Lemma 46.4.1. Let $A$ be a ring. For every module-valued functor $F$ on $\textit{Alg}_ A$ there exists a morphism $Q(F) \to F$ of module-valued functors on $\textit{Alg}_ A$ such that (1) $Q(F)$ is adequate and (2) for every adequate functor $G$ the map $\mathop{\mathrm{Hom}}\nolimits (G, Q(F)) \to \mathop{\mathrm{Hom}}\nolimits (G, F)$ is a bijection.
Proof. Choose a set $\{ L_ i\} _{i \in I}$ of linearly adequate functors such that every linearly adequate functor is isomorphic to one of the $L_ i$. This is possible. Suppose that we can find $Q(F) \to F$ with (1) and (2)' or every $i \in I$ the map $\mathop{\mathrm{Hom}}\nolimits (L_ i, Q(F)) \to \mathop{\mathrm{Hom}}\nolimits (L_ i, F)$ is a bijection. Then (2) holds. Namely, combining Lemmas 46.3.6 and 46.3.11 we see that every adequate functor $G$ sits in an exact sequence
with $K$ and $L$ direct sums of linearly adequate functors. Hence (2)' implies that $\mathop{\mathrm{Hom}}\nolimits (L, Q(F)) \to \mathop{\mathrm{Hom}}\nolimits (L, F)$ and $\mathop{\mathrm{Hom}}\nolimits (K, Q(F)) \to \mathop{\mathrm{Hom}}\nolimits (K, F)$ are bijections, whence the same thing for $G$.
Consider the category $\mathcal{I}$ whose objects are pairs $(i, \varphi )$ where $i \in I$ and $\varphi : L_ i \to F$ is a morphism. A morphism $(i, \varphi ) \to (i', \varphi ')$ is a map $\psi : L_ i \to L_{i'}$ such that $\varphi ' \circ \psi = \varphi $. Set
There is a natural map $Q(F) \to F$, by Lemma 46.3.12 it is adequate, and by construction it has property (2)'. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)