Let f:A->B and g:C->D be functions. Disprove if g o f is 1-1 then g is 1-1. Solu
ID: 3082684 • Letter: L
Question
Let f:A->B and g:C->D be functions. Disprove if g o f is 1-1 then g is 1-1.Explanation / Answer
Both statements are true. (1) Let z be any element of set C. Since the function g o f : A --> C is onto, there is an element x of set A such that (g o f)( x ) = z. This is equivalent to g( f( x )) = z. Let y = f( x ). y is in set B and we have g( y ) = z. We have shown that for every z in C, there is a y in B such that g( y ) = z. Therefore, g is onto. ( 2 ) Let y be any element of set B. Let z = g( y ) , an element of C. Since g o f is onto, there is an x in set A for which ( g o f )( x ) = z. We have g( f( x )) = z. We also have g( y ) = z, hence g( f( x )) = g( y ). Since g is one-to-one, we conclude that f( x ) = y. We have shown that for any y in B, there is an x in A such that f( x ) = y. Therefore, f is onto.
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