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Suppose that $A$ is (complex) unital algebra. We will consider $A-A$ bimodules $M$: such a bimodule is the same as (say) left $A \otimes A^{op}$ module. Let us define $C_n(A,M)$ as $M \otimes A^{\otimes n}$ and $C^n(A,M)$ as $Hom(A^{\otimes n},M)$ ($C^0(A,M):=M$) and define maps $b$ and $\delta$ with the formulas $$b(m \otimes a_1 \otimes ... \otimes a_n)=ma_1 \otimes a_2 \otimes ... \otimes a_n+\sum_{j=1}^{n-1}(-1)^j m \otimes a_1 \otimes ... \otimes a_ja_{j+1} \otimes ... \otimes a_n+(-1)^n a_nm \otimes a_0 \otimes ... \otimes a_{n-1}$$ (on $C_0(A,M)=M$ we define $b=0$) $(\delta m)(a)=ma-am$ and $$(\delta f)(a_1,...,a_{n+1})=a_1f(a_2,...,a_{n+1}+\sum_{j=1}^{n}(-1)^jf(a_1,...,a_ja_{j+1},...,a_{n+1})+(-1)^{n+1}f(a_1,...,a_n)a_{n+1}$$ One checks that $b^2=0$ and $\delta^2=0$. Therefore we get (co)chain complexes $C^*(A,M)$ and $C_*(A,M)$. The underlying (co)homology is called Hochschild (co)homology. In order to prove that Hochschild (co)homology could be computed as the derived functor I have a problem with the following thing: according to J. Varilly, H. Figureoa, J.M Bracia-Bondia book one can identify $M \otimes_{A \otimes A^{op}} A^{\otimes (n+2)} \cong M \otimes A^{\otimes n}$ via the map: $m \otimes a_0 \otimes ... \otimes a_{n+1} \mapsto a_{n+1}ma_0 \otimes a_1 \otimes ... \otimes a_n$.

Question 1: Why is this map an isomorphism?

What is evident, is that this map is onto: but why it is one-to-one and why it is corerctly defined (autohrs didn't comment how $M \otimes_{A \otimes A^{op}} A^{\otimes (n+2)}$ and $M \otimes A^{\otimes n}$ are made into $A \otimes A^{op}$ modules: the problem is of course with $A \otimes A^{op}$ multiplication).
For cohomology authors claim that $Hom_{A \otimes A^{op}}(A^{\otimes (n+2)},M) \cong Hom(A^{\otimes n}, M)$ via the map $\alpha$ where $(\alpha T)(a_1,...,a_n):=T(1,a_1,...,a_n,1)$. Again, I'm not sure how $Hom_{A \otimes A^{op}}(A^{\otimes (n+2)},M)$ and $Hom(A^{\otimes n}, M)$ are made into $A \otimes A^{op}$ my second qestion is the following:

Question 2: Why is this map an isomorphism?

I always tried to be very careful checking all the properties which most of the authors finds trivial (I'm very careful especially when I meet something for the first time) so I would like to have a clear picture. Therefore I would be grateful for any help.

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    $\begingroup$ $M$ and $A^{\otimes n}$ are both $(A,A)$ bimodules. The tensor product of $(R,S)$ and $(S,T)$ bimodules is naturally an $(R,T)$ bimodule, so $M\otimes A^{\otimes n}$ is naturally an $(A,A)$ bimodule, i.e. the left multiplicaiton is via $M$ and the right multiplication via $A^{\otimes n}$. Consideration of the case $M=A,n=2$ should clarify the injectivity claim. Hom into a left module (out of any vector space) is always a left module via $(af)(m)=a(f(m))$ and similarly for right modules. $\endgroup$ Feb 23 '15 at 0:43
  • $\begingroup$ So you prefer to think of $M \otimes A^{\otimes n}$ as of $A-A$ bimodule. But what about $M \otimes_{A \otimes A^{op}} A^{\otimes (n+2)}$? While we tensor over $A \otimes A^{op}$ and both $M$ and $A^{\otimes (n+2)}$ are $(A,A)$ bimodules, therefore left $A \otimes A^{op}$ (as I explained) I guess that we take a tensor product as left $A \otimes A^{op}$ modules. So how we make $M \otimes_{A \otimes A^{op}} A^{\otimes (n+2)}$ into left $A \otimes A^{op}$ module (or $A-A$ bimodule)? $\endgroup$
    – truebaran
    Feb 23 '15 at 1:26
  • $\begingroup$ It seems all the definitions of (Co)homology are irrelevant in this post. $\endgroup$
    – Pedro Tamaroff
    Feb 23 '15 at 3:27
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First, the $A\otimes A^{op}$-module structure on $A^{\otimes(n+2)}$ is of course by multiplication on the left on the first factor and multiplication on the right on the last. More precisely : $$(a\otimes b).(a_0\otimes\dots\otimes a_{n+2})=aa_0\otimes\dots a_{n+2}b$$

Hence, it is also isomorphic as an $A\otimes A^{op}$-module to $(A\otimes A^{op})\otimes A^{\otimes n}$ via $$ a_0\otimes\dots\otimes a_{n+1}\mapsto (a_0\otimes a_{n+1})\otimes (a_1\otimes\dots\otimes a_n)$$

Now, $M$ is a right $A\otimes A^{op}$-module via $$m.(a\otimes b)=b.m.a$$

I claim that for any ring $B$, any right-module $M$ and any left-module $N$ $$M\otimes_B (B\otimes N)=M\otimes N$$ via the isomorphism $m\otimes (b\otimes n)\mapsto mb \otimes n$ whose inverse is $m\otimes n\mapsto m\otimes (1\otimes n)$. With $B=A\otimes A^{op}$ and $N=A^{\otimes n}$, this is the isomorphism you seek. Explicitely, the inverse of the morphism you wrote is $$m\otimes a_1\otimes\dots\otimes a_n\mapsto m\otimes 1\otimes a_1\otimes\dots\otimes a_n\otimes 1$$

I let you check the dual statement, this is exactly the same module structures. Just note that we don't need any module structure on $M\otimes A^{\otimes n}$ nor do we need on $Hom(A^{\otimes n},M)$.

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  • $\begingroup$ Thank you for your answer. Just to be sure: you view $M \otimes_B (B \otimes N)$ and $M \otimes N$ only as vector spaces? I tell you the whole context: one starts with a resolution of $A$ which is viewed as $A \otimes A^{op}$ left module (so each term of this resolution is also an $A \otimes A^{op}$ module) and then we tensor with bimodule (therefore equivalently left $A \otimes A^{op}$ module) $M$: so it seems that we want to stay within the category of left $A \otimes A^{op}$ modules? $\endgroup$
    – truebaran
    Feb 23 '15 at 15:54
  • $\begingroup$ @truebaran If $M$ is a right $B$-module and $N$ a left $B$ module, $M\otimes_B N$ does not have any $B$-module structure any more. However, if $M$ or $N$ have other structures, those extends to $M\otimes_B N$. Moreover, the isomorphism I wrote is compatible with any of these structures. In your context, $M\otimes A^{\otimes n}$ have in fact many structure : it is an $A^{\otimes (n+1)}\otimes (A^{op})^{\otimes(n+1)}$-module, we just forget many of them. $\endgroup$
    – Roland
    Feb 23 '15 at 16:26
  • $\begingroup$ But, as I said, there is indeed an $A\otimes A^{op}$-module structure on $A^{\otimes(n+2)}$ but it does not exist any more on $M\otimes_{A\otimes A^{op}} A^{\otimes(n+2)}$. Hence, even if everything is an $A\otimes A^{op}$-module, we don't stay in the category of $A\otimes A^{op}$-modules. $\endgroup$
    – Roland
    Feb 23 '15 at 16:33
  • $\begingroup$ I should probably add that, unless $A$ is commutative, Hochshild homology and cohomology are only vector spaces. $\endgroup$
    – Roland
    Feb 23 '15 at 16:50
  • $\begingroup$ So these two isomorphisms about which I asked the question has to be understood as isomorphism of vector spaces and after tensoring (in the case of homology) or taking $Hom$ functor (in the case of cohomology) we get a complex of vector spaces? And one more thing, just to be sure: we write $Tor^{A \otimes A^{op}}_n(A,M)$ or $Ext^n_{A \otimes A^{op}}(A,M)$ (I mean that we write upper or lower script $A \otimes A^{op}$) just to underline the fact that we apply functors of tensoring over $A \otimes A^{op}$ or $Hom_{A \otimes A^{op}}(\cdot,M)$? $\endgroup$
    – truebaran
    Feb 23 '15 at 22:33

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