# The Stacks Project

## Tag 010M

Lemma 12.7.1. Let $\mathcal{A}$ and $\mathcal{B}$ be additive categories. Let $F : \mathcal{A} \to \mathcal{B}$ be a functor. The following are equivalent

1. $F$ is additive,
2. $F(A) \oplus F(B) \to F(A \oplus B)$ is an isomorphism for all $A, B \in \mathcal{A}$, and
3. $F(A \oplus B) \to F(A) \oplus F(B)$ is an isomorphism for all $A, B \in \mathcal{A}$.

Proof. Additive functors commute with direct sums by Lemma 12.3.7 hence (1) implies (2) and (3). On the other hand (2) and (3) are equivalent because the composition $F(A) \oplus F(B) \to F(A \oplus B) \to F(A) \oplus F(B)$ is the identity map. Assume (2) and (3) hold. Let $f, g : A \to B$ be maps. Then $f + g$ is equal to the composition $$A \to A \oplus A \xrightarrow{\text{diag}(f, g)} B \oplus B \to B$$ Apply the functor $F$ and consider the following diagram $$\xymatrix{ F(A) \ar[r] \ar[rd] & F(A \oplus A) \ar[rr]_{F(\text{diag}(f, g))} & & F(B \oplus B) \ar[r] \ar[d] & F(B) \\ & F(A) \oplus F(A) \ar[u] \ar[rr]^{\text{diag}(F(f), F(g))} & & F(B) \oplus F(B) \ar[ru] }$$ We claim this is commutative. For the middle square we can verify it separately for each of the for induced maps $F(A) \to F(B)$ where it follows from the fact that $F$ is a functor (in other words this square commutes even if $F$ does not satisfy any properties beyond being a functor). For the triangle on the left, we use that $F(A \oplus A) \to F(A) \oplus F(A)$ is an isomorphism to see that it suffice to check after composition with this map and this check is trivial. Dually for the other triangle. Thus going around the bottom is equal to $F(f + g)$ and we conclude. $\square$

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 12.7.2. 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 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 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 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. If $F$ is left exact, i.e., $F$ commutes with finite limits, then $F$ sends products to products, hence $F$ preserved direct sums, hence $F$ is additive by Lemma 12.7.1. On the other hand, suppose that 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. Let $A, B$ be two objects. Then we have a short exact sequence $$0 \to A \to A \oplus B \to B \to 0$$ see for example Lemma 12.3.10. By assumption, the lower row in the commutative diagram $$\xymatrix{ 0 \ar[r] & F(A) \ar[d] \ar[r] & F(A) \oplus F(B) \ar[r] \ar[d] & F(B) \ar[d] \ar[r] & 0 \\ 0 \ar[r] & F(A) \ar[r] & F(A \oplus B) \ar[r] & F(B) }$$ is exact. Hence by the snake lemma (Lemma 12.5.17) we conclude that $F(A) \oplus F(B) \to F(A \oplus B)$ is an isomorphism. Hence $F$ is additive in this case as well. Thus for the rest of the proof we may assume $F$ is additive.

Denote $f : B \to C$ a map from $B$ to $C$. Exactness of $0 \to A \to B \to C$ just means that $A = \mathop{\rm Ker}(f)$. 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 $F(\mathop{\rm Ker}(f)) = \mathop{\rm Ker}(F(f))$ 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 $\mathop{\rm Ker}(f) = \mathop{\rm Ker}(f')$. Also $0 \to \mathop{\rm Ker}(f') \to A \to I \to 0$ and $0 \to I \to B \to B/I \to 0$ are short exact. By the condition imposed on $F$ we see that $0 \to F(\mathop{\rm Ker}(f')) \to F(A) \to F(I)$ and $0 \to F(I) \to F(B) \to F(B/I)$ are exact. Hence it is also the case that $F(\mathop{\rm Ker}(f'))$ 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 12.7.3. 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 $$\mathop{\rm Ext}\nolimits_\mathcal{A}(B, A) \longrightarrow \mathop{\rm Ext}\nolimits_\mathcal{B}(F(B), F(A))$$ 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 12.7.4. 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 $K = \mathop{\rm Ker}(B_1 \to B_2)$, 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 $Q = \mathop{\rm Coker}(B_1 \to B_2)$, 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 $$\mathop{\rm Ker}(\psi) = b\mathop{\rm Ker}(a\psi), \quad\text{and}\quad \mathop{\rm Coker}(\psi) = b\mathop{\rm Coker}(a\psi).$$ Now we still have to show that $\mathop{\rm Coim}(\psi)= \mathop{\rm Im}(\psi)$. 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*} \mathop{\rm Coim}(\psi) & = \mathop{\rm Coker}(\mathop{\rm Ker}(\psi) \to A_1) & \text{by definition} \\ & = b\mathop{\rm Coker}(a(\mathop{\rm Ker}(\psi) \to A_1)) & \text{by formula above} \\ & = b\mathop{\rm Coker}(\mathop{\rm Ker}(a\psi) \to aA_1)) & a\text{ preserves kernels} \\ & = b\mathop{\rm Coim}(a\psi) & \text{by definition} \\ & = b\mathop{\rm Im}(a\psi) & \mathcal{B}\text{ is abelian} \\ & = b\mathop{\rm Ker}(aA_2 \to \mathop{\rm Coker}(a\psi)) & \text{by definition} \\ & = \mathop{\rm Ker}(baA_2 \to b\mathop{\rm Coker}(a\psi)) & b\text{ preserves kernels} \\ & = \mathop{\rm Ker}(A_2 \to b\mathop{\rm Coker}(a\psi)) & ba = \text{id}_\mathcal{A} \\ & = \mathop{\rm Ker}(A_2 \to \mathop{\rm Coker}(\psi)) & \text{by formula above} \\ & = \mathop{\rm Im}(\psi) & \text{by definition} \end{align*} Thus the lemma holds. $\square$

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

\noindent
First a completely silly lemma characterizing additive functors

\begin{lemma}
Let $\mathcal{A}$ and $\mathcal{B}$ be additive categories.
Let $F : \mathcal{A} \to \mathcal{B}$ be a functor.
The following are equivalent
\begin{enumerate}
\item $F$ is additive,
\item $F(A) \oplus F(B) \to F(A \oplus B)$ is an isomorphism for
all $A, B \in \mathcal{A}$, and
\item $F(A \oplus B) \to F(A) \oplus F(B)$  is an isomorphism for
all $A, B \in \mathcal{A}$.
\end{enumerate}
\end{lemma}

\begin{proof}
Additive functors commute with direct sums by
implies (2) and (3). On the other hand (2) and (3)
are equivalent because the composition
$F(A) \oplus F(B) \to F(A \oplus B) \to F(A) \oplus F(B)$
is the identity map. Assume (2) and (3) hold.
Let $f, g : A \to B$ be maps. Then $f + g$ is equal to
the composition
$$A \to A \oplus A \xrightarrow{\text{diag}(f, g)} B \oplus B \to B$$
Apply the functor $F$ and consider the following diagram
$$\xymatrix{ F(A) \ar[r] \ar[rd] & F(A \oplus A) \ar[rr]_{F(\text{diag}(f, g))} & & F(B \oplus B) \ar[r] \ar[d] & F(B) \\ & F(A) \oplus F(A) \ar[u] \ar[rr]^{\text{diag}(F(f), F(g))} & & F(B) \oplus F(B) \ar[ru] }$$
We claim this is commutative. For the middle square we can verify it
separately for each of the for induced maps $F(A) \to F(B)$
where it follows from the fact that $F$ is a functor (in other words
this square commutes even if $F$ does not satisfy any properties
beyond being a functor). For the triangle on the left, we use that
$F(A \oplus A) \to F(A) \oplus F(A)$ is an isomorphism
to see that it suffice to check after composition with
this map and this check is trivial. Dually for the other triangle.
Thus going around the bottom is equal to $F(f + g)$ and we conclude.
\end{proof}

\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 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 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 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}
If $F$ is left exact, i.e., $F$ commutes with finite limits, then
$F$ sends products to products, hence $F$ preserved direct sums,
hence $F$ is additive by Lemma \ref{lemma-additive-functor}.
On the other hand, suppose that 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. Let $A, B$ be two objects. Then we have a short
exact sequence
$$0 \to A \to A \oplus B \to B \to 0$$
By assumption, the lower row in the commutative diagram
$$\xymatrix{ 0 \ar[r] & F(A) \ar[d] \ar[r] & F(A) \oplus F(B) \ar[r] \ar[d] & F(B) \ar[d] \ar[r] & 0 \\ 0 \ar[r] & F(A) \ar[r] & F(A \oplus B) \ar[r] & F(B) }$$
is exact. Hence by the snake lemma (Lemma \ref{lemma-snake})
we conclude that $F(A) \oplus F(B) \to F(A \oplus B)$ is an
isomorphism. Hence $F$ is additive in this case as well.
Thus for the rest of the proof we may assume $F$ is additive.

\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
$$\Ext_\mathcal{A}(B, A) \longrightarrow \Ext_\mathcal{B}(F(B), F(A))$$
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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