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Let $A, B$ be commutative (unital) rings and $f\colon A \to B$ an $A$-algebra. There then exists a canonical functor $f_*\colon \mathbf{Mod}_B \to \mathbf{Mod}_A$ such that, for every morphism of $B$-module $g\colon N \to N'$, it associates the morphism of $A$-modules $f_*(g)\colon f_*(N) \to f_*(N')$, where, if $\rho_N\colon B\times N\to N$ (resp. $\rho_{N'}\colon B\times N' \to N'$) is the action of $B$ on $N$ (resp. on $N'$), $f_*(N)$ (resp. $f_*(N')$) has the induced $A$-module structure

$\rho_N\circ(f, \text{id}_N)\colon A\times N \to B\times N \to N\quad \text{(resp. }\rho_{N'}\circ(f, \text{id}_{N'})\text{)}$

and $f_*(g)$ is the same morphism of abelian group canonically $A$-linear with these induced structure of $A$-modules on $f_*(N)$ and $f_*(N')$.

The functor $f_*$ is faithful and has both left and right adjoints, resp. $f^*$ and $f^!$, such that, for every morphism $h\colon M\to M'$ of $A$-modules,

$f^*(g)= \text{id}_B \otimes h\colon B\otimes_A M \to B\otimes_A M'$

(the base change) and

$f^! = h\circ -\colon \text{Hom}_A(B, M) \to \text{Hom}_A(B, M'): u\mapsto h\circ u.$


Question. When is $f_*$ also full as functor, i.e., for every $B$-modules $N, N'$, when is $f_*\bigl(\text{Hom}_B(N, N')\bigr) \cong \text{Hom}_A\bigl(f_*(N), f_*(N')\bigr)$ as $A$-modules?

Equivalently, we can ask when the counit

$(\epsilon_f)_N\colon f^*(f_*(N)) = B\otimes_A f_*(N) \to N: b\otimes n \mapsto bn$

of the adjunction $(f^*\dashv f_*)$ is an isomorphism of $B$-modules for every $B$-module $N$.

Remark. For every multiplicative subset $S$ of $A$, it holds for the canonical $A$-algebra $i^S_A\colon A \to S^{-1}A$, but I cannot say much more that this.

Any reference is welcome.

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    $\begingroup$ $f_*$ is full iff $A \to B$ is an epi (formal exercise). There are lots of non-surjective epis. $\endgroup$ May 14, 2014 at 10:19
  • $\begingroup$ Nice! Formal because $\colon A \to B$ is epi iff $b\otimes 1_B = 1_B \otimes b$ in $B\otimes_A B$? Or it's even simpler? $\endgroup$ May 14, 2014 at 10:25
  • $\begingroup$ Oh, I think I see: $f$ is also epi iff $B \to B \otimes_A B$ is iso; tensoring by $N\otimes_B-$ (which, as functor, preserves isos) and using the associativity of tensor product we are done. $\endgroup$ May 14, 2014 at 10:33
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    $\begingroup$ @AndreaGagna If you found the answer to your question, please consider self-answering your question, so that it gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. $\endgroup$
    – gebruiker
    Nov 10, 2015 at 11:15

1 Answer 1

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A basic fact about category theory states that a morphism $f \colon A \to B$ is an epimorphism in a category $\mathcal C$ if and only if the pushout of it with itself (i.e., the cobase change of $f$ along itself) exists and $\require{AMScd}$ \begin{CD} A @>f>> B\\ @V f V V @VV \text{id}_B V\\ B @>>\text{id}_B> B \end{CD} is a cocartesian square.

In the case in which $\mathcal C = \mathbf{CRing}$, the pushout square of a morphism $f$ along itself is given by \begin{CD} A @>f>> B\\ @V f V V @VV p_2 V\\ B @>>p_1> B\otimes_A B \end{CD} where $p_1$ sends $b\mapsto b\otimes 1_B$ and $p_2$ maps $b \mapsto 1_B \otimes b$. If $g, g'\colon B \to C$ are two morphisms such that $gf = g'f$, then the universal morphism $B\otimes_A B \to C$ maps $b\otimes b'$ to $g(b)g'(b')$.

Therefore $f$ is a epimorphism in $\mathbf{CRing}$ if and only if the canonical morphism $\phi\colon B\otimes_A B \to B$ sending $b\otimes b' \mapsto bb'$ is an isomorphism.

The functor $N\otimes_B -$ preserves and reflects isomorphisms, so we get that $N\otimes_B \phi$ from $N\otimes_B (B\otimes_A B)$ to $N\otimes_B B$ is an isomorphism if and only if $\phi$ is so.

But, by the associativity of tensor product in $\mathbf{CRing}$, we have that $N\otimes_B (B\otimes_A B)$ is canonically isomorphic to $(N\otimes_B B)\otimes_A B$ (the mapping being the obvious one).

Further, there is a canonical isomorphism $N\otimes_B B \to N$ given by $n\otimes b\mapsto bn$.

Finally, we can conclude that the morphism $N\otimes_B \phi$ is isomorphic to the counit $(\epsilon_f)_N$ and hence the latter is an isomorphism if and only if the former is so, and this holds if and only if $f$ is an epimorphism.

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