# Thread: Set Theory: Functions (injectivity/surjectivity)

1. ## Set Theory: Functions (injectivity/surjectivity)

Suppose f : A -> B and g : B -> C are functions.
(a) Show that if f is surjective and g is not injective then g o f is not injective.
(Hint: draw pictures to get an idea.)
(b) Show that if f is not surjective and g is injective then g o f is not surjective.

hey, I have a general idea how to solve questions like this, but these ones seem more challenging than any ive done so far, so any help is is really appreciated

2. Originally Posted by zukias
Suppose f : A -> B and g : B -> C are functions.
(a) Show that if f is surjective and g is not injective then g o f is not injective.
(Hint: draw pictures to get an idea.)
(b) Show that if f is not surjective and g is injective then g o f is not surjective.
From the given we can get the following:
$\displaystyle \left( {\exists b_1 \in B} \right)\left( {\exists b_2 \in B} \right)\left[ {b_1 \ne b_2 \wedge g(b_1 ) = g(b_2 )} \right]$.

ALSO $\displaystyle \left( {\exists a_1 \in A} \right)\left[ {f(a_1 ) = b_1 } \right] \wedge \left( {\exists a_2 \in A} \right)\left[ {f(a_2 ) = b_2 } \right]$.

There is an easy contradiction to injectivity there for $\displaystyle g\circ f$.