This document discusses properties of composite functions and provides examples to prove statements about one-to-one and onto functions. It states that if f o g is one-to-one, f is not necessarily one-to-one but g is one-to-one. If f o g is onto, then f is onto but g is not necessarily onto. It is proven that if both f and g are one-to-one, then their composite f o g is one-to-one. Counter-examples are provided to support statements about one-to-one and onto functions.

1. Let g be a function from A to B and f be a function from B to C.
1.) If f o g is one-to-one, then is f one-to-one? Is g one-to-one? Prove or give a counter-
example in each case.
2.) If f o g is onto, then is f onto? Is g onto? Prove or give a counter-example in each case
3.) Prove that if both f and g are one-to-one, then f o g is one-to-one.
Solution
i)False.
Consider f: R-> R defined by f(x) = x2, and consider g: R-> R defined by g(x) = x.
Then (f o g)(x) = f(g(x)) = f(x) = x , fog is one-one
But f is not one to one because f(-1) = f(1).
ii) True
For all y C , there exisits a x A such that f(g(x)) = y since fog is onto
==> For all y C , there exists a z B , such that g(x) = z and f(z) = y ,
Therefore , f is onto
iii) If x1 is not equal to x2 , then g(x1) is not equal to g(x2) since g is one -one
Since, g(x1) is not equal to g(x2) , then f(g(x1)) is not equal to f(g(x2)) , since f is one-one.
==> If x1 is not equal to x2 , then f(g(x1)) is not equal to f(g(x2))
==> fog is one-one