Left (Right) Exact Functor Definition
一般来说, left (right) exact functor 的定义是把 exact sequence 变成 left (right) exact sequence 的 functor. 我们可以把这个定义稍微加强.
$$
\DeclareMathOperator{\coker}{coker}
\DeclareMathOperator{\img}{im}
\DeclarePairedDelimiter{\abs}{\lvert}{\rvert}
\DeclarePairedDelimiter{\norm}{\lVert}{\rVert}
\newcommand{\ol}{\overline}
$$
Let $\mathcal{P}$ and $\mathcal{Q}$ be abelian categories. Let $F: \mathcal{P} \to \mathcal{Q}$ be additive functor (so that, in particular $F(0) = 0$).
Proposition
The following are equivalent
- If $0 \to A \to B \to C \to 0$ is exact, then $0 \to F(A) \to F(B) \to F(C)$ is exact.
- If $0 \to A \to B \to C$ is exact, then $0 \to F(A) \to F(B) \to F(C)$ is exact.
Proof.
(2) $\Rightarrow$ (1) 是显然的. 下面我们说明 (1) $\Rightarrow$ (2).
关键在于注意到 $F$ preserve monomorphism: if $i: A \to B$ is a monomorphism, then $0 \to A \xrightarrow{i} B \to \coker(i) \to 0$ is exact, hence $0 \to F(A) \to F(B) \to F(\coker(i))$ is exact. In particular, $F(i)$ is a monomorphism.
Now if $0 \to A \to B \xrightarrow{f} C$ is exact, then $0 \to A \to B \to \img(f) \to 0$ is exact. Hence by assumption $0 \to F(A) \to F(B) \to F(\img(f))$ is exact. Since $F(\img(f)) \to F(C)$ is a monomorphism, it follows that $0 \to F(A) \to F(B) \to F(C)$ is exact. $\square$
关键在于注意到 $F$ preserve monomorphism: if $i: A \to B$ is a monomorphism, then $0 \to A \xrightarrow{i} B \to \coker(i) \to 0$ is exact, hence $0 \to F(A) \to F(B) \to F(\coker(i))$ is exact. In particular, $F(i)$ is a monomorphism.
Now if $0 \to A \to B \xrightarrow{f} C$ is exact, then $0 \to A \to B \to \img(f) \to 0$ is exact. Hence by assumption $0 \to F(A) \to F(B) \to F(\img(f))$ is exact. Since $F(\img(f)) \to F(C)$ is a monomorphism, it follows that $0 \to F(A) \to F(B) \to F(C)$ is exact. $\square$
如果 $F$ 满足任意一条性质, 我们称 $F$ is a left exact functor. 同理可以定理 right exact functor (对偶的, 它总是 preserve epimorphism). 对于反变函子, 我们依然是要求最后得到的序列 $0 \to F(C) \to F(B) \to F(A)$ 是从左边开始的.