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Shameless upfloat of 1-year old question - the motivation is that in general the corresponding Banach version is false, so I am trying to see where the proof breaks down, and what (if anything) can be salvaged.

Rather embarrassingly, the question is on how to prove something I used to know how to do... but I thought that the collective wisdom of MO might be able to quickly sketch how the argument should go, or direct me to a reference.

Throughout all algebras are over a field $k$ of characteristic zero (in fact ${\mathbb C})$ and have identity elements. $\newcommand{\rhc}{\overline{HC}}$

Given such $A$ and $B$, let $A\ast B$ be their free product (coproduct in the category of unital algebras). I seem to remember convincing myself, or reading, that $$ \rhc_n (A\ast B) \cong \rhc_n(A) \oplus \rhc_n(B) $$ where $\rhc$ denotes reduced cyclic homology (so $\rhc_\bullet(k)=0$ rather than having non-zero contributions in even degrees).

Unfortunately I can't remember how the proof goes. My vague recollection is that it goes via the identification of $\rhc$ as the derived functor associated to the comonad ${\rm Vect}_k \leftrightarrow {\rm Alg}_k$ (which follows from the fact that for any vector space $V$, the reduced cyclic homology of the tensor algebra $T(V)$ vanishes in all positive degrees). But how to proceed? It should be some kind of acyclic complex argument, related to the fact that $T(V_1\oplus V_2) \cong T(V_1)\ast T(V_2)$, but I can no longer recall how these kinds of argument work in detail.

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Would whoever voted to close please leave a brief comment explaining why? I'd like to know whether you find this too vague, or off-topic, or too trivial, or some combination. Cf. the old MO purpose if "ask a question you don't have the background/details for, but which should be easily explainable and answered by a specialist with the right expertise". (I am not a homological or homotopical algebraiat) – Yemon Choi Jun 28 '13 at 18:12
Im thinking (in the periodic case) use the fact on tensor algebras you mention, then try to conclude via goodwillie? – CSA Apr 6 '14 at 3:12
@CSA Hmm, I'm away from my books right now & occupied with other projects, so can't check this right now. But thanks for the suggestion – Yemon Choi Apr 6 '14 at 4:31
no problemo. :) – CSA Apr 6 '14 at 14:59

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