• December 12th 2009, 03:08 PM
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
• December 12th 2009, 03:18 PM
tonio
Quote:

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
• December 12th 2009, 04:33 PM
Quote:

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
• December 12th 2009, 04:54 PM
Jameson
Quote: