# Surjective functions

• Nov 3rd 2009, 02:18 PM
l888l888l888
Surjective functions
let f: From A to B. let g: From B to A.

How do I prove f(g(x)) = x (indentity function) implies f is onto (surjective).
• Nov 4th 2009, 02:23 AM
HallsofIvy
Quote:

Originally Posted by l888l888l888
let f: From A to B. let g: From B to A.

How do I prove f(g(x)) = x (indentity function) implies f is onto (surjective).

Suppose f is NOT surjective. Then there exist y in B such that, for all x in A, $\displaystyle f(x)\ne y$. Now, what is f(g(y))?
• Nov 4th 2009, 06:53 AM
l888l888l888
f(g(y))= y because f(g(x))=x
• Nov 4th 2009, 03:51 PM
Drexel28
More explicitly (like HallsOfIvy said)

Problem: Suppose that $\displaystyle f:X\mapsto Y$ and $\displaystyle g:Y\mapsto X$ are functions such that $\displaystyle f\circ g=\iota_Y$ (identity mapping on Y). Prove that $\displaystyle f:X\mapsto Y$ is surjective.

Proof: Let $\displaystyle y\in Y$. Since $\displaystyle g:Y\mapsto X$ we know that $\displaystyle g(y)\in X$. Therefore $\displaystyle f(g(y))=y$ since $\displaystyle f\circ g=\iota_Y$. And since $\displaystyle y$ was arbitrary this proves surjectivity.