# Proof verification: If $gf$ is surjective and $g$ is injective, then $f$ is surjective

Proof. Let $$f:A\rightarrow B$$ and $$g:B\rightarrow C$$. Therefore, $$gf:A\rightarrow C$$.

According to definition, for all $$c\in C$$ there exists $$a\in A$$ such that $$g(f(a))=c$$.

Because $$gf$$ is surjective, two different values of $$a$$ can map the same value, hence assuming that $$g(f(a_1))=g(f(a_2))$$.

We know that $$g$$ is injective, therefore we get:

$$g(f(a_1))=g(f(a_2)) \overset{def}{\Longrightarrow} f(a_1)=f(a_2)$$ and thus $$f$$ is surjective. QED.

Is the proof correct? I have a strong sensation that I am missing something crucial...

• Your proof is not clear : for example, I don't understand how do you derive your conclusion about the surjectivity of $f$. It seems you make a confusion between surjectivity and non-injectivity ?? – TheSilverDoe Feb 24 at 17:56

You did not even try to prove that $$f$$ is surjective. That means that, for each $$b\in B$$, there is a $$a\in A$$ such that $$f(a)=b$$. Where did you prove that?

If $$b\in B$$, then, since $$g\circ f$$ is surjective, there is a $$a\in A$$ such that $$g\bigl(f(a)\bigr)=g(b)$$. But then, since $$g$$ is injective, $$f(a)=b$$.

• Okey, I understand it now (about time...) Thank you for your reply. – user615771 Feb 24 at 21:20
• I'm glad I could help. – José Carlos Santos Feb 24 at 22:01