Stone-Čech compactification of a dense subset of compact Hausdorff space Let $X$ be a dense subset of compact Hausdorff space Y. 
Every continous function $f:X \to [0,1]$ extends to a continous function $\bar{f} : Y\to [0,1]$ ($\bar{f}(x) = f(x) \; for \; x \in X$). I want to 
show that $\bar{k} : \beta X \to Y $ is a homemorphism  where $k:X->Y$ is inclusion and $\beta X$ is the Stone-Čech compactification of $X$.
Here is my trial.
Let $i :X-> \beta X$ is the embedding for Stone-Čech compactification, i.e, 
$$i(x)(f)= f(x), i(x)=eval_x$$
Inclusion $k:X \to Y$ ($k(x)=x$) is continous and Y is given as compact Hausdorff space, thus universal property of Stone-Čech compactification gaurantee
$$\exists \bar{k}: \beta X \to Y \; s.t \; k = \bar{k} \circ i $$
$j : Y \to \beta X$ is given as 
$$j(y) (f) = \bar{f} (y) $$
I want to show that $j$ is the continuous inverse for $\bar{k}$. First,
$$ \pi_f \circ j = \bar{f} $$
is continous for each $f$ Hence $j$ is continous. Moreover
$$j(x) (f) = \bar{f} (x) = f(x) = i(x) (f) $$, $$j(x) = i(x) \; for \; x \in X$$
Hence
$$ j \circ \bar{k} (i(x)) = j( \bar{k} \circ i (x) ) = j( k (x) ) = j(x) = i(x) ,\; for \; x \in X$$
Therfore $ j \circ \bar{k}$ is identity on $i(X)$ 
$$\bar{ i(X)} = \beta X$$
.......
It is quite daunting for me dealing with 'function space' of 'functions' on 'function space'. This is too abstract! 
I will appreciate a lot for any help. 
Thanks.
 A: A different way. 
(1).Theorem. (I give the proof of it at the end.) Any continuous $g:X\to A,$ where $A$ is compact Hausdorff, is extendable to a continuous $\bar g:\beta X\to A .$
(2). Let $h:\beta X\to Y$  be a continuous surjection with $h|_X=id|_X$ and $h(\beta X$ \ $X)=Y$ \ $X.$ 
Suppose $p_1,p_2 \in \beta X$ \ $X$ with $p_1\ne p_2.$ By Urysohnn's Lemma let  $\bar f:{\beta X}\to [0,1]$ be continuous with $\bar f(p_1)\ne \bar f(p_2).$  Let $f$ be the restriction of $\bar f$ to the domain $X.$ Let $f_Y:Y\to [0,1]$ be a continuous extension of $f$. 
Consider the composite function $F=f_Y\cdot h:\beta X\to [0,1]$ and the function $G=\bar f:\beta X\to [0,1].$ Since $F$ and $G$ are continuous and agree on $X,$ which is a dense subset of $\beta X,$ and since $[0,1]$ is Hausdorff, we have $F=G.$
Therefore $h(p_1)\ne h(p_2)$. (Otherwise $F(p_1)=F(p_2)$ and $G(p_1)\ne G(p_2)$.) 
Therefore $h:\beta X\to Y$ is a bijection. A continuous bijection from one compact Hausdorff space to another is a homeomorphism.  So $h^{-1}:Y\to \beta X$ is also a homeomorphism . QED.
Proof of Theorem: Let $B$ be the closure of $g(X)$ in $A.$ Let $C$ be the closure of $\{(x,g(x)):x\in X\}$ in $\beta X\times B.$ By the Diagonal Theorem, the function $e(x)=(x,g(x))$ is a homeomorphic embedding of $X$ into $C.$ And $e(X) $ is dense in $C,$ so $e:X\to C$ is a compactification of $X.$ 
So let $j:\beta X\to C$ be a continuous surjection with $j(x)=e(x)=(x,g(x))$ for $x\in X,$ and $j(\beta X \setminus X)=C$ \ $e(X).$  The projection $p_2:C\to A$ where $p_2((u,v))=v,$ is continuous, so the composite function $p_2\cdot j:\beta X\to A$ is continuous. And of course for $x\in X$ we have $p_2(j(x))=p_2(\;(x,g(x))\;)=g(x).$ 
This clever proof of the Theorem is  in General Topology, by R. Engelking (Chapter 3, Section 3.5) . His thorough cross-referencing tells you in the proof where to find the Diagonal Theorem in the book.
