# Set Theory: Functions (injectivity/surjectivity)

• Dec 10th 2010, 03:05 PM
zukias
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 :)
• Dec 10th 2010, 03:33 PM
Plato
Quote:

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 $\displaystyle g\circ f$.