Suppose we have two functions, $f:X\rightarrow Y$ and $g:Y\rightarrow Z$. If both of these functions are onto, how can we show that $g\circ f:X\rightarrow Z$ is also onto?
Functions – Surjectivity of Composition of Surjective Functions
elementary-set-theoryfunctions
Best Answer
Note that $(g\circ f)(x)=g(f(x))$. So if $f$ is onto, then it means for all $y \in Y$ there exists an $x \in X$ such that $ y=f(x)$. Since $g$ is onto, it also meas that for all $z \in Z$ there exists a $ y \in Y$ such that $g(y)=g(f(x))=z$. Thus, for all $z\in Z$ there exists an $x \in X$ such that $g(f(x))=z$. Hence $g\circ f$ is onto.
One important point you should know from the construction above is that $g\circ f$ is still onto even if $f$ is not onto but $g$ is onto. In other words $g$ must necessarily be onto for $g\circ f$ to be onto.