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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.

Mike Pierce
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Jared
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1 Answers1

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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.

silvascientist
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  • That's just a fancy and pretentious version of what I wrote! – mathreadler Feb 10 '15 at 08:28
  • Honestly, I had no idea how what you wrote had anything to do with the question. Looking at it now, I can start to see it, but it's still confusing. – silvascientist Feb 11 '15 at 02:45
  • So I'm given a book which uses like 3 pages to explain that. I had trouble understanding, as I was losing focus while reading. Came here, understood in 5 seconds. Thank you very much. – student Oct 25 '17 at 00:59