# The Stacks Project

## Tag 05TA

A functor on an Abelian categories is extended to the (bounded below or above) derived category by resolving with a complex that is acyclic for that functor.

Proposition 13.17.8. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor of abelian categories.

1. If every object of $\mathcal{A}$ injects into an object acyclic for $RF$, then $RF$ is defined on all of $K^{+}(\mathcal{A})$ and we obtain an exact functor $$RF : D^{+}(\mathcal{A}) \longrightarrow D^{+}(\mathcal{B})$$ see (13.15.9.1). Moreover, any bounded below complex $A^\bullet$ whose terms are acyclic for $RF$ computes $RF$.
2. If every object of $\mathcal{A}$ is quotient of an object acyclic for $LF$, then $LF$ is defined on all of $K^{-}(\mathcal{A})$ and we obtain an exact functor $$LF : D^{-}(\mathcal{A}) \longrightarrow D^{-}(\mathcal{B})$$ see (13.15.9.1). Moreover, any bounded above complex $A^\bullet$ whose terms are acyclic for $LF$ computes $LF$.

Proof. Assume every object of $\mathcal{A}$ injects into an object acyclic for $RF$. Let $\mathcal{I}$ be the set of objects acyclic for $RF$. Let $K^\bullet$ be a bounded below complex in $\mathcal{A}$. By Lemma 13.16.4 there exists a quasi-isomorphism $\alpha : K^\bullet \to I^\bullet$ with $I^\bullet$ bounded below and $I^n \in \mathcal{I}$. Hence in order to prove (1) it suffices to show that $F(I^\bullet) \to F((I')^\bullet)$ is a quasi-isomorphism when $s : I^\bullet \to (I')^\bullet$ is a quasi-isomorphism of bounded below complexes of objects from $\mathcal{I}$, see Lemma 13.15.15. Note that the cone $C(s)^\bullet$ is an acyclic bounded below complex all of whose terms are in $\mathcal{I}$. Hence it suffices to show: given an acyclic bounded below complex $I^\bullet$ all of whose terms are in $\mathcal{I}$ the complex $F(I^\bullet)$ is acyclic.

Say $I^n = 0$ for $n < n_0$. Setting $J^n = \mathop{\rm Im}(d^n)$ we break $I^\bullet$ into short exact sequences $0 \to J^n \to I^{n + 1} \to J^{n + 1} \to 0$ for $n \geq n_0$. These sequences induce distinguished triangles $(J^n, I^{n + 1}, J^{n + 1})$ in $D^+(\mathcal{A})$ by Lemma 13.12.1. For each $k \in \mathbf{Z}$ denote $H_k$ the assertion: For all $n \leq k$ the right derived functor $RF$ is defined at $J^n$ and $R^iF(J^n) = 0$ for $i \not = 0$. Then $H_k$ holds trivially for $k \leq n_0$. If $H_n$ holds, then, using Proposition 13.15.8, we see that $RF$ is defined at $J^{n + 1}$ and $(RF(J^n), RF(I^{n + 1}), RF(J^{n + 1}))$ is a distinguished triangle of $D^+(\mathcal{B})$. Thus the long exact cohomology sequence (13.11.1.1) associated to this triangle gives an exact sequence $$0 \to R^{-1}F(J^{n + 1}) \to R^0F(J^n) \to F(I^{n + 1}) \to R^0F(J^{n + 1}) \to 0$$ and gives that $R^iF(J^{n + 1}) = 0$ for $i \not \in \{-1, 0\}$. By Lemma 13.17.1 we see that $R^{-1}F(J^{n + 1}) = 0$. This proves that $H_{n + 1}$ is true hence $H_k$ holds for all $k$. We also conclude that $$0 \to R^0F(J^n) \to F(I^{n + 1}) \to R^0F(J^{n + 1}) \to 0$$ is short exact for all $n$. This in turn proves that $F(I^\bullet)$ is exact.

The proof in the case of $LF$ is dual. $\square$

The code snippet corresponding to this tag is a part of the file derived.tex and is located in lines 5653–5679 (see updates for more information).

\begin{proposition}
\label{proposition-enough-acyclics}
\begin{slogan}
A functor on an Abelian categories is extended to the (bounded below or above)
derived category by resolving with a complex that is acyclic for that functor.
\end{slogan}
Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor of
abelian categories.
\begin{enumerate}
\item If every object of $\mathcal{A}$ injects into an object acyclic
for $RF$, then $RF$ is defined on all of $K^{+}(\mathcal{A})$
and we obtain an exact functor
$$RF : D^{+}(\mathcal{A}) \longrightarrow D^{+}(\mathcal{B})$$
see (\ref{equation-everywhere}). Moreover, any bounded below complex
$A^\bullet$ whose terms are acyclic for $RF$ computes $RF$.
\item If every object of $\mathcal{A}$ is quotient of
an object acyclic for $LF$, then $LF$ is defined on all of
$K^{-}(\mathcal{A})$ and we obtain an exact functor
$$LF : D^{-}(\mathcal{A}) \longrightarrow D^{-}(\mathcal{B})$$
see (\ref{equation-everywhere}). Moreover, any bounded above complex
$A^\bullet$ whose terms are acyclic for $LF$ computes $LF$.
\end{enumerate}
\end{proposition}

\begin{proof}
Assume every object of $\mathcal{A}$ injects into an object acyclic
for $RF$. Let $\mathcal{I}$ be the set of objects acyclic for $RF$.
Let $K^\bullet$ be a bounded below complex in $\mathcal{A}$. By
Lemma \ref{lemma-subcategory-right-resolution}
there exists a quasi-isomorphism $\alpha : K^\bullet \to I^\bullet$ with
$I^\bullet$ bounded below and $I^n \in \mathcal{I}$. Hence in order to
prove (1) it suffices to show that
$F(I^\bullet) \to F((I')^\bullet)$ is a quasi-isomorphism when
$s : I^\bullet \to (I')^\bullet$ is a quasi-isomorphism of bounded
below complexes of objects from $\mathcal{I}$, see
Lemma \ref{lemma-find-existence-computes}.
Note that the cone $C(s)^\bullet$ is an acyclic bounded below complex
all of whose terms are in $\mathcal{I}$.
Hence it suffices to show: given an acyclic bounded below complex
$I^\bullet$ all of whose terms are in $\mathcal{I}$ the complex
$F(I^\bullet)$ is acyclic.

\medskip\noindent
Say $I^n = 0$ for $n < n_0$. Setting $J^n = \Im(d^n)$ we break
$I^\bullet$ into short exact sequences
$0 \to J^n \to I^{n + 1} \to J^{n + 1} \to 0$
for $n \geq n_0$. These sequences induce distinguished triangles
$(J^n, I^{n + 1}, J^{n + 1})$ in $D^+(\mathcal{A})$ by
Lemma \ref{lemma-derived-canonical-delta-functor}.
For each $k \in \mathbf{Z}$ denote $H_k$ the assertion:
For all $n \leq k$ the right derived functor
$RF$ is defined at $J^n$ and $R^iF(J^n) = 0$ for $i \not = 0$.
Then $H_k$ holds trivially for $k \leq n_0$. If $H_n$ holds,
then, using Proposition \ref{proposition-derived-functor},
we see that $RF$ is defined at $J^{n + 1}$ and
$(RF(J^n), RF(I^{n + 1}), RF(J^{n + 1}))$ is a distinguished
triangle of $D^+(\mathcal{B})$. Thus the long exact cohomology sequence
(\ref{equation-long-exact-cohomology-sequence-D})
associated to this triangle gives an exact sequence
$$0 \to R^{-1}F(J^{n + 1}) \to R^0F(J^n) \to F(I^{n + 1}) \to R^0F(J^{n + 1}) \to 0$$
and gives that $R^iF(J^{n + 1}) = 0$ for $i \not \in \{-1, 0\}$.
By Lemma \ref{lemma-negative-vanishing} we see that $R^{-1}F(J^{n + 1}) = 0$.
This proves that $H_{n + 1}$ is true hence $H_k$ holds for all $k$.
We also conclude that
$$0 \to R^0F(J^n) \to F(I^{n + 1}) \to R^0F(J^{n + 1}) \to 0$$
is short exact for all $n$. This in turn proves that $F(I^\bullet)$ is exact.

\medskip\noindent
The proof in the case of $LF$ is dual.
\end{proof}

Comment #1274 by JuanPablo on January 27, 2015 a 1:45 am UTC

Hello

In the second paragraph in this proof it says that the short exact sequence $0\rightarrow \text{Im}(d^n)\rightarrow I^n\rightarrow \text{Im}(d^{n+1})\rightarrow 0$ produces a distinguished triangle $(\text{Im}(d^n), I^n, \text{Im}(d^{n+1}))$. If I understand this right the distinguished triangle comes from the canonical delta functor, so it is distinguished in $D^+(\mathcal{B})$ and the reason is that the mapping from the mapping cone of $\text{Im}(d^n)[0]\rightarrow I^n[0]$ into $\text{Im}(d^{n+1})[0]$ is a quasi-isomorphism.

The problem I have is that the lemma 13.15.12 (tag 05SZ) applies to distinguised triangles in $\mathcal{D}$ which in this case $\mathcal{D}=K^+(\mathcal{B})$, so it does not seem to apply here.

Comment #1275 by JuanPablo on January 27, 2015 a 2:17 am UTC

In the second paragraph of this proof, in two places in the enunciate and in the comment above, $\mathcal{B}$ and $\mathcal{A}$ should be exchanged. (As $F:\mathcal{A}\rightarrow \mathcal{B}$).

Comment #1276 by JuanPablo on January 27, 2015 a 3:12 pm UTC

Ok. This problem can be fixed as follows:

$(\text{Im}(d^n)[0],I^n[0],\text{Im}(d^{n+1})[0])$ is distinguished in $D^{+}(\mathcal{A})$ so by Lemmas 13.15.4 and 13.15.6 (tags 05SB and 05SC) we get $(RF(\text{Im}(d^n)[0]),I^n[0],RF(\text{Im}(d^{n+1})[0])$ is distinguished in $D^{+}(\mathcal{B})$.

Now using the long exact cohomology sequence, obtain by induction in $n$, that $0\rightarrow R^0F(\text{Im}(d^n))\rightarrow F(I^n)\rightarrow R^0F(\text{Im}(d^{n+1}))\rightarrow 0$ is exact and $R^iF(\text{Im}(d^n))=0$ for $i>0$. So the sequence $0\rightarrow F(I^0)\rightarrow F(I^1)\rightarrow \dots$ is exact.

Comment #1301 by Johan (site) on February 10, 2015 a 1:29 am UTC

Yes, this was a kind of subtle mistake. I actually had to move the proposition a bit later because using $R^0F$ requires this. I also changed the approach slightly, using Proposition 05SE instead of a bunch of lemmas at one point, which required me to improve the statement of said proposition. Hopefully it is correct now. Here is the change.

Comment #2601 by Rogier Brussee on June 6, 2017 a 8:55 pm UTC

Suggested slogan: A functor on an Abelian categories is extended to the (bounded below or above) derived category by resolving with a complex that is acyclic for that functor.

Comment #2626 by Johan (site) on July 7, 2017 a 12:36 pm UTC

OK, the slogans were added. See here.

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