Remark 22.37.10. Let $R$ be a ring. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded $R$-algebras which are derived equivalent, i.e., such that there exists an $R$-linear equivalence $D(A, \text{d}) \to D(B, \text{d})$ of triangulated categories. We would like to show that certain invariants of $(A, \text{d})$ and $(B, \text{d})$ coincide. In many situations one has more control of the situation. For example, it may happen that there is an equivalence of the form

$- \otimes _ A \Omega : D(A, \text{d}) \longrightarrow D(B, \text{d})$

for some differential graded $(A, B)$-bimodule $\Omega$ (this happens in the situation of Proposition 22.37.6 and is often true if the equivalence comes from a geometric construction). If also the quasi-inverse of our functor is given as

$- \otimes _ B^\mathbf {L} \Omega ' : D(B, \text{d}) \longrightarrow D(A, \text{d})$

for a differential graded $(B, A)$-bimodule $\Omega '$ (and as before such a module $\Omega '$ often exists in practice). In this case we can consider the functor

$D(A^{opp} \otimes _ R A, \text{d}) \longrightarrow D(B^{opp} \otimes _ R B, \text{d}),\quad M \longmapsto \Omega ' \otimes ^\mathbf {L}_ A M \otimes _ A^\mathbf {L} \Omega$

on derived categories of bimodules (use Lemma 22.28.3 to turn bimodules into right modules). Observe that this functor sends the $(A, A)$-bimodule $A$ to the $(B, B)$-bimodule $B$. Under suitable conditions (e.g., flatness of $A$, $B$, $\Omega$ over $R$, etc) this functor will be an equivalence as well. If this is the case, then it follows that we have isomorphisms of Hochschild cohomology groups

$HH^ i(A, \text{d}) = \mathop{\mathrm{Hom}}\nolimits _{D(A^{opp} \otimes _ R A, \text{d})}(A, A[i]) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{D(B^{opp} \otimes _ R B, \text{d})}(B, B[i]) = HH^ i(B, \text{d}).$

For example, if $A = H^0(A)$, then $HH^0(A, \text{d})$ is equal to the center of $A$, and this gives a conceptual proof of the result mentioned in Remark 22.37.9. If we ever need this remark we will provide a precise statement with a detailed proof here.

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