$g\circ f$ is open, $f$ continuous and surjective $\Rightarrow g$ is open Let $(X, \tau_{1}), (Y, \tau_{2}), (Y,\tau_{3})$ be topological spaces. Let $f:(X, \tau_{1}) \rightarrow (Y, \tau_{2}), \ g:(Y, \tau_{2}) \rightarrow (Y,\tau_{3})$ be two functions. Show that:
If $g\circ f$ is open, $f$ continuous and surjective, then $g$ is open.
Let $ X \supset U \in \tau_{1} $. Then $ g \circ f = g(f(U)) \in \tau_{3}$ because $ g \circ f $ is open. So, $ f^{-1}(g(f(u)) \in \tau_{3} $ because $ f $ is continuous. Here I get stuck when I try to use the fact that $ f $ is surjective.
 A: We want to show that $g(U)$ is open whenever $U$ is.  
First, since $U$ is open, so is $f^{-1}(U)$, since $f$ is continuous.
Now, since $g \circ f$ is open, $g \circ f(f^{-1}(U))$ is open.
We claim that $g \circ f \circ f^{-1}(U) = g(U)$. This will prove from above that $g(U)$ is open, hence $g$ is open.
To  go one way, if $x \in g(U)$, then there exists $y\in U$ so  that $g(y) = x$. Note that $f$ is surjective, hence there exists some $z \in f^{-1}U$ so that $f(z) = y$. Hence, $y = f(z) \in f(f^{-1}(U))$. Note that surjective is used very crucially : if $f$ is not surjective, then $f^{-1}(y)$  could have been empty, so $f(f^{-1}(U))$ would not have contained $y$.
Now, $x = g(y) \in g \circ f \circ f^{-1}(U)$, so one containment is clear.
On the other hand, if $x \in g \circ f \circ f^{-1}(U)$, then there is some $y \in U$ so that $x = g \circ f \circ f^{-1}(y)$. Of course, this simplifies  to $x = g(y)$, so $x \in g(U)$. Hence, $g(U) = g \circ f \circ f^{-1} (U)$ and the conclusion follows.
