# Existence and uniqueness of a certain continuous map

Can you help me with this?

Suppose $X$ is a completely regular space, and $Y$ is a compact and Hausdorff space. Let $$A(X) = \{f : X \to [0,1] \text{ continuous}\}$$ and let $i(X) : X\to [0,1]^{A(X)}$ defined as $i(X)=(f(x))$ where $f$ belongs to $A(X)$. Finally, let $B(X)$ be the closure of $\mathrm{Im}(i(X))$ and let $g:X\to Y$ be a continuous map. The aim is to show that there exists a unique continuous map $G:B(X)\to Y$ such that $g(x) = [G \circ i(X)](x)$.

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You can format this with LaTeX and you have all standard notations avaialble. –  ncmathsadist Jul 1 '12 at 12:58
thanks for the editing ! –  thetruth Jul 1 '12 at 13:12

First embed $Y$ in $[0,1]^{A(Y)}$ by the map $$i_Y:Y\to[0,1]^{A(Y)}:y\mapsto\Big\langle f(y):f\in A(Y)\Big\rangle\;,$$ exactly as $X$ is embedded in $[0,1]^{A(X)}$ by the map $i_X$. An arbitrary point of $[0,1]^{A(X)}$ is simply a family of real numbers in the interval $[0,1]$ indexed by the functions in $A(X)$; I’ll denote such a point by $\big\langle r_f:f\in A(X)\big\rangle$. Similarly, an arbitrary point of $[0,1]^{A(Y)}$ is $\big\langle r_f:f\in A(Y)\big\rangle$.

Now define

$$H:[0,1]^{A(X)}\to[0,1]^{A(Y)}:\Big\langle r_f:f\in A(X)\Big\rangle\mapsto\Big\langle r_{f\circ g}:f\in A(Y)\Big\rangle\;;$$

for each $f\in A(Y)$, $f\circ g\in A(X)$, so this makes sense. Moreover, each projection $$\pi_h:[0,1]^{A(Y)}\to[0,1]:\Big\langle r_f:f\in A(Y)\Big\rangle\mapsto r_h$$ is easily seen to be continuous, so $H$ is continuous. You now have the following continuous maps:

$$\begin{array}{c} &X&\underset{g}\longrightarrow&Y\\ &\;\;\;\;\;\downarrow i_X&&\;\;\;\;\downarrow i_Y\\ &[0,1]^{A(X)}&\underset{H}\longrightarrow&[0,1]^{A(Y} \end{array}$$

Consider a point $i_X(x)\in B(X)$: $i_X(x)=\big\langle f(x):f\in A(X)\big\rangle$, so

$$H\big(i_X(x)\big)=\Big\langle(f\circ g)(x):f\in A(Y)\Big\rangle=\Big\langle f\big(g(x)\big):f\in A(Y)\Big\rangle=i_Y\big(g(x)\big)\;.$$

Thus, $H\big[i_X[X]\big]\subseteq i_Y[Y]$. Moreover, $i_X[X]$ is dense in $B(X)$, so $i_Y[Y]=H\big[i_X[X]\big]$ is dense in $H[B(X)]$. But $Y$ is compact, so $i_Y[Y]$ is closed in $[0,1]^{A(Y)}$, and therefore $H[B(X)]\subseteq i_Y[Y]$. That is, $H[B(X)]$ is a subset of the domain of $i_Y^{-1}$, so $G=i_Y^{-1}\circ H:B(X)\to Y$ makes sense and of course is continuous.

Now you have only to verify that $G\circ i_X=g$, which is straightforward.

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