Surjective functions

• Nov 3rd 2009, 03:18 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, 03:23 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, $f(x)\ne y$. Now, what is f(g(y))?
• Nov 4th 2009, 07:53 AM
l888l888l888
f(g(y))= y because f(g(x))=x
• Nov 4th 2009, 04:51 PM
Drexel28
More explicitly (like HallsOfIvy said)

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

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