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i'm working on this question here but don't exactly know how I am supposed to answer this.

"Let f and g be two functions. Prove that if f and g are both surjective, f ◦ g is also surjective."

Am I supposed to assume $f: A ⇒ B$ and $g: A ⇒ B$?

From here I don't know how to proceed.

M.G
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  • If $f$ and $g$ are both $A \rightarrow B$, then $f ◦ g$ makes no sense. – Paul Jun 13 '16 at 16:01
  • Choose a x in the co-domain of f ◦ g. If f is surjective, it maps to a point in the domain of f, which is also in the co-domain of g. – Doug M Jun 13 '16 at 16:33

1 Answers1

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Hint: In order for $f\circ g$ to make sense, you must have the image of $g$ contained in the domain of $f$. So you should assume $g:A\longrightarrow B$ and $f:B\longrightarrow C$, so that $f\circ g : A\longrightarrow C$.

This is sometimes written $A \stackrel{g}{\longrightarrow}B\stackrel{f}{\longrightarrow}C$.

MPW
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  • Did you mean to write $f: A ⟶ B$ and $g: B ⟶ C$, so that $f o g: A ⟶ C$? – M.G Jun 13 '16 at 20:32
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    No, what I wrote was correct. For $f\circ g$, you evaluate $g$ first, then $f$ (recall "$f\circ g$" is read "$f$ following $g$"). The composition you mention in the comment above would be the map $g\circ f: A\to C$. IN other words, $(f\circ g)(x) = f(g(x))$, and $(g\circ f)(x)=g(f(x))$. – MPW Jun 13 '16 at 20:49
  • ok thanks for the clarification. – M.G Jun 13 '16 at 20:54