Proof explanation: $g\circ f$ surjective $\implies$ $g$ is surjective

Problem:

Let $$X,y,Z$$ be sets and $$f:X\to Y$$, $$g:Y\to Z$$ functions. Prove: $$g\circ f$$ surjective $$\implies$$ $$g$$ is surjective

Proof:

Let $$g\circ f$$ be surjective and $$z\in Z$$. Choose $$x\in X$$ such that $$g(f(x))=z$$. Hence $$g(y)=z$$ with $$y=f(x)$$ and thus $$g$$ is surjective.

I don't understand why $$g(y)=z$$ with $$y=f(x)$$ implies that $$g$$ is surjective???

What does it mean that a function is surjective? $$g$$ is a function from $$Y$$ to $$Z$$. You need to show that for each $$z\in Z$$ there is some $$y\in Y$$ such that $$g(y)=z$$. And this is what appears in the proof. You start from picking any element $$z\in Z$$ (without assuming anything about it) and show it is in the image of $$g$$.

We have

$$g \circ f: X \to Z, \tag 1$$

a surjective map; by definition this means

$$\forall z \in Z \exists x \in X, \; g \circ f(x) = z. \tag 2$$

Now $$g \circ f$$ factors through $$Y$$, thus:

$$X \overset{f}{\to} Y \overset{g}{\to} Z; \tag 3$$

therefore,

$$\forall x \in X, \; f(x) \in Y; \tag 4$$

thus,

$$\exists y = f(x) \in Y, \; g(y) = g(f(x)) = g \circ f(x) = z; \tag 5$$

since in accord with (2) this holds for any $$z \in Z$$,

$$g: Y \to Z \tag 6$$

is itself surjective. $$OE\Delta$$.

Because that is what surjective means.

To prove $$g:Y\to Z$$ is surjective must prove that for every $$z \in Z$$ there is a $$y\in Y$$ so that $$g(y) = z$$.

And we know $$g\circ f: X \to Z$$ is surjective. So for every $$z \in Z$$ there is an $$x \in X$$ so that $$g(f(x)) = z$$.

.... But then $$f(x) \in Y$$ and $$g(f(x)) = z$$! So if we set $$y= f(x)$$ that is exactly the $$y$$ we needed!

In other word: For every $$z \in Z$$ there is an $$x\in X$$ and an $$y=f(x) \in Y$$ so that $$g(y)=g(f(x)) = z$$. And that's the definition of surjection.