most recent 30 from http://mathoverflow.net 2013-05-24T00:06:55Z http://mathoverflow.net/feeds/question/109469 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109469/coinduction-and-corestriction-are-quasi-inverse-equivalences-for-comodules Julian Kuelshammer 2012-10-12T16:55:53Z 2012-10-12T16:55:53Z

<p>I'm reading <a href="http://arxiv.org/abs/math/0310337" rel="nofollow">http://arxiv.org/abs/math/0310337</a>.</p> <p>There the following statement is given without proof: Let $k$ be a field. Let $C$ be a counitary coaugmented coalgebra, i.e. there is $\eta: C\to k$ such that $(1\otimes \eta)\Delta=(\eta\otimes 1)\Delta=1$ and $\varepsilon: k\to C$ which is a counitary coalgebra map, in particular $\eta\varepsilon=id_k$. Define $\overline{C}:=coker \varepsilon$, $p:C\to \overline{C}$ the canonical projection.</p> <p>There are two functors: Corestriction from counital (i.e. $(1\otimes \eta)\Delta_N=id_{N}$) $C$-comodules which sends a $C$-comodule $N$ to the $\overline{C}$-comodule $N$ with comultiplication $(1\otimes p)\Delta_N$ and Coinduction: A Comodule $M$ is sent to the $ker u$, where $u=\Delta_M\otimes 1_C- (1_M\otimes p\otimes 1_C)(1_M\otimes \Delta): M\otimes C\to M\otimes \overline{C}\otimes C$.</p> <p>The author claims that these two are quasi-inverse equivalences.</p> <p>I don't even see what the isomorphisms between $N$ and $\ker u$ should be. A canonical choice would be $\Delta_N: N\to \ker u$ and $1\otimes \eta: \ker u\to N$, then one has $(1\otimes \eta)\Delta_N=id_N$, but where does the other direction come from? Another natural choice for a map $N\to N\otimes C$ would be $1\otimes \varepsilon: N\cong N\otimes k\to N\otimes C$ but I also don't see it there.</p>