The Stacks project

13.34 Derived limits

In a triangulated category there is a notion of derived limit.

Definition 13.34.1. Let $\mathcal{D}$ be a triangulated category. Let $(K_ n, f_ n)$ be an inverse system of objects of $\mathcal{D}$. We say an object $K$ is a derived limit, or a homotopy limit of the system $(K_ n)$ if the product $\prod K_ n$ exists and there is a distinguished triangle

\[ K \to \prod K_ n \to \prod K_ n \to K[1] \]

where the map $\prod K_ n \to \prod K_ n$ is given by $(k_ n) \mapsto (k_ n - f_{n+1}(k_{n + 1}))$. If this is the case, then we sometimes indicate this by the notation $K = R\mathop{\mathrm{lim}}\nolimits K_ n$.

By TR3 a derived limit, if it exists, is unique up to (non-unique) isomorphism. Moreover, by TR1 a derived limit $R\mathop{\mathrm{lim}}\nolimits K_ n$ exists as soon as $\prod K_ n$ exists. The derived category $D(\textit{Ab})$ of the category of abelian groups is an example of a triangulated category where all derived limits exist.

The nonuniqueness makes it hard to pin down the derived limit. In More on Algebra, Lemma 15.86.4 the reader finds an exact sequence

\[ 0 \to R^1\mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits (L, K_ n[-1]) \to \mathop{\mathrm{Hom}}\nolimits (L, R\mathop{\mathrm{lim}}\nolimits K_ n) \to \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits (L, K_ n) \to 0 \]

describing the $\mathop{\mathrm{Hom}}\nolimits $s into a derived limit in terms of the usual $\mathop{\mathrm{Hom}}\nolimits $s.

Lemma 13.34.2. Let $\mathcal{A}$ be an abelian category with exact countable products. Then

  1. $D(\mathcal{A})$ has countable products,

  2. countable products $\prod K_ i$ in $D(\mathcal{A})$ are obtained by taking termwise products of any complexes representing the $K_ i$, and

  3. $H^ p(\prod K_ i) = \prod H^ p(K_ i)$.

Proof. Let $K_ i^\bullet $ be a complex representing $K_ i$ in $D(\mathcal{A})$. Let $L^\bullet $ be a complex. Suppose given maps $\alpha _ i : L^\bullet \to K_ i^\bullet $ in $D(\mathcal{A})$. This means there exist quasi-isomorphisms $s_ i : K_ i^\bullet \to M_ i^\bullet $ of complexes and maps of complexes $f_ i : L^\bullet \to M_ i^\bullet $ such that $\alpha _ i = s_ i^{-1}f_ i$. By assumption the map of complexes

\[ s : \prod K_ i^\bullet \longrightarrow \prod M_ i^\bullet \]

is a quasi-isomorphism. Hence setting $f = \prod f_ i$ we see that $\alpha = s^{-1}f$ is a map in $D(\mathcal{A})$ whose composition with the projection $\prod K_ i^\bullet \to K_ i^\bullet $ is $\alpha _ i$. We omit the verification that $\alpha $ is unique. $\square$

The duals of Lemmas 13.33.6, 13.33.7, and 13.33.9 should be stated here and proved. However, we do not know any applications of these lemmas for now.

Lemma 13.34.3. Let $\mathcal{A}$ be an abelian category with countable products and enough injectives. Let $(K_ n)$ be an inverse system of $D^+(\mathcal{A})$. Then $R\mathop{\mathrm{lim}}\nolimits K_ n$ exists.

Proof. It suffices to show that $\prod K_ n$ exists in $D(\mathcal{A})$. For every $n$ we can represent $K_ n$ by a bounded below complex $I_ n^\bullet $ of injectives (Lemma 13.18.3). Then $\prod K_ n$ is represented by $\prod I_ n^\bullet $, see Lemma 13.31.5. $\square$

Lemma 13.34.4. Let $\mathcal{A}$ be an abelian category with countable products and enough injectives. Let $K^\bullet $ be a complex. Let $I_ n^\bullet $ be the inverse system of bounded below complexes of injectives produced by Lemma 13.29.3. Then $I^\bullet = \mathop{\mathrm{lim}}\nolimits I_ n^\bullet $ exists, is K-injective, and the following are equivalent

  1. the map $K^\bullet \to I^\bullet $ is a quasi-isomorphism,

  2. the canonical map $K^\bullet \to R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K^\bullet $ is an isomorphism in $D(\mathcal{A})$.

Proof. The statement of the lemma makes sense as $R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K^\bullet $ exists by Lemma 13.34.3. Each complex $I_ n^\bullet $ is K-injective by Lemma 13.31.4. Choose direct sum decompositions $I_{n + 1}^ p = C_{n + 1}^ p \oplus I_ n^ p$ for all $n \geq 1$. Set $C_1^ p = I_1^ p$. The complex $I^\bullet = \mathop{\mathrm{lim}}\nolimits I_ n^\bullet $ exists because we can take $I^ p = \prod _{n \geq 1} C_ n^ p$. Fix $p \in \mathbf{Z}$. We claim there is a split short exact sequence

\[ 0 \to I^ p \to \prod I_ n^ p \to \prod I_ n^ p \to 0 \]

of objects of $\mathcal{A}$. Here the first map is given by the projection maps $I^ p \to I_ n^ p$ and the second map by $(x_ n) \mapsto (x_ n - f^ p_{n + 1}(x_{n + 1}))$ where $f^ p_ n : I_ n^ p \to I_{n - 1}^ p$ are the transition maps. The splitting comes from the map $\prod I_ n^ p \to \prod C_ n^ p = I^ p$. We obtain a termwise split short exact sequence of complexes

\[ 0 \to I^\bullet \to \prod I_ n^\bullet \to \prod I_ n^\bullet \to 0 \]

Hence a corresponding distinguished triangle in $K(\mathcal{A})$ and $D(\mathcal{A})$. By Lemma 13.31.5 the products are K-injective and represent the corresponding products in $D(\mathcal{A})$. It follows that $I^\bullet $ represents $R\mathop{\mathrm{lim}}\nolimits I_ n^\bullet $ (Definition 13.34.1). Moreover, it follows that $I^\bullet $ is K-injective by Lemma 13.31.3. By the commutative diagram of Lemma 13.29.3 we obtain a corresponding commutative diagram

\[ \xymatrix{ K^\bullet \ar[r] \ar[d] & R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n} K^\bullet \ar[d] \\ I^\bullet \ar[r] & R\mathop{\mathrm{lim}}\nolimits I_ n^\bullet } \]

in $D(\mathcal{A})$. Since the right vertical arrow is an isomorphism (as derived limits are defined on the level of the derived category and since $\tau _{\geq -n}K^\bullet \to I_ n^\bullet $ is a quasi-isomorphism), the lemma follows. $\square$

Lemma 13.34.5. Let $\mathcal{A}$ be an abelian category having enough injectives and exact countable products. Then for every complex there is a quasi-isomorphism to a K-injective complex.

Proof. By Lemma 13.34.4 it suffices to show that $K \to R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K$ is an isomorphism for all $K$ in $D(\mathcal{A})$. Consider the defining distinguished triangle

\[ R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K \to \prod \tau _{\geq -n}K \to \prod \tau _{\geq -n}K \to (R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K)[1] \]

By Lemma 13.34.2 we have

\[ H^ p(\prod \tau _{\geq -n}K) = \prod \nolimits _{p \geq -n} H^ p(K) \]

It follows in a straightforward manner from the long exact cohomology sequence of the displayed distinguished triangle that $H^ p(R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K) = H^ p(K)$. $\square$

Comments (3)

Comment #8295 by ZL on

Can you explain what is the canonical map in tag 13.34.4? I imagine that it comes from the exact sequence , since the canonical map in the second term will be mapped to in the third term. But I can not get a unique map from to .

Comment #8296 by on

This is a good question, thank you! The answer is that it is constructed in the proof. But let me explain a more general canonical map.

I claim: given an abelian category with countable products and enough injectives, an inverse system of bounded below complexes of , an (unbounded) complex , and maps of complexes compatible with the transition maps in the inverse system, there is a canonical map in .

To construct we argue as in the proof of tag 13.34.4. Namely, we find an inverse system of bounded below complexes of injectives with termwise surjective transition maps and quasi-isomorphisms compatible with transition maps. Then is represented by the complex where as in the proof of 13.34.4. The map is defined to be the one corresponding to the obvious map of complexes . Of course we have to show that this map is independent of choices...

If the transition maps in the system are not termwise surjective, we can still get a unique map easily. Namely, then the Rlim is represented by where is the cone on , see references given in proof of 13.34.4. But now you can use that is zero on the nose, to get a canonical map into , see Lemma 13.9.3 and its proof. The non-uniqueness that your are pointing out corresponds, in the proof of Lemma 13.9.3, to the choice of the homotopy witnessing the vanishing of ; however, since in our case it makes sense to take the homotopy and this does indeed give a well defined map into the cone. Again, one has to show that the map so obtained does not depend on the choice of the inverse system and the quasi-isomorphisms ... but this does not present any problems (as far as I can see at the moment).

The next time I go through all the comments I will make the corresponding edits. Please feel free to complain if you think this doesn't work.

Comment #8922 by on

OK, I fixed this in the opposite way, because actually to prove the "canonicity" would be quite annoying. So I fixed it by making the map maximally noncanonical. See somewhat large set of changes here.

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