$f\subseteq g\Rightarrow \mathscr{D}(f)\subseteq\mathscr{D}(g)\Rightarrow\mathscr{D}(f)=\mathscr{D}(g)$.

从而 $\forall x\in \mathscr{D}(g),f(x)=g(x),\Rightarrow g\subseteq f,\Rightarrow g=f.$

- (1) $$ \begin{aligned} f\circ g=f(g(x))=f(x+2)=(x+2)^2-1=x^2+4x+3,\ g\circ f=g(f(x))=g(x^2-1)=x^2+1. \end{aligned}$$ - (2) $g$ 是双射, $f,f\circ g,g\circ f$ 均不是.

- (1) 取 $f=(1234)$ 即 $f(1)=2,f(2)=3,f(3)=4,f(4)=1$. $$ \begin{aligned} f^2=(13)(24) \ f^3=(4321) \ f^{-1}=f^3 \ f\circ f^{-1}=I_A \end{aligned} $$ - (2) 取 $g=(12)(34)$, 即 $g(1)=2,g(2)=1,g(3)=4,g(4)=3$.

$ P^{-1}=\left(\begin{aligned} 123 \ 231 \end{aligned}\right) ,\quad P\circ P^{-1}=\left(\begin{aligned} 123 \ 123 \end{aligned}\right) $

- (1) 不是, 因为 $A\cap B \subseteq B$. - (2) 是, 因为 $A \subseteq B\cup A$. - (3) 是, 因为 $A$ 无穷那么存在可数集 $C\subseteq A$, 又 $C\backslash B$ 仍为可数集且 $C\backslash B \subseteq A\backslash B$.