open map equivalent definition $f : (X,\tau_X) \to (Y,\tau_Y) $ continuous and surjective. I need to prove that
$f$ is open  $\Longleftrightarrow \forall U\in\tau_X, f^{-1}(f(U))\in \tau_X$
Proof:
$\implies)$ By definition, $f$ is open if $\forall U\in\tau_X, f(U)\in\tau_Y$. As f is continuous, $f^{-1}(f(U))\in \tau_X$
$\Longleftarrow)$ Now we suppose $\forall U\in\tau_X, f^{-1}(f(U))\in \tau_X$
I know that $U\subseteq f^{-1}(f(U))\in \tau_X$ and the equality holds when f is injective, but I just know that f is surjetive
How can I continue? 
Thank you for your help and time

This isn't true.  Let $X$ denote the real line with the discrete topology, let $Y$ denote the real line with the usual topology, and let $f : X \to Y$ be the identity map.  Clearly $f$ is a continuous surjection, and is not an open mapping.  However $f^{-1} [ f [ U ] ] = U$ for all (open) $U \subseteq X$.
