# The Stacks Project

## Tag 0156

### 13.20. Right derived functors and injective resolutions

At this point we can use the material above to define the right derived functors of an additive functor between an abelian category having enough injectives and a general abelian category.

Lemma 13.20.1. Let $\mathcal{A}$ be an abelian category. Let $I \in \mathop{\mathrm{Ob}}\nolimits(\mathcal{A})$ be an injective object. Let $I^\bullet$ be a bounded below complex of injectives in $\mathcal{A}$.

1. $I^\bullet$ computes $RF$ relative to $\text{Qis}^{+}(\mathcal{A})$ for any exact functor $F : K^{+}(\mathcal{A}) \to \mathcal{D}$ into any triangulated category $\mathcal{D}$.
2. $I$ is right acyclic for any additive functor $F : \mathcal{A} \to \mathcal{B}$ into any abelian category $\mathcal{B}$.

Proof. Part (2) is a direct consequences of part (1) and Definition 13.16.3. To prove (1) let $\alpha : I^\bullet \to K^\bullet$ be a quasi-isomorphism into a complex. By Lemma 13.18.6 we see that $\alpha$ has a left inverse. Hence the category $I^\bullet/\text{Qis}^{+}(\mathcal{A})$ is essentially constant with value $\text{id} : I^\bullet \to I^\bullet$. Thus also the ind-object $$I^\bullet/\text{Qis}^{+}(\mathcal{A}) \longrightarrow \mathcal{D}, \quad (I^\bullet \to K^\bullet) \longmapsto F(K^\bullet)$$ is essentially constant with value $F(I^\bullet)$. This proves (1), see Definitions 13.15.2 and 13.15.10. $\square$

Lemma 13.20.2. Let $\mathcal{A}$ be an abelian category with enough injectives.

1. For any exact functor $F : K^{+}(\mathcal{A}) \to \mathcal{D}$ into a triangulated category $\mathcal{D}$ the right derived functor $$RF : D^{+}(\mathcal{A}) \longrightarrow \mathcal{D}$$ is everywhere defined.
2. For any additive functor $F : \mathcal{A} \to \mathcal{B}$ into an abelian category $\mathcal{B}$ the right derived functor $$RF : D^{+}(\mathcal{A}) \longrightarrow D^{+}(\mathcal{B})$$ is everywhere defined.

Proof. Combine Lemma 13.20.1 and Proposition 13.17.8 for the second assertion. To see the first assertion combine Lemma 13.18.3, Lemma 13.20.1, Lemma 13.15.14, and Equation (13.15.9.1). $\square$

Lemma 13.20.3. Let $\mathcal{A}$ be an abelian category with enough injectives. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor.

1. The functor $RF$ is an exact functor $D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$.
2. The functor $RF$ induces an exact functor $K^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$.
3. The functor $RF$ induces a $\delta$-functor $\text{Comp}^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$.
4. The functor $RF$ induces a $\delta$-functor $\mathcal{A} \to D^{+}(\mathcal{B})$.

Proof. This lemma simply reviews some of the results obtained so far. Note that by Lemma 13.20.2 $RF$ is everywhere defined. Here are some references:

1. The derived functor is exact: This boils down to Lemma 13.15.6.
2. This is true because $K^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})$ is exact and compositions of exact functors are exact.
3. This is true because $\text{Comp}^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})$ is a $\delta$-functor, see Lemma 13.12.1.
4. This is true because $\mathcal{A} \to \text{Comp}^{+}(\mathcal{A})$ is exact and precomposing a $\delta$-functor by an exact functor gives a $\delta$-functor.

$\square$

Lemma 13.20.4. Let $\mathcal{A}$ be an abelian category with enough injectives. Let $F : \mathcal{A} \to \mathcal{B}$ be a left exact functor.

1. For any short exact sequence $0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0$ of complexes in $\text{Comp}^{+}(\mathcal{A})$ there is an associated long exact sequence $$\ldots \to H^i(RF(A^\bullet)) \to H^i(RF(B^\bullet)) \to H^i(RF(C^\bullet)) \to H^{i + 1}(RF(A^\bullet)) \to \ldots$$
2. The functors $R^iF : \mathcal{A} \to \mathcal{B}$ are zero for $i < 0$. Also $R^0F = F : \mathcal{A} \to \mathcal{B}$.
3. We have $R^iF(I) = 0$ for $i > 0$ and $I$ injective.
4. The sequence $(R^iF, \delta)$ forms a universal $\delta$-functor (see Homology, Definition 12.11.3) from $\mathcal{A}$ to $\mathcal{B}$.

Proof. This lemma simply reviews some of the results obtained so far. Note that by Lemma 13.20.2 $RF$ is everywhere defined. Here are some references:

1. This follows from Lemma 13.20.3 part (3) combined with the long exact cohomology sequence (13.11.1.1) for $D^{+}(\mathcal{B})$.
2. This is Lemma 13.17.3.
3. This is the fact that injective objects are acyclic.
4. This is Lemma 13.17.6.

$\square$

The code snippet corresponding to this tag is a part of the file derived.tex and is located in lines 6476–6632 (see updates for more information).

\section{Right derived functors and injective resolutions}
\label{section-right-derived-functor}

\noindent
At this point we can use the material above to define the right derived
functors of an additive functor between an abelian category having
enough injectives and a general abelian category.

\begin{lemma}
\label{lemma-injective-acyclic}
Let $\mathcal{A}$ be an abelian category.
Let $I \in \Ob(\mathcal{A})$ be an injective object.
Let $I^\bullet$ be a bounded below complex of injectives in $\mathcal{A}$.
\begin{enumerate}
\item $I^\bullet$ computes $RF$ relative to $\text{Qis}^{+}(\mathcal{A})$
for any exact functor $F : K^{+}(\mathcal{A}) \to \mathcal{D}$
into any triangulated category $\mathcal{D}$.
\item $I$ is right acyclic for any additive functor
$F : \mathcal{A} \to \mathcal{B}$ into any abelian category $\mathcal{B}$.
\end{enumerate}
\end{lemma}

\begin{proof}
Part (2) is a direct consequences of part (1) and
Definition \ref{definition-derived-functor}.
To prove (1) let $\alpha : I^\bullet \to K^\bullet$ be a quasi-isomorphism
into a complex. By
Lemma \ref{lemma-morphisms-lift}
we see that $\alpha$ has a left inverse. Hence the category
$I^\bullet/\text{Qis}^{+}(\mathcal{A})$ is essentially constant with value
$\text{id} : I^\bullet \to I^\bullet$. Thus also the ind-object
$$I^\bullet/\text{Qis}^{+}(\mathcal{A}) \longrightarrow \mathcal{D}, \quad (I^\bullet \to K^\bullet) \longmapsto F(K^\bullet)$$
is essentially constant with value $F(I^\bullet)$. This proves (1), see
Definitions \ref{definition-right-derived-functor-defined} and
\ref{definition-computes}.
\end{proof}

\begin{lemma}
\label{lemma-enough-injectives-right-derived}
Let $\mathcal{A}$ be an abelian category with enough injectives.
\begin{enumerate}
\item For any exact functor $F : K^{+}(\mathcal{A}) \to \mathcal{D}$
into a triangulated category $\mathcal{D}$ the right derived
functor
$$RF : D^{+}(\mathcal{A}) \longrightarrow \mathcal{D}$$
is everywhere defined.
\item For any additive functor $F : \mathcal{A} \to \mathcal{B}$ into an
abelian category $\mathcal{B}$ the right derived functor
$$RF : D^{+}(\mathcal{A}) \longrightarrow D^{+}(\mathcal{B})$$
is everywhere defined.
\end{enumerate}
\end{lemma}

\begin{proof}
Combine
Lemma \ref{lemma-injective-acyclic}
and
Proposition \ref{proposition-enough-acyclics}
for the second assertion. To see the first assertion combine
Lemma \ref{lemma-injective-resolutions-exist},
Lemma \ref{lemma-injective-acyclic},
Lemma \ref{lemma-existence-computes},
and Equation (\ref{equation-everywhere}).
\end{proof}

\begin{lemma}
\label{lemma-right-derived-properties}
Let $\mathcal{A}$ be an abelian category with enough injectives.
Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor.
\begin{enumerate}
\item The functor $RF$ is an exact functor
$D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$.
\item The functor $RF$ induces an exact functor
$K^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$.
\item The functor $RF$ induces a $\delta$-functor
$\text{Comp}^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$.
\item The functor $RF$ induces a $\delta$-functor
$\mathcal{A} \to D^{+}(\mathcal{B})$.
\end{enumerate}
\end{lemma}

\begin{proof}
This lemma simply reviews some of the results obtained so far.
Note that by
Lemma \ref{lemma-enough-injectives-right-derived}
$RF$ is everywhere defined. Here are some references:
\begin{enumerate}
\item The derived functor is exact: This boils down to
Lemma \ref{lemma-2-out-of-3-defined}.
\item This is true because $K^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})$
is exact and compositions of exact functors are exact.
\item This is true because
$\text{Comp}^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})$ is
a $\delta$-functor, see
Lemma \ref{lemma-derived-canonical-delta-functor}.
\item This is true because $\mathcal{A} \to \text{Comp}^{+}(\mathcal{A})$
is exact and precomposing a $\delta$-functor by an exact functor gives
a $\delta$-functor.
\end{enumerate}
\end{proof}

\begin{lemma}
\label{lemma-higher-derived-functors}
Let $\mathcal{A}$ be an abelian category with enough injectives.
Let $F : \mathcal{A} \to \mathcal{B}$ be a left exact functor.
\begin{enumerate}
\item For any short exact sequence
$0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0$
of complexes in $\text{Comp}^{+}(\mathcal{A})$ there
is an associated long exact sequence
$$\ldots \to H^i(RF(A^\bullet)) \to H^i(RF(B^\bullet)) \to H^i(RF(C^\bullet)) \to H^{i + 1}(RF(A^\bullet)) \to \ldots$$
\item The functors $R^iF : \mathcal{A} \to \mathcal{B}$
are zero for $i < 0$. Also $R^0F = F : \mathcal{A} \to \mathcal{B}$.
\item We have $R^iF(I) = 0$ for $i > 0$ and $I$ injective.
\item The sequence $(R^iF, \delta)$ forms a universal $\delta$-functor (see
Homology, Definition \ref{homology-definition-universal-delta-functor})
from $\mathcal{A}$ to $\mathcal{B}$.
\end{enumerate}
\end{lemma}

\begin{proof}
This lemma simply reviews some of the results obtained so far.
Note that by
Lemma \ref{lemma-enough-injectives-right-derived}
$RF$ is everywhere defined. Here are some references:
\begin{enumerate}
\item This follows from
Lemma \ref{lemma-right-derived-properties} part (3)
combined with the long exact cohomology sequence
(\ref{equation-long-exact-cohomology-sequence-D}) for
$D^{+}(\mathcal{B})$.
\item This is
Lemma \ref{lemma-left-exact-higher-derived}.
\item This is the fact that injective objects are acyclic.
\item This is
Lemma \ref{lemma-right-derived-delta-functor}.
\end{enumerate}
\end{proof}

There are no comments yet for this tag.

## Add a comment on tag 0156

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 lower-right corner).