📝 题目
例 6 设 $f : X \rightarrow Y,g : Y \rightarrow X$ . 求证:
(1)若 $g\left\lbrack {f\left( x\right) }\right\rbrack = x\left( {\forall x \in X}\right)$ ,则 $f$ 为单射, $g$ 为满射；
(2)若 $g\left\lbrack {f\left( x\right) }\right\rbrack = x\left( {\forall x \in X}\right) ,f\left\lbrack {g\left( y\right) }\right\rbrack = y\left( {\forall y \in Y}\right)$ ,则 $f$ 与 $g$ 互为反函数.
💡 答案与解析
证 (1) $\forall {x}_{1} \in X$ ,由条件得 $g\left\lbrack {f\left( {x}_{1}\right) }\right\rbrack = {x}_{1}$ ,即 $\exists {y}_{1} = f\left( {x}_{1}\right)$ 使得 $g\left( {y}_{1}\right) = {x}_{1}$ ,故 $g$ 为满射.
若 $f\left( {x}_{1}\right) = f\left( {x}_{2}\right)$ ,则由条件推出 ${x}_{1} = g\left\lbrack {f\left( {x}_{1}\right) }\right\rbrack = g\left\lbrack {f\left( {x}_{2}\right) }\right\rbrack =$ ${x}_{2}$ ,即 $f$ 为单射.
评注 只假定 $g\left\lbrack {f\left( x\right) }\right\rbrack = x\left( {\forall x \in X}\right)$ ,一般推不出 $f$ 为满射、 $g$ 为单射. 例如
$$ f : \left\lbrack {0,1}\right\rbrack \rightarrow \left\lbrack {-1,1}\right\rbrack ,\;g : \left\lbrack {-1,1}\right\rbrack \rightarrow \left\lbrack {0,1}\right\rbrack , $$
$$ x \mapsto \sqrt{x},\;x \mapsto {x}^{2}. $$
虽然 $g\left\lbrack {f\left( x\right) }\right\rbrack = x\left( {\forall x \in \left\lbrack {0,1}\right\rbrack }\right)$ ,但是
$$ f\left\lbrack {g\left( x\right) }\right\rbrack = \left| x\right| \;\left( {\forall x \in \left\lbrack {-1,1}\right\rbrack }\right) . $$
由此可见, $f$ 非满射, $g$ 也非单射.
(2)所给条件表明, $f$ , $g$ 为双射. 因此 $f$ 和 $g$ 的反函数都存在. $f\left\lbrack {g\left( y\right) }\right\rbrack = y\left( {\forall y \in Y}\right)$ 意味着 $g\left( y\right)$ 是方程
$$ f\left( x\right) = y \tag{2.3} $$
的解. 又因为 $f$ 是单射,所以 $g\left( y\right)$ 是方程 (2.3) 的惟一解. 按定义即有 $g = {f}^{-1}$ . 同理 $f = {g}^{-1}$ .