Surjective functions

l888l888l888 · #1 $Old$ November 3rd, 2009, 04:18 PM

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).

HallsofIvy · #2 $Old$ November 4th, 2009, 04:23 AM

Quote:

Originally Posted by l888l888l888 $View Post$

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))?

l888l888l888 · #3 $Old$ November 4th, 2009, 08:53 AM

f(g(y))= y because f(g(x))=x

Drexel28 · #4 $Old$ November 4th, 2009, 05:51 PM

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.