## 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

$K \to \prod K_ n \to \prod K_ n \to K$

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

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

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 $H^ p(R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K) = H^ p(K)$. $\square$

Comment #8295 by ZL on

Can you explain what is the canonical map $K^{\bullet}\rightarrow R \lim \tau_{\geq -n} K^{\bullet}$ in tag 13.34.4? I imagine that it comes from the exact sequence $\mathrm{Hom}(K^{\bullet},R \lim \tau_{\geq -n} K^{\bullet})\rightarrow \mathrm{Hom}(K^{\bullet}, \prod \tau_{\geq -n} K^{\bullet})\rightarrow \mathrm{Hom}(K^{\bullet}, \prod \tau_{\geq -n} K^{\bullet})$, since the canonical map $K^{\bullet}\rightarrow \prod \tau_{\geq -n} K^{\bullet}$ in the second term will be mapped to $0$ in the third term. But I can not get a unique map from $K^{\bullet}$ to $R \lim \tau_{\geq -n} K^{\bullet}$.

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 $\mathcal{A}$ with countable products and enough injectives, an inverse system $\ldots \to K_3^\bullet \to K_2^\bullet \to K_1^\bullet$ of bounded below complexes of $\mathcal{A}$, an (unbounded) complex $K^\bullet$, and maps of complexes $K^\bullet \to K_n^\bullet$ compatible with the transition maps in the inverse system, there is a canonical map $\gamma : K^\bullet \to R\lim K_n^\bullet$ in $D(\mathcal{A})$.

To construct $\gamma$ we argue as in the proof of tag 13.34.4. Namely, we find an inverse system $\ldots \to I_3^\bullet \to I_2^\bullet \to I_1^\bullet$ of bounded below complexes of injectives with termwise surjective transition maps and quasi-isomorphisms $K_n^\bullet \to I_n^\bullet$ compatible with transition maps. Then $R\lim K_n$ is represented by the complex $I^\bullet$ where $I^p = \lim I_n^p$ as in the proof of 13.34.4. The map $\gamma$ is defined to be the one corresponding to the obvious map of complexes $K^\bullet \to I^\bullet$. Of course we have to show that this map is independent of choices...

If the transition maps in the system $\ldots \to I_3^\bullet \to I_2^\bullet \to I_1^\bullet$ are not termwise surjective, we can still get a unique map easily. Namely, then the Rlim is represented by $C[-1]$ where $C$ is the cone on $\prod I_n^\bullet \to \prod I_n^\bullet$, see references given in proof of 13.34.4. But now you can use that $K^\bullet \to \prod I_n^\bullet \to \prod I_n^\bullet$ is zero on the nose, to get a canonical map into $C[-1]$, 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 $g \circ f$; however, since in our case $g \circ f = 0$ it makes sense to take the $0$ homotopy and this does indeed give a well defined map into the cone. Again, one has to show that the map $\gamma$ so obtained does not depend on the choice of the inverse system $I_n^\bullet$ and the quasi-isomorphisms $K_n^\bullet \to I_n^\bullet$... 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.

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