Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 :)

share|cite|improve this question
up vote 3 down vote accepted

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....

share|cite|improve this answer
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. – Code-Guru Aug 2 '12 at 0:47

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.

share|cite|improve this answer

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 $c=g(b)$?

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.