# Math Help - proof about composition

1. ## proof about composition

Let f : A to B and g : B to C.
if g o f is surjective then g must be surjective

how do you prove this

2. Originally Posted by leinadwerdna
Let f : A to B and g : B to C.
if g o f is surjective then g must be surjective

how do you prove this

Let $c\in C$ : since $g\circ f$ is surjective there exists a $a\in A\,\,s.t.\,\,\,g\circ f(a)=c\Longrightarrow$ denote $b=f(a)\in B\Longrightarrow g(b)=g(f(a))=c\Longrightarrow g$ is surjective.

Tonio

3. Originally Posted by tonio
Let $c\in C$ : since $g\circ f$ is surjective there exists a $a\in A\,\,s.t.\,\,\,g\circ f(a)=c\Longrightarrow$ denote $b=f(a)\in B\Longrightarrow g(b)=g(f(a))=c\Longrightarrow g$ is surjective.

Tonio
can you elaborate a little more on this

4. Originally Posted by leinadwerdna
can you elaborate a little more on this

Since g o f is surjective and is a transformation from A->C, any c in C must have at least one element a in A such that g o f(a) = c. It could have more, but it has at least one.

Now let b be in B and be the result of f(a), thus f(a)=b. Since g o f(a) = c is surjective, g(b) is by substitution. So for every c in C, there exists b in B such that g(b)=c, thus g is a surjection.

That was just writing tonio's post into words. Hopefully it has no errors, but the idea is there.