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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:

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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.

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    $\begingroup$ Take an element $z$ of $Z$. Since $g\circ f$ is surjective, there is an element $x$ of $X$ such that $g(f(x))=z$. Now consider the element $f(x)$ of $Y$. What can you conclude about $f$? $\endgroup$ – Valerio Jul 30 at 1:02
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You don't really need to argue with elements:

Since $f(X) \subseteq Y$, we have $Z = g(f(X)) \subseteq g(Y) \subseteq Z$, which implies $g(Y)=Z$, and $g$ is surjective.

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Let $f:X \to Y$ and $g:Y\to Z$. We need to prove that if $g\circ f:X\to Z$ is surjective, then so is $g$.

Idea of Proof: We need to find a $y\in Y$ such that $g(y)=z$ for every $z\in Z$ in accordance to the definition of a surjection.

Proof: Suppose $g\circ f:X\to Z$ is surjective. Take any $z\in Z$. Since $g\circ f$ is surjective, there exists some $x\in X$ such that $(g\circ f)(x)=z$.

Therefore, $g(f(x))=z$. Set $y=f(x)\in Y$. Then, $g(y)=g(f(x))=z$. Hence, $g$ must be surjective.

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Let $c \in Z \implies a \in X$ such that $(g\circ f)(a) = c$. This is true since $(g\circ f)$ is onto. But $(g\circ f)(a) = g(f(a))$. So if you put $b = f(a) \in Y$, then $g(b) = c$, proving $g$ onto. I think this notation is easier to remember than yours...

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We need to prove that $$\forall z \in Z \exists y \in Y(z=f(y))$$ Let z in Z.
$f \circ g$ iz onto, so let x in X so that $z=(f \circ g)(x) $
$g(x) \in Y$
Well, we found element y in Y - $g(x)$ - so that $z=f(y)$. And so we proved that $f$ is onto.

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