Let be $\mathfrak{X}=\{(X_j,\mathcal{T}_j):j\in J\}$ a collection of topological not empty spaces and we consider the product space $\Pi_{j\in J}X_j$. Well using the Choice Axiom for any $i\in J$ we can define for some fixed $z\in\Pi_{j\in J}X_j$ the set
$$
Z_i=\{x\in\Pi_{j\in J}: x(j)=z(j), j\neq i \land x(j)=x_h\in X_i, j=i\}_{h\in|X_i|}
$$
and we prove that it is homeomorphic to $X_i$. So we consider the restriction $\pi_i|_{Z_i}$ of the projection $\pi_i$ and we observe that by a previous theorem it is continuous on the subspace topology $\mathcal{T}_Z$ of $Z$; moreover since two elements $x$ and $y$ of $Z_i$ differ only for their values $x(i)$ and $y(i)$ it result that $\pi_i|_{Z_i}$ is bijective and so it is that
$$
\forall A\in\mathcal{T}:\pi_i|_{Z_i}(A\cap Z_i)=\pi_i|_{Z_i}(A)\cap\pi_i|_{Z_i}(Z_i)=\pi_i(A)\cap X_i=\pi_i(A)\in\mathcal{T}_i
$$
from which we can colude that $\pi_i|_{Z_i}$ is open and so it is a homeomorphism between $Y_i$ and $X_i$.
Well I doubt that the prove of the "openness" of $\pi_i|_{Z_i}$ is uncorrect, since it would result that $\pi_i|_{Z_i}(A\cap Z_i)=\pi_i|_{Z_i}(A)\cap\pi_i|_{Z_i}(Z_i)\neq\pi_i(A)\cap X_i$; but anyway I am sure that the proof is generally okay, since the same procedure is indicated here by the professor Brian M. Scott. So could someone help me, please?
Best Answer
I just gave the full argument in the first part of this proof where I think you got the question as a spin-off, as it were.
You don't need to define $Z_j$, just define an embedding $e$ and show its continuous, 1-1 and has a continuous inverse. Openness of $e$ follows (between $X_j$ and its image) automatically.