设
.......u...1...2...3
f1(u)...1...2....3
f2(u)...1....3....2
f3(u)...2.....1....3
f4(u)...2.....3.....1
f5(u)...3.....1......2
f6(u)...3.....2.......1,
f1(u)=u,所以f1[f1(u)]=u,
f2[f2(u)]=u,
f3[f3(u)]=u,
f6[f6(u)]=u,
所以满足f[f(u)]= u的映射有4个.
1、(1,2,3)->(1,2,3)
2、(1,2,3)->(2,1,3)
3、(1,2,3)->(3,2,1)
4、(1,2,3)->(1,3,2)
懂了吗。这里是分别映射
比如第二个:f(f(2))=f(1)=2