This question is a continuation of the discussion Normalization of Hochschild cocycles but this time in the cyclic context. I would like to ask whether the following is true:

The inclusion of normalized cyclic cochains into all cyclic cochains induces an isomorphism in (cyclic) cohomology.

In the discussion linked above I gave the argument which however turns out to be invalid, as indicated in the comment. Let me summarize what I have figured out already after reading carefully Loday's book: let $C^k=C^k(A)$ (where $A$ is an algebra) be a space of $k+1$-linear maps $\varphi: A^{k+1} \to \mathbb{C}$. Since drawing diagrams here is somehow problematic I will try to briefly describe underlying complexes:

**1.** The cyclic bicomplex $\mathcal{C}$ has in zeroth row $C^0$, in the first row $C^1$ and so on: vertical differentials are $b$ and $-b'$ (for even/odd columns) while horizontal differentials are $1-\lambda$ and $N$ where $\lambda$ is cyclic operator and $N=1+\lambda+...+\lambda^k$.

**2.** $(b,B)$-bicomplex $\mathcal{B}$ has in the zeroth row: $C^0,0,0,...$ in the first $C^1,C^0,0,0,...$ in the second $C^2,C^1,C^0,0,...$. Vertical differential is $b$ and horizontal differential is $B=Ns(1-\lambda)$ where for $\varphi \in C^n$ we define
$$(s\varphi)(a_0,...,a_{n-1})=(-1)^{n-1}\varphi(1,a_0,...,a_{n-1})$$

**3.** Normalized version of $\mathcal{B}$ bicomplex, denoted by $\overline{\mathcal{B}}$ where all cochains are normalized, meaning that $\varphi (a_0,a_1,...,a_n)=0$ whenever $a_k=1$ for some $k>0$ ($0$-entry is excluded!). Note that $b$ as well $B$ preserves normalization (unlike $b'$ which prevents us from considering normalized version of the cyclic bicomplex)

**4.** Finally we have ordinary cyclic complex $C^*_{\lambda}$ with differential $b$ together with its normalization $\overline{C^*_{\lambda}}$

Bicomplexes give rise to cohomology via totalization of complexes.

**Theorem 1.** The inclusion $C^*_{\lambda} \to Tot \ \mathcal{C}$, $\varphi \mapsto (0,0,...,0,\varphi)$ is a quasi-isomorphism.

**Theorem 2.** The map $(id,Ns):Tot \ \mathcal{C} \to Tot \ \mathcal{B}$ is also a quasiisomorphism.

Luckily, even the second map is not an inclusion, if we look at the corresponding degrees of cochains, we can see that the composition $C^*_{\lambda} \to Tot \ \mathcal{C} \to Tot \ \mathcal{B}$ is also an inclusion and is clearly a quasi-isomorphism.

Finally we have that the inclusion $Tot \ \overline{\mathcal{B}} \to Tot \ \mathcal{B}$ is a quasi-isomorphism.

If we could somehow show that the inclusion $\overline{C^*_{\lambda}} \to Tot \overline{\mathcal{B}}$ is a quasi-isomorphism we would obtain that the inclusion $\overline{C^*_{\lambda}} \to C^*_{\lambda}$ is also a quasi-iso. However I'm not so sure about this fact: while showing that the inclusion $C^*_{\lambda} \to Tot \ \mathcal{B}$ is quasi-iso we have after all used the auxiliary complex, namely cyclic bicomplex (which is absent in the normalized setting). Therefore I'm stuck and would be very grateful for any help.