For homeomorphism $f:X\to Y$, show that $f^{-1}(\mathcal{B}_{Y})=\mathcal{B}_{X}$ 
My question is that if a function $f: X\to Y$ is homeomorphism, show that $f^{-1}(\mathcal{B}_{Y})=\mathcal{B}_{X}$.
  where $f^{-1}(\mathcal{B}_{Y}):=\{f^{-1}(A): A\in \mathcal{B}_{Y}\}$.

I try to use the continuity of $f$ but I am stuck get the relation between $f^{-1}(\mathcal{B}_{Y})\subset \mathcal{B}_{X}$ or $ \mathcal{B}_{X}\subset f^{-1}(\mathcal{B}_{Y})$. Actually, I konw that $f^{-1}(\mathcal{B}_{Y})\neq \mathcal{B}_{X}.$ 

Prove that $\{f^{-1}(B): B \in \mathcal{B}_Y\}$ is a sigma algebra on $X$ and contains the open sets. 
Every open set $A$ in $X$  has the form $A=f^{-1}(B)$ where $B$ an open subest of $Y$  since $f$ is a homeomorphism, so we have $\mathcal{B}_X \subset f^{-1}(\mathcal{B}_Y)$ because $\mathcal{B}_X$ is the smallest sigma algebra that contains the open subsets of $X$
