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Oct 19, 2016 at 17:55 comment added HeinrichD This is a mistake, sorry. So probably there won't be an adjunction in this generality.
Oct 19, 2016 at 17:54 history edited HeinrichD CC BY-SA 3.0
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Oct 19, 2016 at 15:31 comment added fosco I don't understand why fully faithfulness of $G$ implies $\int^{a' \in \cal A} \hom(G(a'),X) \otimes G(a') \cong \int^{X' \in \mathbf{C}} \hom(X',X) \otimes X'$, can you expand?
Oct 19, 2016 at 6:36 history answered HeinrichD CC BY-SA 3.0