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Let $f: X \rightarrow Y$ and $g: Y \rightarrow X$

If $f \circ g$ is surjective, then f is surjective, too. I think that is true.

Question: How can I proove that? I have so far:

$\forall y \in Y: \exists x \in X : f(x)=y$

$\forall x \in Y : \exists y \in Y : f \circ g(y)= x$

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  • $\begingroup$ Apply the second statement that you wrote, than the number $z$ with $f(z)=x$ that you are looking for is $z=g(y)$ $\endgroup$
    – Momo
    Feb 20, 2017 at 14:31

2 Answers 2

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Set $y\in Y$, then exists $z\in Y$ with $f(g(z))=f\circ g(z)=y$, so take $x=g(z)\in X$ that satisfices $f(x)=f(g(z))=y$

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Let $y \in Y$. Then by assumption there is some $x \in Y$ such that $f\circ g(x) = y$. Note that $g(x) \in X$ by assumption.

Another way is to suppose $f$ is not surjective. Then there is some $y \in Y$ such that there is no $x \in X$ such that $f(x) = y$. So there is no value of $g$ such that $f$ at that value is $=y$, a contradiction.

The two ways are essentially the same in this case.

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