If C is a cocommutative R-coalgebra, R is some commutative semi-simple artinian ring and A and B are C-bicomodules, then is $A\square_C B \cong B \square_C A$?
If not what other conditions are required for this to hold?
2026-03-26 06:26:27.1774506387
Commutativity with cotensor
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The cocommutativity of C is not a requirement or the assumptions on R except it be unital and commutative.
The cotensor of N and M is defiend as: $N\square_C M = ker(p_N \otimes_R 1_M - 1_N\otimes_R p_M)$. Since M,N are symmetric wrt $\otimes_R$ then $p_N \otimes_R 1_M$ may be identified with $1_M \otimes_R p_N $, likewise with $1_N\otimes_R p_M$ and $p_M\otimes_R 1_N$.
The R-linearity of the maps then imply that $ker(p_N \otimes_R 1_M - 1_N\otimes_R p_M) \cong ker( 1_M \otimes_R p_N - p_M\otimes_R 1_N ) \cong ker( - 1_M \otimes_R p_N + p_M\otimes_R 1_N ) = M \square_C N$.
So if R is commutative, then the cotensor of any C-bicomodules is symmetric :)