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It is well known that if $ R $ is a ring, then every $ R $-module $ M $ is semisimple (that is, $ M $ is the direct sum of simple $ R $-modules) if and only if every submodule of $ M $ is a direct summand. Is it true that if $ C $ is a coalgebra over a field $k$, then the analogous result holds: a coalgebra comodule $ N $ is cosemisimple if and only if every subcomodule is a direct summand?

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Yes, if your coalgebra is flat over the base ring $\Bbbk$ (for example when $\Bbbk$ is a field). You can find it in Brzezinski-Wisbauer, Corings and Comodules, §19.13 and results around.

Broadly speaking, if $P$ is a simple subcomodule of $N$ and $M$ is any subcomodule of $N$ then $P\cap M$ is still a subcomodule and it can only be $0$ or $P$. Therefore, if $N$ is cosemisimple then $M$ is the direct sum of the simple subcomodules it contains and hence is a direct summand of $N$ and conversely, if every subcomodule is a direct summand, then you check iteratively that $N$ is the direct sum of its simple subcomodules.

I would dare to say that the only difference is that in this case you don't know that $N$ is the direct sum of a finite number of summands, as it happens for modules over a semisimple ring. However, I am not completely sure of this claim at the present moment so, please, check and forgive me if I am wrong.

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