An other definition of continuity I know that $$f: E\rightarrow F~\text{is continuous}~\Longleftrightarrow \forall V ~\text{open in}~F, f^{-1}(V)~\text{is open in}~ E$$
How to prove that $$f: E\rightarrow F~\text{is continuous}~\Longleftrightarrow \forall V ~\text{is an open of the basis of }~F, f^{-1}(V)~\text{is open in}~ E$$
I think that $\Rightarrow$ commes naturly because if it is satisfyed for all open then it is satisfied for open from the basis no ?
But how to fined the inverse ?
Thank you 
 A: HINT: If $V$ is an open set, write it as a union of basic open sets.
A: Assume that RHS holds (on the bottom). Now we must take an open set $V$ in $F$, and show that $f^{-1}(V)$ is open in $E$. We will do this by the following characterization of open sets: $$U \text{ open in } E \Leftrightarrow \forall \ x\in U, \exists B\text{ open  such that } x\in B\subset U$$
So, let $x\in f^{-1}(V)$. This means that $f(x)= y\in V$ and there exists a basis element $B_y$ such that: $$y\in B_y\subset V$$
Take preimages to get: $$x\in f^{-1}(B_y)\subset f^{-1}(V)$$
Use RHS and the characterization mentioned to conclude.
Edit: Since you have doubts about the characterization I use, I will include the proof. Let $(X,\mathcal T)$ be a topological space. We want to prove: $$\mathcal U\in\mathcal T \iff\forall \ x\in\mathcal U, \ \exists \ \mathcal N\in\mathcal T \text{ such that } x\in\mathcal N\in\mathcal T$$
Proof of $(\Rightarrow)$: $$\text{Let } \mathcal U \in \mathcal T. \text{The desired } \mathcal N \text{can obviously be taken as }\mathcal U \text{ for all } x\in \mathcal U.$$
Proof of $(\Leftarrow)$: $$\text{Let $\mathcal U\subset X$ with the RHS property. Then, for each $x\in\mathcal U$ let $\mathcal N_x$ be such that $x\in\mathcal N_x\in\mathcal T$. } \\ \text{Then we have $\mathcal U = \bigcup_{x\in X} \mathcal N_x$, which is open by definition of $\mathcal N_x$ and the axioms of $\mathcal T$.}$$
