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.


  • $\begingroup$ You are working with an explicit construction of $\beta X$, I think it would be good to mention that explicitly. Then, what problem remains? You have the extension $j \colon Y \to \beta X$ of the inclusion $i \colon X \to \beta X$. You know it is continuous. You know that $(\overline{k}\circ j)\lvert_X = \operatorname{id}_X$. You also know that $(j\circ \overline{k})_{i(X)} = \operatorname{id}_{i(X)}$. What is missing to conclude $j\circ \overline{k} = \operatorname{id}_{\beta X}$ and $\overline{k}\circ j = \operatorname{id}_Y$? $\endgroup$ Commented May 18, 2015 at 13:01
  • $\begingroup$ What I am not clear is.. if we have a continuous function $f : B->B $ which is identity on a dense subset $A$ of $B$, does this imply $f$ is identity on the whole $B$ space? It seems plausible but I've not heard about this ever, nor have any feeling about the truth of this statement.. $\endgroup$
    – Guldam
    Commented May 18, 2015 at 14:42
  • $\begingroup$ If $B$ is Hausdorff, it follows. Slightly more generally, if $H$ is a Hausdorff space, and $T$ an arbitrary topological space, for any two continuous $f,g \colon T \to H$, the set $\{ x \in T : f(x) = g(x)\}$ is closed - since we can write it as $(f\times g)^{-1}(\Delta_H)$, where $\Delta_H = \{ (h,h) : h\in H\}$ is the diagonal of $H\times H$. A topological space $S$ is Hausdorff if and only if the diagonal in $S\times S$ is closed. So in your situation, the set $\{ x\in B : f(x) = x\}$ is closed, since $B$ is Hausdorff, and it's dense by assumption. Hence it's $B$. $\endgroup$ Commented May 18, 2015 at 14:48
  • $\begingroup$ That is exactly what I have missed. Things become clear! Thanks! $\endgroup$
    – Guldam
    Commented May 18, 2015 at 15:07

1 Answer 1


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.


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