Prove $\ g \circ f $ is not one to one

Question

Given f is Onto and g is not one to one

Now $\exists b_1, b_2 \in B $ such that $(b_1,c)$ and $(b_2,c) \in g$. Now since f is onto so that

$\exists a_1, a_2 \in A $ such that $(a_1,b_1)$ and $(a_2,b_2) \in f$ and so $(a_1,c)$ and $(a_2,c) \in g \circ f$ . So its not one to one .

Is this correct ? Thanks

Answer 1

The idea is correct, but

• when you wrote $\exists a_1,a_2\in A$, you should have added that $a_1\neq a_2$;
• the final conclusion should be “So it's not one-to-one”, instead of “So its not Onto”.