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Let $f: R \to S$ be a morphism of noncommutative rings. Let $$f_!:= S \otimes_R (-) : R \text{Mod} \to S \text{Mod}$$ denote the extension of scalars functor, and $$f^*: S \text{Mod} \to R \text{Mod}$$ the restriction of scalars functor. These form an adjunction $f_! \dashv f^*$.

Using the natural isomorphism $f^* \cong f^*S \otimes_S (-)$, I get $$f_!f^*W \cong S \otimes_R (f^*S \otimes_S W) \cong (S \otimes_R f^*S) \otimes_S W,$$ The $S$ on the left is an $(S,R)$-bimodule, and $f^*S$ is an $(R,S)$-bimodule, and it would be nice if we had $S \otimes_R f^* S \cong S$ naturally as $(S,S)$-bimodules, but I get the feeling this is not correct, since then we would have $f_! f^* W \cong W$ as $S$-modules, which seems to be wrong.

Can this be "simplified" further, and what would be the correct way of writing down and understanding the counit $$\varepsilon: f_! f^* \Rightarrow \text{id}_{S \text{Mod}}$$ of the adjunction here?

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  • $\begingroup$ What is $f^*$? You have only introduced $f_*$. $\endgroup$ – Tobias Kildetoft Jul 7 '16 at 6:18
  • $\begingroup$ Sorry, I meant to write $f^*$ for restriction of scalars, since a module can be thought of as an additive functor from the delooping of a ring to the category of abelian groups, then restriction of scalars corresponds to precomposition by f. $\endgroup$ – ಠ_ಠ Jul 7 '16 at 6:20
  • $\begingroup$ There seems to be something quite wrong here. If you restrict and then extend again, you certainly don't get back what you started with (not even when applied to the ring $S$). $\endgroup$ – Tobias Kildetoft Jul 7 '16 at 6:25
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    $\begingroup$ For understanding the counit, there is an "obvious" way to define a map from $S\otimes_R S$ to $S$ (just multiply). $\endgroup$ – Tobias Kildetoft Jul 7 '16 at 6:30
  • $\begingroup$ Ah I see so the counit is just given by $(s' \otimes_R s) \otimes_S w \mapsto s\cdot 's \cdot w$ then, right? $\endgroup$ – ಠ_ಠ Jul 7 '16 at 6:35
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First of all, we do not even have an isomorphism between $S\otimes_R f^*S = S\otimes_R S$ and $S$ as left $S$-modules (take for example $S$ to be a field and $R$ to be a subfield to see this).

The counit is fortunately very easy to describe. We have the map from $S\otimes_R S$ to $S$ that sends $x\otimes y$ to $xy$, and as you noted, we can identify $f_!f^*W$ with $(S\otimes_R S)\otimes_S W$, which we can map to $W$ by sending $x\otimes y\otimes w$ to $(xy).w$.

That this map is in fact natural I leave as an (easy) exercise.

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