Assume $f:A→B$ and $g:B→C$ be the two functions? Assume $f:A→B$ and $g:B→C$ be the two functions ?
Then, if $g∘f:A→C$ is onto and $g:B→C$ is one-one ? What can be said about $f$ ?

My try :
Let me suppose $f$ as onto so, I can say $f(x) = y$ where $x∈A$ and $y∈B$.
Its also given that $g$ is onto, then $g(y)∈C$.
$g∘f:A→C$ is onto, then $g∘f(x) = g(y)$ => $f(x) = y$
Hence, what I assumed was right $f$ as onto.

Have I got it right ? Is there anything extra that can be implied from above ?
 A: If you assume $f$ is onto, then you can certainly prove that it is onto, but you have proved nothing about $f$. Besides, you have not used the important hypothesis that $g$ is one-to-one.

Since $g\circ f$ is onto, then $g$ is onto (prove it). Since, by assumption, $g$ is one-to-one, you have that $g$ is bijective. Therefore
$$
f=g^{-1}\circ (g\circ f)
$$
is the composition of two onto functions. Hence…
A: $f$ must be onto:
Suppose it is not. Then there is $y_0 \in B$ such that there is no correspondence in $A$ by $f$.
If we  apply $g$ at $y_0$ we get $g(y_0)$ but $g(f(x))$ is onto, so there is $x_0 \in A$ such that 
$$g(f(x_0))=g(y_0)$$
But $g$ is one-one then 
$$g(f(x_0))=g(y_0) \rightarrow f(x_0)=y_0$$ 
which is a contradiction.
It is easy to find out that $f$ is not necessarily one-one. Can you finish?
A: $g\circ f$ is onto, so $g$ is onto too.
But $g$ is also supposed to be one-to-one; so $g$ is a bijection ...
We can now use the reciprocal map $g^{-1}$ which is also a bijection and -- in particular -- onto.
Finally, you can write $f=g^{-1}\circ(g\circ f)$, so that $f$ is onto (as a composition of two onto maps).
Hope this help.
