Need Help Showing that the Composition of Bijections is a Bijection 
Let $f\colon X \rightarrowtail Y$ and $g\colon Y \rightarrowtail Z$ be bijections. Prove that $g \circ f:X \rightarrowtail Z$ is a bijection.

Since $f$ and $g$ are bijections, they are both one to one and onto. 
To prove $g \circ f$ is injective,
suppose $ g \circ f$ is 1 to 1. $\implies \exists x,y \in X:g(f(x))=g(f(y))$. Since g is one to one, so $f(x)=f(y)$. Since is one to one, $x=y$. So, $g \circ f$ is one to one. 
To prove $g\circ f$ is onto,
suppose $g \circ f$ is onto, $\forall z \in Z,\exists f(x) \in Y: f(x) \mapsto z \implies f(x)=z$. Since g is onto, so this is true. And because f is also onto, $\implies \forall Y,\exists x\in X:x \mapsto y \implies f(x)=y$. therefore, $\forall z,\exists x\in X:x\mapsto z \implies g \circ f(x) \mapsto z$. Therefore, $g \circ f(x)$ is onto. 
Since $g \circ f$ is both one to one, and onto, so it is a bijection. 
Please give comment on my proof, hope I didn't make too many mistakes. Also, how do I write it better?
 A: We do not want to begin the proofs for injectivity or surjectivity by claiming that $ g \circ f$ is one to one or onto, lest we run the risk of circular reasoning. In particular, you write

Suppose $ g \circ f$ is 1 to 1 ... So, $g \circ f$ is one to one. 

which does not make sense because the point is to prove that $g \circ f$ is one to one and this is not something we can assume from the get-go.
For injectivity, we must begin with the assumption that there exists such $x,y \in X$ that $g(f(x))=g(f(y))$, and then proceed to prove that indeed $x = y$. Alternatively, one can prove the contraposition of the statement by assuming that $x \neq y$ and then demonstrating that $g(f(x)) \neq g(f(y))$.
In the case of surjectivity, similarly, we must heed the definition and begin by assuming that for every element in the codomain, there exists an element in the domain that will map to it. You should reexamine your claim that
$$f(x) \mapsto z \implies f(x)=z$$
It is best to suppose that $\forall z \in Z$ (with $g(f(x)) = z$), $\exists \ y \in Y$ s.t. $g(y) = z$, namely, $y = f(x)$. This is true by the surjectivity of $g$. Can you proceed from here?
