Surjectivity and composition in functions

Question

Sorry for this pretty dumb question, but I couldn't find any answer to it.

Supposing we have a set $A$ and functions $f,g:A\to A$. Prove that if $f$ isn't surjective then $f \circ g$ isn't surjective.

Basically as I think, In order for a composite function to be surjective' both the functions that are getting composited (In this case functions $f$ and $g$) should be surjective.

Is that right or wrong? And how can I prove that?

EDIT: Question is solved, Check my answer.

Answer 1

The statement is equivalent to $f\circ g$ surjective implies $f$ surjective, which is rather clear since : Let $y\in A$, there is $x\in A$ s.t. $y=f(g(x))$. In particular, if $u=g(x)$, then $y=f(u)$. We proved that $$\forall y\in A, \exists u\in A: y=f(u),$$ and thus surjectivity of $f$.

Answer 2

Since $f$ is not surjective then

$$\exists y\in A\quad \forall x\in A\quad f(x)\neq y$$

then

$$\forall x\in A\quad z=g(x)\in A\quad f(g(x))=f(z)\neq y$$

therefore $f \circ g$ is not surjective.

Answer 3

It is given that $f$ isn't surjective. Because of that, There is $y \in A$ in such way that for each $x\in A$ exists $f(x)\neq y$.

So if, $f \circ g$ isn't surjective, then for the same element $y$ from beforehand, there is an origin in relation to the function $f \circ g$.

So there exists $t\in A$ in such way that $f(g(t))=y$.

But then, if we mark in $x$ the element $g(t)$ we will receive that $f(x)=y$ which is the opposite for what we have said at first.