Showing that the Stone Cech compatificacion is equivalent to Alexandroff under continuous maps. Let X and Y be Tychonoff spaces and $f:X\rightarrow Y $ a continuous and surjective function. Prove that if X is locally compact and $\alpha X\equiv_{X} \beta X$ then Y is locally compact and $\alpha Y\equiv_{Y} \beta Y$.
I have already proved that Y is locally compact by showing that f is perfect but I got stuck when I tried to prove than $\alpha Y \geq \beta Y$. I was trying by proving that if I have $Z_1,Z_2\subset \beta Y$ disjoint closed subsets then $cl_{\alpha Y}(Y\cap Z_1)$ is disjoint to $cl_{\alpha Y}(Y\cap Z_2)$ and the only thing that I have already proved is that if there exists one point in the intersection then that point is the extra point of the alexandroff compactification.
Every suggestion is deeply appreciated

Recall that for a Tychonoff space $X$ and a compactification $\gamma X$ of $X$, if every pair of completely separated closed subsets of $X$ have disjoint closures in $\gamma X$, then $\gamma X$ is the Stone-Čech compactification of $X$.

Let $\alpha X = X \cup \{ \star \}$ and $\alpha Y = Y \cup \{ \star \}$ be the one-point compactifications of $X$, $Y$, respectively.
Recall that by the definition of the one-point compactification for a closed subset $E \subseteq X$ we have that
$$
\operatorname{cl}_{\alpha X} ( E ) = \begin{cases}
E, &\text{if $E$ is compact} \\
E \cup \{ \star \}, &\text{if $E$ is not compact.}
\end{cases}
$$
Let $E, F \subseteq Y$ be completely separated closed subsets of $Y$. Then there is a continuous function $h : Y \to [0,1]$ such that $E \subseteq h^{-1} \{ 0 \}$ and $F \subseteq h^{-1} \{ 1 \}$. Note that $hf : X \to [0,1]$ is continuous, and $f^{-1} [ E ] \subseteq (hf)^{-1} \{ 0 \}$ and $f^{-1} [ F ] \subseteq (hf)^{-1} \{ 1 \}$, and so $f^{-1} [ E ]$ and $f^{-1} [ F ]$ are completely separated closed subsets of $X$. Since $\alpha X \equiv \beta X$ it follows that $\operatorname{cl}_{\alpha X} ( f^{-1} [ F ] ) \cap \operatorname{cl}_{\alpha X} ( f^{-1} [ E ] ) = \emptyset$, and so in particular either $\star \notin \operatorname{cl}_{\alpha X} ( f^{-1} [ F ] )$ or $\star \notin \operatorname{cl}_{\alpha X} ( f^{-1} [ E ] )$. 
Without loss of generality, assume that $\star \notin \operatorname{cl}_{\alpha X} ( f^{-1} [ F ] )$. Note that by the above observation it follows that $f^{-1} [F]$ is compact, and so by continuity and surjectivity of $f$ it follows that $F$ is a compact subset of $Y$. In particular $\star \notin \operatorname{cl}_{\alpha Y} ( F )$, and since $F , E$ are disjoint closed subsets of $Y$ we have that $\operatorname{cl}_{\alpha Y} ( F ) \cap \operatorname{cl}_{\alpha Y} ( E ) = \emptyset$.
