Homeomorphic function in a product topology. Let $\{(X_\alpha,\mathscr{T}_\alpha):\alpha\in\Lambda\}$ be a collection of topological spaces, and let $\mathscr{T}$ be the product topology on $X=\prod_{\alpha\in\Lambda}X_\alpha$. Let $p\in X$, let $\beta\in\Lambda$, and let $H_{p\beta}=\{x\in X;\mbox{if }\alpha\neq\beta\mbox{, then }x_\alpha=p_\alpha\}$. Define the function $f:X_\beta\to H_{p\beta}$ as follows: for each $x_\beta\in X_\beta$ let $f(x_\beta)$ be the member of $H_{p\beta}$ defined by $[f(x_\beta)]_\beta=x_\beta$ and for $\alpha\neq\beta,[f(x_\beta)]_\alpha=p_\alpha$. Then $f$ is a homeomorphism.


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*I would like see a proof of this theorem please.

 A: Just work through the definitions.  The product topology has a base given by the sets $\prod_{\alpha \in \Lambda} U_\alpha$, where each $U_\alpha$ is open in $X_\alpha$ and for all but finitely many $\alpha$, $U_\alpha = X_\alpha$.  Since $H_{p\beta}$ is a subspace of $X$, it has a base given by the intersections of these basic open sets with $H_{p\beta}$.  Given such a set, its inverse image along $f$ is open. it's just the factor $U_\beta$ of the original basic open set.  This shows $f$ is continuous.
Conversely, given an open set $U$ in $X_\beta$, its image along $f$ is $H_{p\beta} \cap \prod_\alpha U_\alpha$, where $U_\alpha = X_\alpha$ for $\alpha \ne \beta$ and $U$ for $\alpha = \beta$.  This shows $f$ is open.  $f$ is clearly injective, and it should also be clear that its image is $H_\beta$, completing the proof.
To get an intuitive understanding for what's going on here, try visualizing this when the product is finite.  You're just including one factor into the product by fixing all the other coordinates at some constant value.  For example, if $\Lambda = \{1,2\}$, $\beta = 1$, $X_1 = X_2 = \mathbb{R}$, and $p = (0,0)$, this is the inclusion of the $x$-axis into the plane.
