I am trying to prove that the Stone Cech Compactification map is a homeomorphism. I have most the proof finished, but I am stuck on showing that the inverse function is continuous. Here is what I have so far.
Let $X$ be a $T_{3 \frac{1}{2}}$ space. "completely regular" and let $C(X)$ be the space of all bounded continuous functions on $X$. Then for $f_\alpha \in C(X)$ indexed by some index set $\Lambda$, choose $$ I_\alpha = [\inf f_\alpha(x), \sup f_\alpha(x)]. $$ Define $\displaystyle h : X \rightarrow \prod_{\alpha \in \Lambda} I_\alpha$ by $$ h(x) = (f_\alpha(x))_{\alpha \in \Lambda} $$
Then $\overline{h(x)}$ is a compactification of $X$, called the Stone Cech compactification of $X$. Commonly denoted $\beta(X)$.
We can first note that Tychonoff's theorem implies that $\prod I_\alpha$ is compact. So, $\overline{h(x)}$ will be compact as well. We need to show that $h$ provides us with a homeomorphism. In order to do so we show three things. $h$ is continuous, $h$ is injective, and $h^{-1}$ is continuous with respect to its image.
We start with continuity, consider the projection mapping $p_\alpha$ and the composition $p_\alpha \circ h(x): X \rightarrow I_\alpha$, and observe that $p_\alpha \circ h(x) = f_\alpha(x)$. Since $f_\alpha$ is continuous by assumption $h$ is also continuous.
To show that $h$ is injective, we note that since $X$ is $T_{3 \frac{1}{2}}$ that there exists a continuous function $f : X \rightarrow [0,1]$ such that $f(x_1) = 0$ and $f(x_2)=1$. Which implies that $h(x_1)$ and $h(x_2)$ must be different since their values at $f$ namely $h(x_1) = f(x_1) = 0$ and $h(x_2) = f(x_2) = 1$, are different. We conclude $h$ is injective.
All that remains is to show that $h^{-1}$ is continuous with respect to its image. I am pretty sure I need to take an arbitrary open set $V = h^{-1}(U) \subset X$, and show that $U = \prod O_\alpha$ is open, but not really sure how to make this connection. Any help is appreciated.
Since you already know that $h$ is a continuous bijection, you need only show that $h$ is an open map, i.e., that $h[U]$ is open in $h[X]$ for each open $U\subseteq X$. Let $U$ be a non-empty open set in $X$; to show that $h[U]$ is open in $h[X]$, it suffices to show that for each $h(x)\in h[U]$ there is an open $V$ in $h[X]$ such that $h(x)\in V\subseteq h[U]$. HINT: There is a continuous $f:X\to\Bbb R$ such that $f(x)=0$, and $f(y)=1$ for $y\in X\setminus U$.