How to prove that the composition of two surjective functions is surjective [duplicate]

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I know that the map $f:A\to B$ is a surjective function (onto) if for all $b$ in $B$, there exists an $a$ in $A$ such that $f(a)=b$

But I am having trouble getting started with this proof since it involves the composition of two surjective functions.

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1 Answer

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Ok, so let $f:A \to B$ and $g:B \to C$ be surjective functions. We wish to show that $g \circ f:A \to C$ is also surjective. Now, you know that for every $c \in C$, there is a $b \in B$ such that $c = g(b)$, by surjectivity of $g$. And you know that for that same $b$, there exists an $a \in A$ such that $b=f(a)$, by surjectivity of $f$. So what we have shown is that for every $c \in C$, there is an $a \in A$ such that $c=g(f(a))=g \circ f(a)$, i.e., $g \circ f$ is surjective.

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