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Statement

For any collection $\mathfrak{X}:=\{X_i:i\in I\}$ if $J\subseteq I$ then $X_J:=\Pi_{j\in J}X_j$ is embeddable in $X_I:=\Pi_{i\in I}X_i$

To prove the statement I tried to proceed as follows.

For a fixed $\xi\in X_I$ we define the function $f:X_J\rightarrow X_I$ through the condiction $$ [f(x)](i):=\begin{cases}x(i),\,\,\,\text{if}\,\,\,i\in J\\\xi(i),\,\,\,\text{otherwise}\end{cases} $$ for any $x\in X_J$. So we observe that if $f(x)=f(y)$ for any $x,y\in X_J$ then $x(j)=[f(x)](j)=[f(y)](j)=y(j)$ for any $j\in J$ and so $x=y$ thus $f$ is injective. Then we observe that $(\pi_i\circ f)$ is equal or to $\pi_i$ if $i\in J$ or to a constant function $\xi_i$ each of which is continuous so that by universal mapping theorem for products we conclude that $f$ is continuous too.

Now unfortunately I can't prove that the function $f^{-1}$ is continuous that is the function $f$ is open so I ask to do it. Then I ask to prove that $f[X_J]=\Pi_{i\in I}Y_i$ where $Y_i=X_i$ if $i\in J$ and otherwise $Y_i=\{\xi_i\}$. Finally I ask if using the preceding result is possible to prove that if $X_I:=\Bbb R^{i-1}\times\Bbb R\times\Bbb R^{n-i}$ and $X_J:=\Bbb R^{i-1}\times\Bbb R^{n-i}$ then for any open $U$ there exist an open set $U'$ in $\Bbb R^{i-1}$ and an open set $U''$ in $\Bbb R^{n-i}$ such that $U\cap f[X_J]=U'\times\{\xi_i\}\times U''$. So could someone help me, please?

Antonio Maria Di Mauro
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1 Answers1

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It is clear that $f[X_J] = \prod_{i \in I} Y_j$ where $Y_j$ is as you specified: Let $y \in f(x)$ for $x \in X_J$ so that $y(i)=f(x)(i)=\xi(i)$ for $i \in I\setminus J$ is by definition, and so the left to right inclusion is clear. And if $y \in \prod_{i \in I} Y_j$, $\pi'_I(y)$ is the (unique) preimage of $y$ under $f$.

A basic open set in $X_J$ is of the form $U=\bigcap_{j \in F} \pi_j^{-1}[O_j]$ where $F \subseteq J$ is finite and $O_j \subseteq X_j$ is open for all $j \in F$. Then $f[U]= f[\bigcap_{j \in F} \pi_j^{-1}[O_j]]= \bigcap_{j \in F} f[\pi_j^{-1}[O_j]]$ (as $f$ is 1-1 it preserves intersections in forward images) and $f[\pi_j^{-1}[O_j]] = {\pi'_j}^{-1}[O_j] \cap f[X_I]$, where $\pi'_j$ are the projections on $X_I$. So that intersection is also relatively open, showing that $f: X_I \to f[X_I]$ is open.

Henno Brandsma
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  • Okay, finally using the results that $\Bbb R^{i-1}\times{\xi_i}\times\Bbb R^{n-i}$ is homeomorphic to $\Bbb R^{i-1}\times\Bbb R^{n-i}$ is it possible to prove that $U\cap X_i=U'\times{\xi}\times U''$ as I stated? – Antonio Maria Di Mauro Aug 09 '20 at 10:26
  • Yes, but unfortunately I don't be able to prove this. So could you help me, please? – Antonio Maria Di Mauro Aug 09 '20 at 10:29
  • I discovered the above embedding just to prove this. – Antonio Maria Di Mauro Aug 09 '20 at 10:31
  • @AntonioMariaDiMauro it’s false as stated because e.g. not every open set in the plane is a product of two open sets in the axes. – Henno Brandsma Aug 09 '20 at 10:37
  • I think I made a mistake. Actually I mean that if $X_I:=\Bbb R^{i-1}\times\Bbb{R}\times\Bbb R^{n-i}$ and $X_J:=\Bbb R^{i-1}\times\Bbb R^{n-i}$ then $U\cap f[X_J]=U'\times{\xi_i}\times U''$ where $U'$ and $U''$ are open respectively in $\Bbb R^{i-1}$ and in $\Bbb R^{n-i}$. – Antonio Maria Di Mauro Aug 09 '20 at 10:47
  • I claim this because if $X_I:=\Bbb R^{n-1}\times\Bbb R$ and $X_J:=\Bbb R^{n-1}$ then the function $f^{-1}$ argree with the projection $\tilde{\pi}$ ${\Bbb R^{n-1}}$ of $f[X_J]=\Bbb R^{n-1}\times{\xi}$ to $\Bbb R^{n-1}\times{\xi}$ and so if $U\xi:=\pi_{\Bbb R^{n-1}}\big[U\cap f[X_j]\big]$ then $U\cap f[X_J]=\big(\pi_{\Bbb R^{n-1}}\big)^{-1}[U_\xi]=U_\xi\times{\xi}$, is this false? – Antonio Maria Di Mauro Aug 09 '20 at 10:59
  • And this last thing is stated by Munkres. – Antonio Maria Di Mauro Aug 09 '20 at 11:00
  • So what can you say about? – Antonio Maria Di Mauro Aug 09 '20 at 11:03
  • @AntonioMariaDiMauro the last case where $X_I = \Bbb R^{n-1}$, $X_J= \Bbb R^n$, $f[X_I]= \Bbb R^{n-1} \times {\xi}$ then for every open $U \subset \Bbb R^n$ we have that $U':=\pi_J[U]$ is open in $\Bbb R^{n-1}$ and $U' \times {\xi} = U \cap f[X_I]$. That works fine. – Henno Brandsma Aug 09 '20 at 11:24