The Stacks project

Lemma 13.16.4. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{I} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ be a subset containing $0$ such that every object of $\mathcal{A}$ is a subobject of an element of $\mathcal{I}$. Let $a \in \mathbf{Z}$.

  1. Given $K^\bullet $ with $K^ n = 0$ for $n < a$ there exists a quasi-isomorphism $K^\bullet \to I^\bullet $ with $K^ n \to I^ n$ injective and $I^ n \in \mathcal{I}$ for all $n$ and $I^ n = 0$ for $n < a$,

  2. Given $K^\bullet $ with $H^ n(K^\bullet ) = 0$ for $n < a$ there exists a quasi-isomorphism $K^\bullet \to I^\bullet $ with $I^ n \in \mathcal{I}$ and $I^ n = 0$ for $n < a$.

Proof. Proof of part (1). Consider the following induction hypothesis $IH_ n$: There are $I^ j \in \mathcal{I}$, $j \leq n$ almost all zero, maps $d^ j : I^ j \to I^{j + 1}$ for $j < n$ and injective maps $\alpha ^ j : K^ j \to I^ j$ for $j \leq n$ such that the diagram

\[ \xymatrix{ \ldots \ar[r] & K^{n - 1} \ar[d]^\alpha \ar[r] & K^ n \ar[d]^\alpha \ar[r] & K^{n + 1} \ar[r] & \ldots \\ \ldots \ar[r] & I^{n - 1} \ar[r] & I^ n & & } \]

is commutative, such that $d^ j \circ d^{j - 1} = 0$ for $j < n$ and such that $\alpha $ induces isomorphisms $H^ j(K^\bullet ) \to \mathop{\mathrm{Ker}}(d^ j)/\mathop{\mathrm{Im}}(d^{j - 1})$ for $j < n$. Note that this implies
\begin{equation} \label{derived-equation-new-star} \alpha (\mathop{\mathrm{Im}}(d_ K^{n - 1})) \subset \alpha (\mathop{\mathrm{Ker}}(d_ K^ n)) \cap \mathop{\mathrm{Im}}(d^{n - 1}) \subset \alpha (K^ n) \cap \mathop{\mathrm{Im}}(d^{n - 1}). \end{equation}

If these inclusions are not equalities, then choose an injection

\[ I^ n \oplus K^ n/\mathop{\mathrm{Im}}(d_ K^{n - 1}) \longrightarrow I \]

with $I \in \mathcal{I}$. Denote $\alpha ' : K^ n \to I$ the map obtained by composing $\alpha \oplus 1 : K^ n \to I^ n \oplus K^ n/\mathop{\mathrm{Im}}(d_ K^{n - 1})$ with the displayed injection. Denote $d' : I^{n - 1} \to I$ the composition $I^{n - 1} \to I^ n \to I$ of $d^{n - 1}$ by the inclusion of the first summand. Then $\alpha '(K^ n) \cap \mathop{\mathrm{Im}}(d') = \alpha '(\mathop{\mathrm{Im}}(d_ K^{n - 1}))$ simply because the intersection of $\alpha '(K^ n)$ with the first summand of $I^ n \oplus K^ n/\mathop{\mathrm{Im}}(d_ K^{n - 1})$ is equal to $\alpha '(\mathop{\mathrm{Im}}(d_ K^{n - 1}))$. Hence, after replacing $I^ n$ by $I$, $\alpha $ by $\alpha '$ and $d^{n - 1}$ by $d'$ we may assume that we have equality in Equation ( Once this is the case consider the solid diagram

\[ \xymatrix{ K^ n/\mathop{\mathrm{Ker}}(d_ K^ n) \ar[r] \ar[d] & K^{n + 1} \ar@{..>}[d] \\ I^ n/(\mathop{\mathrm{Im}}(d^{n - 1}) + \alpha (\mathop{\mathrm{Ker}}(d_ K^ n))) \ar@{..>}[r] & M } \]

The horizontal arrow is injective by fiat and the vertical arrow is injective as we have equality in ( Hence the push-out $M$ of this diagram contains both $K^{n + 1}$ and $I^ n/(\mathop{\mathrm{Im}}(d^{n - 1}) + \alpha (\mathop{\mathrm{Ker}}(d_ K^ n)))$ as subobjects. Choose an injection $M \to I^{n + 1}$ with $I^{n + 1} \in \mathcal{I}$. By construction we get $d^ n : I^ n \to I^{n + 1}$ and an injective map $\alpha ^{n + 1} : K^{n + 1} \to I^{n + 1}$. The equality in Equation ( and the construction of $d^ n$ guarantee that $\alpha : H^ n(K^\bullet ) \to \mathop{\mathrm{Ker}}(d^ n)/\mathop{\mathrm{Im}}(d^{n - 1})$ is an isomorphism. In other words $IH_{n + 1}$ holds.

We finish the proof of by the following observations. First we note that $IH_ n$ is true for $n = a$ since we can just take $I^ j = 0$ for $j < a$ and $K^ a \to I^ a$ an injection of $K^ a$ into an element of $\mathcal{I}$. Next, we note that in the proof of $IH_ n \Rightarrow IH_{n + 1}$ we only modified the object $I^ n$, the map $d^{n - 1}$ and the map $\alpha ^ n$. Hence we see that proceeding by induction we produce a complex $I^\bullet $ with $I^ n = 0$ for $n < a$ consisting of objects from $\mathcal{I}$, and a termwise injective quasi-isomorphism $\alpha : K^\bullet \to I^\bullet $ as desired.

Proof of part (2). The assumption implies that the morphism $K^\bullet \to \tau _{\geq a}K^\bullet $ (Homology, Section 12.14) is a quasi-isomorphism. Apply part (1) to find $\tau _{\geq a}K^\bullet \to I^\bullet $. The composition $K^\bullet \to I^\bullet $ is the desired quasi-isomorphism. $\square$

Comments (0)

There are also:

  • 5 comment(s) on Section 13.16: Derived functors on derived categories

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05T6. Beware of the difference between the letter 'O' and the digit '0'.