>>696
f is bijective
∵f is injective
　∵[f(a)=f(a')]→[a=a']
　　∵a=g(f(a))=g(f(a'))=a'
∵f is surjective
　∵[b∈B]→[b=f(a) for at least one a∈A]
　　∵∀b∈B, ∃h(b)∈A, f(h(b))=b
　　　∵h:B→A
g=h=f^(-1)
∵g=h
　∵∀b∈B, ∃a∈A, g(b)=h(b)=a
　　∵∀b∈B, ∃a∈A, g(f(h(b)))=g(b)=a∧g(f(h(b)))=h(b)=a
∵g=f^(-1)
　∵∀a∈A, g(f(a))=a
　∵∀b∈B, f(g(b))=f(h(b))=b
　　∵g=h