推 l12203685 :李華倫老師的嗎= =? 09/27 19:18
→ k6416337 :?? 09/27 19:47
推 WWWZZZXXXMMM:謝謝 09/28 01:33
※ 引述《WWWZZZXXXMMM (WZXM)》之銘言：
: Let f:S→T and g:T→U be bijections. Show that g。f is a bijection and
: -1 -1 -1
: (g。f) = f 。 g .
1.prove g。f:S->U is bijective.
i.one to one:
Let x,y lie in S.Then (g。f)(x)=(g。f)(y) => g(f(x))=g(f(y))
=> f(x)=f(y) [∵g is 1-1] => x=y [∵f is 1-1].
By the definition,g。f is 1-1.
ii.onto:
Let y lie in U.We must find an "x" in S s.t. (g。f)(x)=y.
∵g is onto in U,∴there's a "z" in T s.t. g(z)=y.
∵f is onto in T,∴there's an "x" in S s.t. f(x)=z.
=> (g。f)(x)=g(f(x))=g(z)=y.
By i & ii,g。f is bijective.
2.prove f^-1。g^-1 is an inverse of g。f.
i.e.,prove that (g。f)(f^-1。g^-1)(x)=(f^-1。g^-1)(g。f)(x)=x for all x.
Let g^-1(x)=y.Then f(f^-1(g^-1(x)))=f(f^-1(y))=y.
=> (g。f)(f^-1。g^-1)(x)=g(f(f^-1(g^-1(x))))=g(y)=g(g^-1(x))=x.
Repeat the same discussion above on (f^-1。g^-1)(g。f)(x)=x.
So,f^-1。g^-1 is an inverse of g。f.
By the uniqueness of inverse,(g。f)^-1=f^-1。g^-1.
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