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# am_lect_16 - Another proof involving composition If f A B...

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Another proof involving composition If f : A B is onto and g : B C is onto then g o f : A C is onto Proof : Let y C . Since g is onto, z B such that g ( z ) = y Also since f is onto, x A such that f ( x ) = z Now g o f ( x ) = g ( f ( x )) = g ( z ) = y Hence, x A such that g o f ( x ) = y and hence g o f is onto Are the converse statements true? — If g o f : A C is 1-to-1, is f : A B 1-to-1 and g : B C 1-to-1? — If g o f : A C is onto, is f : A B onto and g : B C onto? (No, for both questions!)

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Without loss of generality Recall that we sometimes do proofs by cases . Consider an example where two cases are very similar: Function f : A B is increasing if x A , y A , x < y f ( x ) < f ( y ) Claim : Any increasing function is one-to-one Proof : We need to show that x A , y A , f ( x ) = f( y ) x = y or equivalently, x A , y A , x y
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am_lect_16 - Another proof involving composition If f A B...

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