Topological semiconjugacy preserves topological transitivity Let $(X, f )$, where $X$ is a compact metric space and $f : X → X $ is a continuous function and let $(Y, g)$  where $Y$ is a compact metric space and $g : Y → Y$ is a continuous function. Suppose they are topologically semiconjugate, i.e., there is a continuous surjection $h : X → Y$ such that $h ◦ f = g ◦ h$, then show that the transitivity of $f$ implies the transitivity of $g$.
Definition of topological transitivity: Let $X$ be a metric space and $f : X → X$ continuous. $f$ is said to be topologically transitive if for every pair of nonempty open sets $U$ and $V$ in $X$, there is a positive integer $n$ such that $f^n(U) ∩ V ≠ ∅$.
I really don't know how to start this demonstration, a little help would be much appreciated. Thank you in advance for your answers! 

Let $U,V$ be non empty open subsets of $Y$. Since $h$ is surjective, $h^{-1}(U)$ and $h^{-1}(V)$ are not empty. Since $f$ is topological transitive, there exists $n$ such that $f^n(h^{-1}(U))\cap h^{-1}(V)$ is not empty. Let $x\in (h^{-1}(U))$ such that $f^{n}(x)\in h^{-1}(V)$, $h(x)=y\in U$ and $g^n(y)=g^n(h(x))=h(f^n(x))$, since $f^n(x)\in h^{-1}(V)$, we deduce that $g^n(y)=h(f^n(x))\in V$.
