prove/disprove if $g(x) $surjective and $g(f(x)) $surjective then$ f(x)$ surjective prove/disprove if $g(x)$ surjective and $g(f(x))$ surjective then $f(x)$ surjective
I'm having a difficulty approabhing this. Thanks in advance.
 A: Suppose that $g:\mathbb{R}\to\{0\}$ so that $g(x)=0$ for all $x$. Then if $f:\mathbb{R}\to\mathbb{R}$, say $f(x)=0$ for all $x$, $f$ clearly isn't surjective, but $g$ and $g\circ f$ are surjective by definition.
Supposing that $g$ isn't a trivial map, suppose $g:[-1,1]\to[0,1]$ defined by $g(x)=|x|$. Define $f:[0,1]\to[-1,1]$ by $f(x)=x$. Again, we satisfy everything, but $f$ isn't surjective.
If we want $f$ to be surjective, we have to have that $g$ is injective. The above example I think is clear enough to explain where we need injectivity.
A: To see that your claim fails, you can take $f(x)=e^x:\mathbb R\to\mathbb R$ and $g(x)=\ln(x):\mathbb R^+\to\mathbb R$ as well. Of course, what @Nameless noted above is definitely right.
A: You have
$A \to^f B \to^g C$
$\forall c \in C, \exists b\in C,g(b)=c$
$\forall c \in C, \exists a\in C,g(f(a))=c$

Now assume you want to prove it's true, you'll try to prove
$\forall b \in B, \exists a\in A,g(a)=b$

Let's take $b \in B$ and try to prove there is an $a\in A$ such as $f(a) = b$
We're in $B$ and can't go directly to $A$ so the only thing we can do is going to $C$
Let $c = g(b)$
Again, what can we do with that $c$ ? Going back to $B$ would be pointless and we're trying to reach $A$ so let's use the sujectivity of $g\circ f$.
We have an $a \in A$ such as $g(f(a))=c$
And then what can we do with that $c$ ? Try to go back to $B$ but can we?
So far we have $g(f(a))=c=g(b)$ Does that imply $b=f(a)$
The answer is no and from that problem you can try to derive a counter-example.
A good idea when searching for counter-examples about sujectivity / injectivity is to try constant function and sets with one element.
In $B$, we need to have $f(a)$ and $b$ two distincts elements. So let's take $A=\{0\}=C$ and see what happens.
$\{0\} \to^f B \to^g \{0\}$
We can see that $g$ is obviously surjective (given that $B$ is not empty).
Same goes for $g\circ f$.
Let's take $B=\mathbb{R}$ (any set with more than one element would do).
Any $b\not=f(0)$ has no $a\in A$ such as $f(a)=b$.

Here's a concrete example:
$\{0\} \to^{x\mapsto x} \mathbb{R} \to^{x\mapsto x} \{0\}$
$\forall c \in \{0\}, c = 0$ so there is an $a\in A$ such as $g(f(a))=0$: $a=0$
$\forall c \in \{0\}, c = 0$ so there is a $b\in B$ such as $g(b)=0$: $b=0$
But if you take $b=1\in B$, you can not find $a\in A$ such as $f(a)=1$ because $\forall a \in A, f(a)=0 \not= 1$
