Can someone show me how to prove that $f:X \rightarrow Y$ is onto if and only if it has a right inverse?

Question

$f:X \rightarrow Y$ is onto if and only if it has a right inverse: that is, a function $g:Y \rightarrow X$ such that $f \circ g = 1_y$

Answer 1

You probably mean the map of sets. If $f: X\to Y$ is onto then for any $y\in Y$ there exists $x\in X$ such that $f(x) = y$. Choose for every $y\in Y$ such an $x$ and call it $x_y$. Define $g:Y\to X$ as $g(y) = x_y$. It obviously follows that $(f\circ g) (y) = y$.

If there exists a right inverse $g:Y\to X$ then for any $y\in Y$ take $x = g(y)\in X$. Since $f\circ g = 1_Y$ $f(x) = f(g(y)) = y$ and hence $f$ is onto.