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Suppose that $f:A \longrightarrow B$ and $g:B \longrightarrow C$ are functions.

If $g \circ f$ is onto and $g$ is one-to-one, then prove that $f$ is onto.

How do I go about proving this?

From $g \circ f$ is onto, I know that there exists an $a \in A$ such that $g(f(a))=c$ and I also know that if $g(b_1)=g(b_2)$ then $b_1=b_2$ from the definition of one-to-one, but I'm not sure where to go from there. Any help would be great :)

4 Answers 4

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Take $b$ in $B$. Let $c=g(b)$. Since $g\circ f$ is onto, you know.... Since $g$ is one-one, you can conclude from $(g\circ f)(a)=c$ and $g(b)=c$ that....

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    Note the general strategy that @Gerry used here: He first notes that he wants to prove that a function is *onto*. In otherwords, you should first look at what you want to prove, not what you are given as premeses. From there, look at the definition of *onto* and pay special attention to the quantifiers as these tell you how to begin your proof.2012-08-02
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Suppose $f$ is not onto. Then there exists a $b \in B$ such that $f(a) \neq b$ for all $a \in A$. Since $(g \circ f)$ is onto, there exists an $a$ such that $(g \circ f)(a) = g(b)$. By the injectivity of $g$, one has that $f(a) = b$. Contradiction.

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You want to show that $f$ is onto, i.e. that for all $b\in B$ there is some $a\in A$ such that $f(b)=a$. So to start fix some $b\in B$. You know that $g$ is one-to-one, so if $g(f(a))=g(b)$ then $f(a)=b$. And you know that $g\circ f$ is onto, so for all $c\in C$ there is some $a\in A$ such that $g(f(a))=c$. In particular, what happens if you let $c=g(b)$?

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Suppose $f$ is not onto. Then there is $x \in B$ such that $x \notin f(A)$, but $g(x) \in C$, so there exists $y \in A$ such that $g(f(y)) = g(x)$, but $f(y) \neq x$, contradicting our assumption that $g$ is one-to-one.