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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 ?

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    Your proof is flawed. You have not proven "$f$ is onto"; rather, you have proven "if $f$ is onto, then $f$ is onto" which isn't very interesting. – 2017-01-13

3 Answers 3

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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…

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$f$ 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?

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    @Jon Garrick: is it clear? – 2017-01-21
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$g\circ f$ 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.