Let $f : X \to Y$ and $g: Y \to Z$ be two functions, such that $(g \circ f) : X \to Z$ is bijective
Prove or refute that $f$ is surjective.
For $f$ to be surjective, $\forall b \in Y, \exists a \in X : f(a) = b$
Because $(g \circ f)$ is bijective, the sets X and Z have to have the same number of elements, otherwise the function composition wouldn’t be bijective.
The set Y can have the same number of elements than X and Z or more. It can’t have less, because given that $g \circ f $ is bijective, in other words that X and Z have the same number of elements, the elements in X can map to a smaller number of elements in Y, but the elements in Y can’t map to more elements in Z than there are in Y.
So the function $f$ isn’t necessarily surjective. It could be, but it isn’s necessarily the case.
Are my thoughts correct or not ? If not, why ?? Also, if I am right, how could I write a mathematical proof ?
Thanks for your help !
(P.S. : English isn't my mother tongue, sorry if I made a grammar mistake)