# Prooving surjective and injective functions

• Oct 20th 2009, 11:33 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, 02:45 PM
p00ndawg
b.

since f is surjective, rng f = C. That is, for any c $\in$C, there exists b $\in$ B such that f(b) = c. Now since g is surjective, there exists an a $\in$ 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, 02:51 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 $f\circ g$ to be injective $f(g(x))=f(g(y))\implies x=y$. Assume $f(g(x))=f(g(y))$. Since $f$ is injective...

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

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

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