MAT102H5 Lecture Notes - Lecture 17: Rational Number, Surjective Function, Bijection
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MAT102H5 Full Course Notes
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The composition of g with f, denoted g o f, is the function from a to c, given by (g o f)(x)=g(f(x)) for x a. )=e f o g:r r,f o g(x)=f(g(x))=f(e )= f o f:r r, f(f(x))= Conclusion: in general, f o g g o f f:n n z, f(n,m)=n-m, g:z r, g(x)= Then g o f :n n r, g o f(n,m)=g(f(n,m))=g(n,m)= f o g is not defined because no r n n. Prove that if g o f is injective, then so is f. Proof: let x ,x g o f (x )=g o f (x ) Then g(f(x ))=g(f(x )) x =x as g o f is injective. If f:[0, injective, but g is not injective, so the answer is no. The composition of two injections is an injection. The composition of two surjections is a surjection. The composition of two bijections is a bijection.
