hochschild and cyclic theoriesShowing induced action of G by conjugation on Hochschild-Serre $H_i(G/N, H_j(N,M))$ is trivialDualizing cochain complex and Hochschild complexHomology of free loop space and Hochschild cohomologyChain complex homotopy equivalent to its homology
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I am working through Lodays book "Cyclic Homology". For an unital $k$-Algebra $A $ ($k$ being some ring) and a two-sided ideal $I$ of $A$ he defines the relative Hochschild homology $HH_ast(A,I)$ as the homology of the complex $C(A,I):=ker(C(A)twoheadrightarrow C(A/I))$, with $C(A)$ denoting the Hochschild complex.
Now exercise 1.1.6 is to define the birelative Hochschild homology $HH_ast(A;I,J)$ for two two-sided ideals such that we get a long exact sequence
$$ldotsrightarrow HH_n(A,I)rightarrow HH_n(A/J,I+J/J)rightarrow HH_{n-1}(A;I,J)rightarrow HH_{n-1}(A,I)rightarrowldots$$
I tried to get this from a short exact sequence of chaincomplexes
$$0rightarrow C(A;I,J)rightarrow C(A,I) rightarrow C(A/J,I+J/J)rightarrow 0$$
with $C(A,I)rightarrow C(A/J,I+J/J)$ being induced by the map $C(A)rightarrow C(A/I)$ and $C(A;I,J)$ being the kernel of this map. The only problem is, that I really do not see why this map should be surjective. After thinking about it for a while it seems to me, that (if this approach is correct at all) this has to come from properties of the tensorproduct, not from homological algebra, but I am really stuck here.
homology-cohomology homological-algebra
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$begingroup$
I am working through Lodays book "Cyclic Homology". For an unital $k$-Algebra $A $ ($k$ being some ring) and a two-sided ideal $I$ of $A$ he defines the relative Hochschild homology $HH_ast(A,I)$ as the homology of the complex $C(A,I):=ker(C(A)twoheadrightarrow C(A/I))$, with $C(A)$ denoting the Hochschild complex.
Now exercise 1.1.6 is to define the birelative Hochschild homology $HH_ast(A;I,J)$ for two two-sided ideals such that we get a long exact sequence
$$ldotsrightarrow HH_n(A,I)rightarrow HH_n(A/J,I+J/J)rightarrow HH_{n-1}(A;I,J)rightarrow HH_{n-1}(A,I)rightarrowldots$$
I tried to get this from a short exact sequence of chaincomplexes
$$0rightarrow C(A;I,J)rightarrow C(A,I) rightarrow C(A/J,I+J/J)rightarrow 0$$
with $C(A,I)rightarrow C(A/J,I+J/J)$ being induced by the map $C(A)rightarrow C(A/I)$ and $C(A;I,J)$ being the kernel of this map. The only problem is, that I really do not see why this map should be surjective. After thinking about it for a while it seems to me, that (if this approach is correct at all) this has to come from properties of the tensorproduct, not from homological algebra, but I am really stuck here.
homology-cohomology homological-algebra
$endgroup$
add a comment |
$begingroup$
I am working through Lodays book "Cyclic Homology". For an unital $k$-Algebra $A $ ($k$ being some ring) and a two-sided ideal $I$ of $A$ he defines the relative Hochschild homology $HH_ast(A,I)$ as the homology of the complex $C(A,I):=ker(C(A)twoheadrightarrow C(A/I))$, with $C(A)$ denoting the Hochschild complex.
Now exercise 1.1.6 is to define the birelative Hochschild homology $HH_ast(A;I,J)$ for two two-sided ideals such that we get a long exact sequence
$$ldotsrightarrow HH_n(A,I)rightarrow HH_n(A/J,I+J/J)rightarrow HH_{n-1}(A;I,J)rightarrow HH_{n-1}(A,I)rightarrowldots$$
I tried to get this from a short exact sequence of chaincomplexes
$$0rightarrow C(A;I,J)rightarrow C(A,I) rightarrow C(A/J,I+J/J)rightarrow 0$$
with $C(A,I)rightarrow C(A/J,I+J/J)$ being induced by the map $C(A)rightarrow C(A/I)$ and $C(A;I,J)$ being the kernel of this map. The only problem is, that I really do not see why this map should be surjective. After thinking about it for a while it seems to me, that (if this approach is correct at all) this has to come from properties of the tensorproduct, not from homological algebra, but I am really stuck here.
homology-cohomology homological-algebra
$endgroup$
I am working through Lodays book "Cyclic Homology". For an unital $k$-Algebra $A $ ($k$ being some ring) and a two-sided ideal $I$ of $A$ he defines the relative Hochschild homology $HH_ast(A,I)$ as the homology of the complex $C(A,I):=ker(C(A)twoheadrightarrow C(A/I))$, with $C(A)$ denoting the Hochschild complex.
Now exercise 1.1.6 is to define the birelative Hochschild homology $HH_ast(A;I,J)$ for two two-sided ideals such that we get a long exact sequence
$$ldotsrightarrow HH_n(A,I)rightarrow HH_n(A/J,I+J/J)rightarrow HH_{n-1}(A;I,J)rightarrow HH_{n-1}(A,I)rightarrowldots$$
I tried to get this from a short exact sequence of chaincomplexes
$$0rightarrow C(A;I,J)rightarrow C(A,I) rightarrow C(A/J,I+J/J)rightarrow 0$$
with $C(A,I)rightarrow C(A/J,I+J/J)$ being induced by the map $C(A)rightarrow C(A/I)$ and $C(A;I,J)$ being the kernel of this map. The only problem is, that I really do not see why this map should be surjective. After thinking about it for a while it seems to me, that (if this approach is correct at all) this has to come from properties of the tensorproduct, not from homological algebra, but I am really stuck here.
homology-cohomology homological-algebra
homology-cohomology homological-algebra
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TakirionTakirion
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