Homology of free loop space and Hochschild cohomology I am looking for honest proof of the following isomorphism.
For a simply connected space $X$, let $LX$ be its free loop space.
Then
$$H_{\ast}(LX) \simeq HH^\ast(C^\ast(X),C_\ast(X))$$
I have looked Jones' approach on ''Cyclic homology and equivariant homology'' that
given a map
$$f_k : \Delta^k \times LX \to X^{k+1}$$
defined by
$$(x_1,\cdots,x_k)\times \gamma \mapsto (\gamma(0),\gamma(x_1),\cdots,\gamma(x_k))$$
its cochain map
$$f_k^\ast : C^\ast(X)^{\otimes k+1} \to C^{\ast -k}(LX)$$
fit together to define a chain map from the Hochschild complex to the free loop space
$$f^\ast : CH_\ast(C^\ast(X)) \to C^\ast(LX)$$
which is a chain homotopy equivalence, so that we get an isomorphism
$$HH_\ast(C^\ast(X),C^\ast(X)) \simeq H^\ast(LX)$$
Many people simply say that ''dualizing'' the statement we get the relation between homology of loop space and Hochschild cohomology, but what is the honest procedure to dualize?
On Cohen and Voronov's Notes on String topology, they just take dual of $f^\ast_k$ and they claim that
$$C_\ast(LX) \simeq Hom(C^{\ast}(LX),\mathbb{Z})$$
and
$$Hom(C^\ast(X)^{\otimes k+1},\mathbb{Z}) \simeq Hom(C^\ast(X)^{\otimes k},C_\ast(X))$$
which also means 
$$C_\ast(X) \simeq Hom(C^{\ast}(X),\mathbb{Z})$$
However, there is no reason to be true that those groups are either isomorphic or quasi-isomorphic.
Please somebody help me.
 A: Let’s use the following grading:

*

*On a dual complex, $$((C_*)^\vee)^i=Hom(C_*,\mathbb{Z})^i=Hom(C_i,\mathbb{Z}).$$

*On the Hochschild chain complex of $C^*(X)$, $$CH_i(C^*(X),C^*(X))=\bigoplus_{i_1+\cdots+ i_k=i+k} C^{i_1}(X)\otimes\cdots\otimes C^{i_k}(X).$$
Therefore $CH_*$ is honestly $\mathbb{Z}$-graded, and the Hochschild differential increases degree: $\partial^*+\delta: CH_*\to CH_{*+1}$.

With the above grading, there is an isomorphism
$$H^i(LX)\cong HH_i(C^*(X),C^*(X))\ \ \ \forall i$$ in case $X$ is simply-connected.
The validity of dualizing the above isomorphism could come from the Universal Coefficient Theorem. For simplicity let’s work over $\mathbb{Z}$, and assume that homology groups are finitely generated in each degree.
For a degree-decreasing complex $\{\cdots \to C_*\to C_{*-1}\to \cdots\}$, form its dual complex $C^*=(C_*)^\vee$, then
$$F(H^i)\cong F(H_i),\ \ \ T(H^i)\cong T(H_{i-1}),$$
where $F,T$ denotes free part and torsion part, respectively. Similarly, if $\{\cdots \to C^*\to C^{*+1}\to \cdots\}$ is a degree-increasing complex, its dual complex $\bar{C}_*=(C^*)^\vee$ has
$$F(\bar{H}_i)\cong F(H^i),\ \ \ T(\bar{H}_i)\cong T(H^{i+1}).$$
As a consequence, take $\bar{C}_*= C^{\vee\vee}_* $ to be the cocochain complex obtained by doubly dualizing $C_*$, we have
$$F(\bar{H}_i)\cong F(H_i),\ \ \ T(\bar{H}_i)\cong T(H^{i+1})\cong T(H_i).$$
Therefore $C_*\to C_*^{\vee\vee}$ is a quasi-isomorphism.
By this MO question I think the finite generation assumption is satisfied in our situation.
Since $CH^*(C^*(X),C^*(X)^\vee)$ is exactly the dual complex of $CH_*(C^*(X),C^*(X))$, where the differential on the latter is degree-increasing,
we have
\begin{align*}
T(H_i(LX))&\cong T(H^{i+1}(LX))\\ &\cong T(HH_{i+1}(C^*(X),C^*(X)))\\ &\cong T(HH^{i}(C^*(X),C^*(X)^\vee)).
\end{align*}
The free parts are of course identified in the same degree, so
$$H_i(LX)\cong HH^{i}(C^*(X),C^*(X)^\vee).$$
Finally, the $C^*(X)$-bimodule quasi-isomorphism $C_*(X)\to C_*(X)^{\vee\vee}=C^*(X)^{\vee}$ induces an isomorphism
$$HH^{i}(C^*(X),C_*(X)))\cong HH^{i}(C^*(X),C^*(X)^{\vee}),$$ and we are done.
