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I have the following question. Let's denote by $\Omega$ the cobar functor, by $C$ some DG-coalgebra (not co-commutative) over a field $k$. Also suppose $V=k\times\dots\times k$ is just the product of $n$ copies of the base field $k$, and $V^*$ is the linear dual of $V$, which is a coalgebra in a natural way.

Q: How can I describe the algebra $\Omega(C\otimes_k V^*)$?

The question boils down to describing $\Omega(C\oplus C)$. The first guess was that $\Omega(C\oplus C)$ is just the free product of copies $\Omega(C)$. I doubt this is true but I am struggling with proving or disproving this.

What can be said about the cobar construction of a direct sum af any two coalgebras?

Sorry if this question is silly! I might be overlooking something obvious.

Thanks a lot for your help!

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Notice that $V^*$ is the direct sum of $1$-dimensional coalgebras, so by induction, you are asking, more generally, what is the cobar construction of a direct sum of coalgebras. –  Mariano Suárez-Alvarez Jul 7 '13 at 19:53
    
@MarianoSuárez-Alvarez Yes, you're right! Thanks! Do you think I should edit the question to state it in the more general form? –  Sasha Patotski Jul 7 '13 at 20:09
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up vote 2 down vote accepted

I am sorry, I guess I was endeed overlooking some simple things.

The cobar functor is left adjoint, so it commutes with colimits. In particular it commutes with coproducts. The part I was missing was that in the category of coalgebras the direct sum of underlying vector spaces gives the coproduct. The coproduct in the category of algebras is the free product. So indeed $\Omega(C\oplus D)=\Omega(C)*\Omega(D)$.

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