Let $X,Y,Z$ be sets and $f: X\to Y$, $g: Y\to Z$ functions. If $g\circ f$ is surjective, prove that $g$ is surjective.
Here's my sketch:
Since $g\circ f : X\to Z$ is surjective, there are for every $z\in Z$ at least one $x\in X$ with $f(z)=x$. This means that $\#X \geq \# Z$ . Furthermore, we know that every $x\in X$ will mapped to $f(x)=y$. $g$ can only map $g(y)=z$ with $y$ being the elements of the image of $f$. Ultimately, the amount of $y=f(x)$-values is dependent on the amount of values in $Z$. There can never be more Elements $f(x)=y\in Y$ then there are elements in $Z$. Hence $g$ is surjective.
I hope you get what I just wrote since it is really hard to explain in English.