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Let $k$ be a commutative ring and $f: C \to C'$ be a homomorphism of $k$-coalgebras (for simplicity, we can suppose that it is surjective so that $f(C) = C'$). There is a functor: $$f_*: {}^lC-comod \to {}^lC'-comod$$ $$(M, \Delta_M) \mapsto (M, \Delta_M' := (f \otimes id_M) \circ \Delta_M)$$ wherein $\Delta_M$ denotes the left-$C$-coaction on $M$. Does this functor admit left/right-adjoints, and if so, do these left/right-adjoints have concrete descriptions ?

Question edited according to comments by
Mariano Suárez-Álvarez

Dat Minh Ha
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  • Your evident functor is not quite as evident as you seem to think... What is $f(C)\otimes_k M$?! – Mariano Suárez-Álvarez Feb 22 '23 at 06:19
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    There is a functor from C comodules to C' comodules which maps a C comodule M with coaction $\lambda$ to the C' comodule which as a vector space is again M but whose C' coaction is the composition $f\otimes 1_M\circ\lambda$. This is the functor of change-of-coalgebras. – Mariano Suárez-Álvarez Feb 22 '23 at 06:22
  • @MarianoSuárez-Álvarez I'm more or less brand new to comodules. Please feel free to edit the question if you think there's a mistake or if I was being unclear. Thank you! – Dat Minh Ha Feb 22 '23 at 06:24
  • You are being unclear, and since I cannot tell what you are asking I cannot edit your question, for i don't know what you want to ask! You apparently have a functor in mind: explain what it does to objects better. – Mariano Suárez-Álvarez Feb 22 '23 at 06:24
  • @MarianoSuárez-Álvarez I have edited my question according to your comment. Hopefully it's clearer now. – Dat Minh Ha Feb 22 '23 at 06:34
  • Ok. Now, there is a similar functor for modules: do you know if it has adjoints and what do they look like? Not everything works the same with comodules as with modules, but it always helps to see what happens with modules... – Mariano Suárez-Álvarez Feb 22 '23 at 06:35
  • If $\varphi: A \to A'$ is a $k$-algebra homomorphism (and let's say we're regarding $A'$ as an $(A', A)$-bimodule via $f$) then the functor you're referring to is that one that sends $(M, \pi)$ (with $\pi$ being the left-$A$-action on $M$) to $(M, \pi')$, where $\pi'$ is the left-$A'$-action on $M$ given by $\alpha_1 \otimes (\alpha_2 \otimes m) = \alpha_1 \alpha_2 \otimes m$, right ? It has a right-adjoint, given by $Hom_{A'}(A', -)$. – Dat Minh Ha Feb 22 '23 at 07:03
  • Notice that that Hom functor is essentially the identity. – Mariano Suárez-Álvarez Feb 22 '23 at 07:58

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