# Prooving surjective and injective functions

• Oct 20th 2009, 10:33 AM
neelpatel89
Prooving surjective and injective functions
Let g : A -> B and f : B -> C be functions.
Show that if f and g are injective, then f o g is injective.
Show that if f and g are surjective, then f o g is surjective.

i don't know how to proove it
• Oct 20th 2009, 01:45 PM
p00ndawg
b.

since f is surjective, rng f = C. That is, for any c $\displaystyle \in$C, there exists b $\displaystyle \in$ B such that f(b) = c. Now since g is surjective, there exists an a $\displaystyle \in$ A such that g(a) = b. but then (f o g)(a) = f(g(a)) = f(b) = c, so
f o g is surjective.
• Oct 20th 2009, 01:51 PM
redsoxfan325
Quote:

Originally Posted by neelpatel89
Let g : A -> B and f : B -> C be functions.
Show that if f and g are injective, then f o g is injective.
Show that if f and g are surjective, then f o g is surjective.

i don't know how to proove it

For $\displaystyle f\circ g$ to be injective $\displaystyle f(g(x))=f(g(y))\implies x=y$. Assume $\displaystyle f(g(x))=f(g(y))$. Since $\displaystyle f$ is injective...

Spoiler:
Since $\displaystyle f$ is injective, $\displaystyle g(x)=g(y)$. Since $\displaystyle g$ is injective, $\displaystyle x=y$, so $\displaystyle f(g(x))=f(g(y))\implies x=y$, and $\displaystyle f\circ g$ is injective. $\displaystyle \square$

For $\displaystyle f\circ g$ to be surjective, $\displaystyle \forall~y\in C$, $\displaystyle \exists~x\in A$ such that $\displaystyle f(g(x))=y$. Let $\displaystyle y\in C$. Because $\displaystyle f$ is surjective...

Spoiler:
Because $\displaystyle f$ is surjective, there exists $\displaystyle z\in B$ such that $\displaystyle f(z)=y$. Because $\displaystyle g$ is surjective, there exists $\displaystyle x\in A$ such that $\displaystyle f(x)=z$. So $\displaystyle f(g(x))=y$ and $\displaystyle f\circ g$ is surjective. $\displaystyle \square$