Composition of Function

JellyFish

Junior Member
Joined
Jan 12, 2009
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51
I am struggling to disprove that if g of f is 1-1, then g is 1-1. (For sets A b and C such that f: A to B and g: B to c) I tried to think of some counterexamples but was not successful.
 
if g(f) is 1-1 then given g(f(c1))= g(f(c2)) in C, we have f(c1)=f(c2) in B. The function f(x)=x^2 has this property for c1=-c2.

However, g does not know what f is doing or where it is coming from. All g knows is that f is handing over positive numbers.

Think of a 1-1 function on the positive numbers to set for g and you're done.
 
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