# 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.

• I would like see a proof of this theorem please.
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General suggestion: when things get as abstract as $\alpha\in \Lambda$, try writing $i\in \{1,2\}$ instead. The problem becomes: show that the map $f:X_1\to X_1\times X_2$ defined by $f(x)=(x,p)$ with fixed $p$ is a homeomorphism. – user53153 Dec 25 '12 at 7:08

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.
Hi, thanks for your answer, why the inverse image of the bass elements of $H_{p\beta}$ along $f$ is open? – Fernando Valle Dec 24 '12 at 20:02
Like I said, a basis element of $H_{p\beta}$ is of the form $V = H_{p\beta} \cap (\prod U_\alpha)$, and then $f^{-1}(V) = U_\beta$, which is open in $X_\beta$ by definition. What about this don't you understand? – Paul VanKoughnett Dec 24 '12 at 20:37
I dont understand why $f^{-1}(V)=U_\beta$ – Fernando Valle Dec 24 '12 at 20:49
Ah, OK, I messed up. If each $U_\alpha$ contains $p_\alpha$, then $f(x)$ (which is the point whose $\beta$th coordinate is $x$ and whose $\alpha$th coordinate is $p_\alpha$ for $x \ne \beta$) is in $V$ iff $x \in U_\beta$. If some $U_\alpha$ does not contain $p_\alpha$, then $f(x)$ is never in $V$, since the points in the image of $f$ always have $\alpha$th coordinate $p_\alpha$. So in this case, $f^{-1}(V) = \emptyset$, which is open. – Paul VanKoughnett Dec 25 '12 at 1:27