# The Stacks Project

## Tag 010M

Recall that we defined, in Categories, Definition 4.23.1 the notion of a ''right exact'', ''left exact'' and ''exact'' functor in the setting of a functor between categories that have finite (co)limits. Thus this applies in particular to functors between abelian categories.

Lemma 12.7.1. Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories. Let $F : \mathcal{A} \to \mathcal{B}$ be a functor.

1. If $F$ is either left or right exact, then it is additive.
2. If $F$ is additive then it is left exact if and only if for every short exact sequence $0 \to A \to B \to C \to 0$ the sequence $0 \to F(A) \to F(B) \to F(C)$ is exact.
3. If $F$ is additive then it is right exact if and only if for every short exact sequence $0 \to A \to B \to C \to 0$ the sequence $F(A) \to F(B) \to F(C) \to 0$ is exact.
4. If $F$ is additive then it is exact if and only if for every short exact sequence $0 \to A \to B \to C \to 0$ the sequence $0 \to F(A) \to F(B) \to F(C) \to 0$ is exact.

Proof. Let us first note that if $F$ commutes with the empty limit or the empty colimit, then $F(0) = 0$. In particular $F$ applied to the zero morphism is zero. We will use this below without mention.

Suppose that $F$ is left exact, i.e., commutes with finite limits. Then $F(A \times A) = F(A) \times F(A)$ with projections $F(p)$ and $F(q)$. Hence $F(A \oplus A) = F(A) \oplus F(A)$ with all four morphisms $F(i), F(j), F(p), F(q)$ equal to their counterparts in $\mathcal{B}$ as they satisfy the same relations, see Remark 12.3.6. Then $f = F(p + q)$ is a morphism $f : F(A) \oplus F(A) \to F(A)$ such that $f \circ F(i) = F(p \circ i + q \circ i) = F(\text{id}_A) = \text{id}_{F(A)}$. And similarly $f \circ F(j) = \text{id}_A$. We conclude that $F(p + q) = F(p) + F(q)$. For any pair of morphisms $a, b : B \to A$ the map $g = F(i \circ a + j \circ b) : F(B) \to F(A) \oplus F(A)$ is a morphism such that $F(p) \circ g = F(p \circ (i \circ a + j \circ b)) = F(a)$ and similarly $F(q) \circ g = F(b)$. Hence $g = F(i) \circ F(a) + F(j) \circ F(b)$. The sum of $a$ and $b$ is the composition $$\xymatrix{ B \ar[rr]^-{i \circ a + j \circ b} & & A \oplus A \ar[r]^-{p + q} & A. }$$ Applying $F$ we get $$\xymatrix{ F(B) \ar[rrr]^-{F(i) \circ F(a) + F(j) \circ F(b)} & & & F(A) \oplus F(A) \ar[rr]^-{F(p) + F(q)} & & A. }$$ where we used the expressions for $f$ and $g$ obtained above. Hence $F$ is additive.1

Denote $f : B \to C$ a map from $B$ to $C$. Exactness of $0 \to A \to B \to C$ just means that $A = \text{Ker}(f)$. Clearly the kernel of $f$ is the equalizer of the two maps $f$ and $0$ from $B$ to $C$. Hence if $F$ commutes with limits, then $F(\text{Ker}(f)) = \text{Ker}(F(f))$ which exactly means that $0 \to F(A) \to F(B) \to F(C)$ is exact.

Conversely, suppose that $F$ is additive and transforms any short exact sequence $0 \to A \to B \to C \to 0$ into an exact sequence $0 \to F(A) \to F(B) \to F(C)$. Because it is additive it commutes with direct sums and hence finite products in $\mathcal{A}$. To show it commutes with finite limits it therefore suffices to show that it commutes with equalizers. But equalizers in an abelian category are the same as the kernel of the difference map, hence it suffices to show that $F$ commutes with taking kernels. Let $f : A \to B$ be a morphism. Factor $f$ as $A \to I \to B$ with $f' : A \to I$ surjective and $i : I \to B$ injective. (This is possible by the definition of an abelian category.) Then it is clear that $\text{Ker}(f) = \text{Ker}(f')$. Also $0 \to \text{Ker}(f') \to A \to I \to 0$ and $0 \to I \to B \to B/I \to 0$ are short exact. By the condition imposed on $F$ we see that $0 \to F(\text{Ker}(f')) \to F(A) \to F(I)$ and $0 \to F(I) \to F(B) \to F(B/I)$ are exact. Hence it is also the case that $F(\text{Ker}(f'))$ is the kernel of the map $F(A) \to F(B)$, and we win.

The proof of (3) is similar to the proof of (2). Statement (4) is a combination of (2) and (3). $\square$

Lemma 12.7.2. Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories. Let $F : \mathcal{A} \to \mathcal{B}$ be an exact functor. For every pair of objects $A, B$ of $\mathcal{A}$ the functor $F$ induces an abelian group homomorphism $$\text{Ext}_\mathcal{A}(B, A) \longrightarrow \text{Ext}_\mathcal{B}(F(B), F(A))$$ which maps the extension $E$ to $F(E)$.

Proof. Omitted. $\square$

The following lemma is used in the proof that the category of abelian sheaves on a site is abelian, where the functor $b$ is sheafification.

Lemma 12.7.3. Let $a : \mathcal{A} \to \mathcal{B}$ and $b : \mathcal{B} \to \mathcal{A}$ be functors. Assume that

1. $\mathcal{A}$, $\mathcal{B}$ are additive categories, $a$, $b$ are additive functors, and $a$ is right adjoint to $b$,
2. $\mathcal{B}$ is abelian and $b$ is left exact, and
3. $ba \cong \text{id}_\mathcal{A}$.

Then $\mathcal{A}$ is abelian.

Proof. As $\mathcal{B}$ is abelian we see that all finite limits and colimits exist in $\mathcal{B}$ by Lemma 12.5.5. Since $b$ is a left adjoint we see that $b$ is also right exact and hence exact, see Categories, Lemma 4.24.5. Let $\varphi : B_1 \to B_2$ be a morphism of $\mathcal{B}$. In particular, if $K = \text{Ker}(B_1 \to B_2)$, then $K$ is the equalizer of $0$ and $\varphi$ and hence $bK$ is the equalizer of $0$ and $b\varphi$, hence $bK$ is the kernel of $b\varphi$. Similarly, if $Q = \text{Coker}(B_1 \to B_2)$, then $Q$ is the coequalizer of $0$ and $\varphi$ and hence $bQ$ is the coequalizer of $0$ and $b\varphi$, hence $bQ$ is the cokernel of $b\varphi$. Thus we see that every morphism of the form $b\varphi$ in $\mathcal{A}$ has a kernel and a cokernel. However, since $ba \cong \text{id}$ we see that every morphism of $\mathcal{A}$ is of this form, and we conclude that kernels and cokernels exist in $\mathcal{A}$. In fact, the argument shows that if $\psi : A_1 \to A_2$ is a morphism then $$\text{Ker}(\psi) = b\text{Ker}(a\psi), \quad\text{and}\quad \text{Coker}(\psi) = b\text{Coker}(a\psi).$$ Now we still have to show that $\text{Coim}(\psi)= \text{Im}(\psi)$. We do this as follows. First note that since $\mathcal{A}$ has kernels and cokernels it has all finite limits and colimits (see proof of Lemma 12.5.5). Hence we see by Categories, Lemma 4.24.5 that $a$ is left exact and hence transforms kernels (=equalizers) into kernels. \begin{align*} \text{Coim}(\psi) & = \text{Coker}(\text{Ker}(\psi) \to A_1) & \text{by definition} \\ & = b\text{Coker}(a(\text{Ker}(\psi) \to A_1)) & \text{by formula above} \\ & = b\text{Coker}(\text{Ker}(a\psi) \to aA_1)) & a\text{ preserves kernels} \\ & = b\text{Coim}(a\psi) & \text{by definition} \\ & = b\text{Im}(a\psi) & \mathcal{B}\text{ is abelian} \\ & = b\text{Ker}(aA_2 \to \text{Coker}(a\psi)) & \text{by definition} \\ & = \text{Ker}(baA_2 \to b\text{Coker}(a\psi)) & b\text{ preserves kernels} \\ & = \text{Ker}(A_2 \to b\text{Coker}(a\psi)) & ba = \text{id}_\mathcal{A} \\ & = \text{Ker}(A_2 \to \text{Coker}(\psi)) & \text{by formula above} \\ & = \text{Im}(\psi) & \text{by definition} \end{align*} Thus the lemma holds. $\square$

1. I'm sure there is an infinitely slicker proof of this.

The code snippet corresponding to this tag is a part of the file homology.tex and is located in lines 1212–1437 (see updates for more information).

\section{Additive functors}
\label{section-functors}

\noindent
Recall that we defined, in
Categories, Definition \ref{categories-definition-exact}
the notion of a right exact'', left exact'' and
exact'' functor in the setting of a functor between
categories that have finite (co)limits. Thus this
applies in particular to functors between abelian
categories.

\begin{lemma}
\label{lemma-exact-functor}
Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories.
Let $F : \mathcal{A} \to \mathcal{B}$ be a functor.
\begin{enumerate}
\item If $F$ is either left or right exact, then it is additive.
\item If $F$ is additive then it is left exact if and only if
for every short exact sequence
$0 \to A \to B \to C \to 0$
the sequence $0 \to F(A) \to F(B) \to F(C)$
is exact.
\item If $F$ is additive then it is right exact if and only if
for every short exact sequence
$0 \to A \to B \to C \to 0$
the sequence $F(A) \to F(B) \to F(C) \to 0$
is exact.
\item If $F$ is additive then it is exact if and only if
for every short exact sequence
$0 \to A \to B \to C \to 0$
the sequence $0 \to F(A) \to F(B) \to F(C) \to 0$
is exact.
\end{enumerate}
\end{lemma}

\begin{proof}
Let us first note that if $F$ commutes with the empty limit or
the empty colimit, then $F(0) = 0$. In particular $F$ applied
to the zero morphism is zero. We will use this below without mention.

\medskip\noindent
Suppose that $F$ is left exact, i.e., commutes with finite limits.
Then $F(A \times A) = F(A) \times F(A)$ with
projections $F(p)$ and $F(q)$. Hence
$F(A \oplus A) = F(A) \oplus F(A)$ with all
four morphisms $F(i), F(j), F(p), F(q)$ equal to their
counterparts in $\mathcal{B}$ as they satisfy the same
relations, see Remark \ref{remark-direct-sum}.
Then $f = F(p + q)$ is a morphism $f : F(A) \oplus F(A) \to F(A)$
such that $f \circ F(i) = F(p \circ i + q \circ i) = F(\text{id}_A) = \text{id}_{F(A)}$. And similarly $f \circ F(j) = \text{id}_A$.
We conclude that $F(p + q) = F(p) + F(q)$. For
any pair of morphisms $a, b : B \to A$ the map
$g = F(i \circ a + j \circ b) : F(B) \to F(A) \oplus F(A)$
is a morphism such that $F(p) \circ g = F(p \circ (i \circ a + j \circ b)) = F(a)$ and similarly
$F(q) \circ g = F(b)$. Hence $g = F(i) \circ F(a) + F(j) \circ F(b)$.
The sum of $a$ and $b$ is the composition
$$\xymatrix{ B \ar[rr]^-{i \circ a + j \circ b} & & A \oplus A \ar[r]^-{p + q} & A. }$$
Applying $F$ we get
$$\xymatrix{ F(B) \ar[rrr]^-{F(i) \circ F(a) + F(j) \circ F(b)} & & & F(A) \oplus F(A) \ar[rr]^-{F(p) + F(q)} & & A. }$$
where we used the expressions for $f$ and $g$ obtained above.
Hence $F$ is additive.\footnote{I'm sure there is an infinitely
slicker proof of this.}

\medskip\noindent
Denote $f : B \to C$ a map from $B$ to $C$.
Exactness of $0 \to A \to B \to C$ just means that
$A = \Ker(f)$. Clearly the kernel of $f$ is
the equalizer of the two maps $f$ and $0$ from $B$ to $C$.
Hence if $F$ commutes with limits, then $F(\Ker(f)) = \Ker(F(f))$ which exactly means that
$0 \to F(A) \to F(B) \to F(C)$ is exact.

\medskip\noindent
Conversely, suppose that $F$ is additive and
transforms any short exact sequence $0 \to A \to B \to C \to 0$ into
an exact sequence $0 \to F(A) \to F(B) \to F(C)$.
Because it is additive it commutes with direct sums
and hence finite products in $\mathcal{A}$. To show
it commutes with finite limits it therefore
suffices to show that it commutes with
equalizers. But equalizers in an abelian category
are the same as the kernel of the difference map,
hence it suffices to show that $F$ commutes with
taking kernels. Let $f : A \to B$ be a morphism.
Factor $f$ as $A \to I \to B$ with $f' : A \to I$ surjective
and $i : I \to B$ injective. (This is possible by the
definition of an abelian category.) Then it is
clear that $\Ker(f) = \Ker(f')$. Also
$0 \to \Ker(f') \to A \to I \to 0$
and
$0 \to I \to B \to B/I \to 0$
are short exact. By the condition imposed on $F$
we see that
$0 \to F(\Ker(f')) \to F(A) \to F(I)$
and
$0 \to F(I) \to F(B) \to F(B/I)$
are exact. Hence it is also the case that
$F(\Ker(f'))$ is the kernel of the map
$F(A) \to F(B)$, and we win.

\medskip\noindent
The proof of (3) is similar to the proof of (2).
Statement (4) is a combination of (2) and (3).
\end{proof}

\begin{lemma}
\label{lemma-exact-functor-ext}
Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories.
Let $F : \mathcal{A} \to \mathcal{B}$ be an exact functor.
For every pair of objects $A, B$ of $\mathcal{A}$ the
functor $F$ induces an abelian group homomorphism
$$\text{Ext}_\mathcal{A}(B, A) \longrightarrow \text{Ext}_\mathcal{B}(F(B), F(A))$$
which maps the extension $E$ to $F(E)$.
\end{lemma}

\begin{proof}
Omitted.
\end{proof}

\noindent
The following lemma is used in the proof that the category of abelian
sheaves on a site is abelian, where the functor $b$ is sheafification.

\begin{lemma}
Let $a : \mathcal{A} \to \mathcal{B}$ and $b : \mathcal{B} \to \mathcal{A}$
be functors. Assume that
\begin{enumerate}
\item $\mathcal{A}$, $\mathcal{B}$ are additive categories,
$a$, $b$ are additive functors, and $a$ is right adjoint to $b$,
\item $\mathcal{B}$ is abelian and $b$ is left exact, and
\item $ba \cong \text{id}_\mathcal{A}$.
\end{enumerate}
Then $\mathcal{A}$ is abelian.
\end{lemma}

\begin{proof}
As $\mathcal{B}$ is abelian we see that all finite limits and colimits
exist in $\mathcal{B}$ by Lemma \ref{lemma-colimit-abelian-category}.
Since $b$ is a left adjoint we see that $b$ is also right exact
and hence exact, see
Let $\varphi : B_1 \to B_2$ be a morphism of $\mathcal{B}$.
In particular, if $K = \Ker(B_1 \to B_2)$, then $K$ is
the equalizer of $0$ and $\varphi$ and hence
$bK$ is the equalizer of $0$ and $b\varphi$, hence
$bK$ is the kernel of $b\varphi$. Similarly, if
$Q = \Coker(B_1 \to B_2)$, then $Q$ is
the coequalizer of $0$ and $\varphi$ and hence
$bQ$ is the coequalizer of $0$ and $b\varphi$, hence
$bQ$ is the cokernel of $b\varphi$. Thus we see that every morphism
of the form $b\varphi$ in $\mathcal{A}$ has a kernel and a cokernel.
However, since $ba \cong \text{id}$ we see that every morphism of
$\mathcal{A}$ is of this form, and we conclude that kernels and
cokernels exist in $\mathcal{A}$. In fact, the argument shows that
if $\psi : A_1 \to A_2$ is a morphism then
$$\Ker(\psi) = b\Ker(a\psi), \quad\text{and}\quad \Coker(\psi) = b\Coker(a\psi).$$
Now we still have to show that $\Coim(\psi)= \Im(\psi)$.
We do this as follows.
First note that since $\mathcal{A}$ has kernels and cokernels it
has all finite limits and colimits (see proof of
Lemma \ref{lemma-colimit-abelian-category}).
Hence we see by Categories, Lemma \ref{categories-lemma-exact-adjoint}
that $a$ is left exact and
hence transforms kernels (=equalizers) into kernels.
\begin{align*}
\Coim(\psi)
& =
\Coker(\Ker(\psi) \to A_1)
& \text{by definition} \\
& =
b\Coker(a(\Ker(\psi) \to A_1))
& \text{by formula above} \\
& =
b\Coker(\Ker(a\psi) \to aA_1))
& a\text{ preserves kernels} \\
& =
b\Coim(a\psi)
& \text{by definition} \\
& =
b\Im(a\psi)
& \mathcal{B}\text{ is abelian} \\
& =
b\Ker(aA_2 \to \Coker(a\psi))
& \text{by definition} \\
& =
\Ker(baA_2 \to b\Coker(a\psi))
& b\text{ preserves kernels} \\
& =
\Ker(A_2 \to b\Coker(a\psi))
& ba = \text{id}_\mathcal{A} \\
& =
\Ker(A_2 \to \Coker(\psi))
& \text{by formula above} \\
& =
\Im(\psi)
& \text{by definition}
\end{align*}
Thus the lemma holds.
\end{proof}

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