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Tag 010N

Chapter 12: Homological Algebra > Section 12.7: Additive functors

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). $\square$

    The code snippet corresponding to this tag is a part of the file homology.tex and is located in lines 1277–1297 (see updates for more information).

    \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
    $$
    0 \to A \to A \oplus B \to B \to 0
    $$
    see for example Lemma \ref{lemma-additive-cat-biproduct-kernel}.
    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}

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