TITLE: How do you prove that Hochschild cohomology is Morita invariant?
QUESTION [7 upvotes]: I am simply trying to show that $HH^\bullet(A)= HH^\bullet(M_r(A))$ for any matrix ring of $A$. 
In Loday's book (Sect 1.5.6) the Morita invariance is explained as follows : it says that if $M$ is an $A$-bimodule, we have
$$HH^\bullet(M_r(A),M_r(M))= HH^\bullet(A,M) $$
If I put $M=A^*$, we get $HH^\bullet(A)=HH^\bullet(A,A^*)$ (Hochschild cohomology of $A$)
As per the formula, we get the left hand side to be $HH^\bullet(M_r(A),M_r(A^*))$. 
We would expect that the left side should give the Hochschild cohomology of $M_r(A)$. Hochschild cohomology of $M_r(A)$ is $HH^\bullet(M_r(A))=HH^\bullet(M_r(A),M_r(A)^*)$. 
So it remains to show that $M_r(A^*)=M_r(A)^*$ as $M_r(A)$-bimodules.
While this seems isomorphic as $k$-vector spaces, the isomorphism does not look like it preserves $M_r(A)$-bimodule structure.
Are we getting something wrong? Is the Morita invariance of Hochschild cohomology to be proved in some other way?

REPLY [5 votes]: It is even a derived invariant. Here is a proof in a special case, but Im not sure whether it works more general (with the same proof?)
Let $A$ and $B$ two noetherian $K$-algebras for a commutative ring $K$ that are projective as $K$-modules.
Then a standard derived equivalence $F=X \otimes_A^L: D^b(A) \rightarrow D^b(B)$ with quasi-inverse $G=Y \otimes_B^L $ induces a derived equivalence between the derived bimodule categories $H=(X \otimes_A^L - ) \otimes_A^L Y : D^b(A^e) \rightarrow D^b(B^e)$ where $A^e=A^{op} \otimes_K A$.
Now the Hochschild cohomology has terms $Ext_{A^e}^l(A,A)$ but $H$ sends $A$ to $B$ and thus preserves Hochschild cohomology. (it also preserves $A^{*}$ in case $A$ is finite dimensional, not sure about the general case)