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
 A: 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 $\operatorname{cl}_{\alpha Y} ( F ) \cap \operatorname{cl}_{\alpha Y} ( E ) = \emptyset$.
