For $H$ a Hopf algebra, with bijective antipode. For a right, and a left, $H$-comodule $(V,\alpha_R)$, and $(W,\alpha_L)$ respectively, the cotensor product of $V$ and $W$ is $$ V \square_H W := \ker(\alpha_R \otimes \text{id} - \text{id} \otimes \alpha_L:V \otimes W \to V \otimes H \otimes W). $$

When does it hold that $$ V \square_H H ~~~ \simeq V? $$

  • $\begingroup$ Transferred from Stack Exchange due to no answer . $\endgroup$ – Alesandro Levi Jan 6 '17 at 11:28
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    $\begingroup$ Always, and you just need $H$ to be a coalgebra with a counit $\varepsilon$: use the maps $\mathrm{id}\otimes\varepsilon:V\otimes H\to V$ and $\mathrm{id}\otimes\varepsilon:V\otimes H\otimes H\to V\otimes H$ to split the equalizer. $\endgroup$ – მამუკა ჯიბლაძე Jan 6 '17 at 13:55
  • $\begingroup$ Can you explain "split the equalizer" please? $\endgroup$ – Alesandro Levi Jan 6 '17 at 14:28
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    $\begingroup$ It is a standard abstract nonsense trick, see e. g. in nLab $\endgroup$ – მამუკა ჯიბლაძე Jan 6 '17 at 14:33

For any $H$-comodule $V$ you have its structure map $\rho:V\to V\otimes H$. The image of $\rho$ is the equalizer defining cotensor product.


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