# Surjective Functions [duplicate]

Possible Duplicate:
Surjectivity of Function Compositions

I am trying to prove this statement:

If $f: A \rightarrow B$ and $g: B \rightarrow C$ are both surjective functions, show that $g \circ f : A \rightarrow C$ is also a surjective function.

I know that some elements B have corresponding elements in A and likewise, some elements in C have corresponding elements in B. Is it enough to say that some elements in C must therefore correspond in A, which is what surjective function, $g \circ f : A \rightarrow C$, would show?

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Start by looking up the definition of "surjective". – Chris Eagle Mar 15 '12 at 20:50
yes, it is an exact duplicate. Sorry I did not see the previous question. feel free to close this or whatever needs to be done – Dominick Gerard Mar 15 '12 at 21:06
Is there a reason why you have a 0% accept rate? – dtldarek Mar 15 '12 at 21:24
@dtldarek sorry I am still new to this website, how can I change that? – Dominick Gerard Mar 15 '12 at 21:26
@DominickGerard I guess you already know ;-) Great! – dtldarek Mar 15 '12 at 21:34

## marked as duplicate by Asaf Karagila, Chris Eagle, Matt N., anon, Benjamin Lim Mar 15 '12 at 21:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 2 Answers

$g$ being surjective means that EVERY element of $C$ is mapped to by (at least) one element of $B$ by $g$, and similarly for $f$. It should now be fairly obvious that if both $g$ and $f$ are surjective then their composition must also be surjective.

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Hint:

Consider three sets of people $A$, $B$, $C$. Assume that for every person $c_1 \in C$ there exists a person $b_1 \in B$ that knows $c_1$, and that for every person $b_2 \in B$ ($b_1 = b_2$ might or might not be true) there exists $a_1 \in A$ that knows $b_2$. Can you conclude that for every $c_2 \in C$ there exists $a_2 \in A$ that (indirectly) knows $c_2$?

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