0
$\begingroup$

I'm reading a paper by Paul Balmer where he supposes to have a nice (let's say noetherian) scheme $X$, with an open $U\subseteq X$, and proves the existence of an equivalence of triangulated categories $$(\mathrm{Pf}(X)/\mathrm{Pf}_Z(U))^{\sim}\longrightarrow \mathrm{Pf}(U)$$

where $\mathrm{Pf}(X)$ is the triangulated category of perfect complexes on $X$, $\mathrm{Pf}_Z(X)$ is the triangulated subcategory of perfect complexes with support contained in $Z$, and the $\sim$ stands for the idempotent completion ($\overset{\sim}{\textbf{C}}$ is the full subcategory of presheaves on $\textbf{C}$ which are retract of a representable ones). The morphism is induces by the restriction functor $\mathrm{Pf}(X)\to \mathrm{Pf}(U)$ and the universal property of the Verider quotient (this gives the functor from the quotient), as well as the fact that $\mathrm{Pf}(U)$ is idempotent complete (giving the functor from the idempotent completion of the quotient).

Now, he first proves that the functor defined on the quotient is fully faithful, and it seems that one can deduce the same for the functor lifted to the idempotent completion (first question, why?). Then , once defined a triangulated structure for the idempotent completion, he claims this triangulated functor to be essentially surjective, and in order to prove that observes that it suffices, given a perfect complex $F$ on $U$, to prove that $F$ is a direct summand of a complex which is the restriction of a perfect complex on $X$. So my question is, why is that true? How is the idempotent completion used to prove that?

In more general terms my questions could be:

(1) Does a fully faithful functor $\textbf{C}\to \textbf{D}$ into an idempotent complete category lift to a fully fraithful functor $\overset{\sim}{\textbf{C}}\to \textbf{D}$?

(2) If a triangulated functor of triangulated categories $G:\textbf{C}\to \textbf{D}$, with $\textbf{D}$ idempotent complete, is such that an object $F$ in $\textbf{D}$ is direct summand of an object $F\oplus W$ isomorphic to $G(X)$ for some $X$ in $\textbf{C}$, is it true that $F$ is isomorphic to $\overset{\sim}G(Y)$ for an object $Y$ in $\overset{\sim}{\textbf{C}}$?

$\endgroup$

1 Answer 1

5
$\begingroup$

These statements are "obvious" if you understand how idempotent completion works. Let me back up and describe that first. Suppose you have an object $c$ in a category and a morphism $f:c\to c$ which satisfies $f^2=f$. Then a splitting of $f$ is an object $d$ together with a pair of morphisms $i:d\to c$ and $p:c\to d$ such that $pi=1_d$ and $ip=f$.

Now, the key point is, when you have a splitting of an idempotent, that uniquely determines what the morphisms to and from $d$ can be, in terms of morphisms to and from $c$. Specifically, for any $e$, morphisms $g:e\to d$ are in natural bijection with morphisms $h:e\to c$ such that $fh=h$ (the bijection sends $g$ to $ig$, and its inverse sends $h$ to $ph$). Dually, morphisms $g:d\to e$ are in natural bijection with morphisms $h:c\to e$ such that $hf=h$.

This means that given a category $C$ with an idempotent $f:c\to c$, we can simply formally adjoin a splitting of it to get a category $C'$ with one more object $d$, where morphisms in and out of $d$ are defined as certain subsets of the morphisms in and out of $c$ (namely, morphisms into $d$ are morphisms into $c$ such that $fh=h$, and morphisms out of $d$ are morphisms out of $c$ such that $hf=h$). This is the only possibility for the morphisms in and out of a splitting of $f$, so if $C$ was a fully faithfully embedded in a category $D$ in which $f$ splits, then $C'$ will be too. Explicitly, suppose $G:C\to D$ is fully faithful and $i:d'\to G(c)$, $p:G(c)\to d'$ is a splitting of $G(f)$. Then we can extend $G$ to $G':C'\to D$ by defining $G'(d)=d'$. We know that morphisms $G(e)\to d'$ are in bijection with morphisms $h:G(e)\to G(c)$ such that $G(f)h=h$. Since $G$ is fully faithful, these are in bijection with morphisms $h:e\to c$ such that $fh=h$. But those are the same as morphisms $e\to d$, so we can extend $G$ to all the morphisms into $d$ and it will remain fully faithful on them. Similarly for morphisms out of $d$. (Of course there are some details I am skipping over here about compatibility with composition but they are straightforward to check.)

Now, idempotent completion is just doing this process for every idempotent in $C$ at once (or if you prefer, doing it iteratively on them one by one by transfinite recursion). So if $D$ is idempotent complete, any fully faithful functor $C\to D$ extends to a fully faithful functor $\tilde{C}\to D$. This is just because every object of $\tilde{C}$ is a formally adjoined splitting of some idempotent in $C$, so you can map it to a splitting of the corresponding idempotent in $D$. This is fully faithful because the new morphisms you get when you split an idempotent are uniquely determined by the morphisms you had in and out of the original object.

Now for your second question, suppose $G:C\to D$ is a fully faithful functor of additive categories and $D$ is idempotent complete. If an object $F$ of $D$ is a direct summand of an object $G(X)$ in the image of $X$, that just means that it splits an idempotent on $G(X)$. (Namely, if $G(X)=F\oplus W$, then the inclusion $i:F\to G(X)$ and the projection $p:G(X)\to F$ split the idempotent $ip:G(X)\to G(X)$.) Since $G$ is fully faithful, we can lift that idempotent up to an idempotent of $X$. This idempotent is split by some object $Y$ of $\tilde{C}$, and then $\tilde{G}(Y)$ splits the same idempotent on $G(X)$ as $F$ does. This implies $\tilde{G}(Y)$ and $F$ are isomorphic. (You can build an isomorphism explicitly, or it follows by Yoneda: morphisms in and out of $\tilde{G}(Y)$ and $F$ are in natural bijection, since they are both in natural bijection with certain morphisms in and out of $G(X)$.)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .