I can't prove the following statements about functions f:A->B and g:B->C
1. If gof is one-to-one then so is f.
2. If gof is onto then so is g.
Here fog and gof are compossitions of functions f and g respectively.
fog=f(g(x)) and gof=g(f(x)) and 1A denotes identity ofn A and 1B denotes identity on B.
1. Let y1= gof(x1), y2 = gof(x2), and x1 = x2, then since gof is one-to-one, then we know y1 = y2, since gof is one-to-one, so we know f(x1) = f(x2), since x1=x2, then we know f is one-to-one
2. Since gof is onto, then for each c in C, there is an a in A, such ...
Composition of functions and isomorphisms are investigated. The solution is detailed and well presented.