$X_i$ is a homeomorphic to a subset of $\prod_{i \in I} X_i$

Question

Let $(X_i)_{i \in I}$ be a family of topological spaces such that $X_i \neq \varnothing$ for all $i \in I$. I want to show that, for any $j \in I$, $X_j$ is homeomorphic to $Y_j = \{ (x_i)_{i \in I} \in \prod_{i \in I} X_i\mid x_i = p_i$ for $i \neq j \}$ for fixed $(p_i)_{i \neq j}$ with $p_i \in X_i$.

The book Topology: An Introduction by Stefan Waldmann attempts to prove it this way: it constructs a map $s_j\colon X_j\to\prod_{i \in I} X_j$ given by $s_j(x) = (x_i)_{i \in I}$ where $x_j = x$ and $x_i = p_i$ for $i \neq j$ which is a right inverse of the projection $p_j\colon\prod_{i \in I} X_i\to X_j, (x_i)_{i \in I} \mapsto x_j$. He then claims that $p_j{\restriction}_{s_j(X_j)}$ is bijective, continuous (which is correct) and open, so it is a homeomorphism. It's not clear to me why $p_j{\restriction}_{s_j(X_j)}$ would be open. I know that $p_j$ is, but it is known that a restriction of an open map if not necessarily open.

Answer 1

There is no need to show that $\pi_j = p_j \mid_{s_j(X_j)}$ is open. In fact $s'_j : X_j \stackrel{s_j}{\rightarrow} s_j(X_j)$ and $\pi_j$ are continuous and by construction we have $\pi_j \circ s'_j = id$ and $s'_j \circ \pi_j = id$. This shows that $s'_j, \pi_j$ are homeomorphims which are inverse to each other.

Edited:

The map $s_j$ occurs in the proof of part (iv) of Proposition 3.1.2. The author says that $s_j$ is a continuous section of $p_j$, but he doesn't explicitly prove continuity. For the sake of completeness let us do it here. By part (iii) of the Proposition it suffices to show that all $p_i \circ s_j$ are continuous. But $p_j \circ s_j = id$ and $p_i \circ s_j$ is constant for $i \ne j$, thus continuity follows.

Edited:

A more general approach is this. You have continuous maps $s : X \to Y$ and $p : Y \to X$ such that $p \circ s = id_X$ (in that case $s$ is called a section of $p$). Note that this implies that $s$ is injective and $p$ is surjective (in the above case with the functions $s_j, p_j$ this is of course obvious from the definition without invoking $p_j \circ s_j = id$). Then the map $s' : X \stackrel{s}{\rightarrow} s(X)$ is a continuous bijection and the map $p' = p \mid_{s(X)} : s(X) \to X$ is continuous (where $s(X)$ carries the subspace topology inherited from $Y$). Let $(s')^{-1} : s(X) \to X$ be the inverse of $s'$. This function is not a priori continuous - a proof is needed. This is very simple. We have $p' \circ s' = id_X$, hence $(s')^{-1} = id_X \circ (s')^{-1} = (p' \circ s') \circ (s')^{-1} = p' \circ (s' \circ (s')^{-1}) = p' \circ id_{s(X)} = p'$. This shows that $s'$ is a homeomorphism whose inverse is $p'$.

This approach is more complicated than the short proof at the top of this answer. However, it was obvious from the definition that $s'_j \circ \pi_j = id$, whereas in the general case it is not obvious that $s' \circ p' = id_{s(X)}$.

Answer 2

Note that you're using $p$ both for points and for the projection. I'll use $\pi_j$ for the projection onto $X_j$.

It is true that this doesn't hold in general, but what's an open set in the sub-space topology $s_j(X_j)?$ Well it's a set of the form $U\cap s_j(X_j)$ for some open $U$ in $\prod_{i\in I} X_i$ in particular, by definition of the product topology, there are boxes $\prod_{i} U_{i,\alpha}$ with each $U_{i,\alpha}$ open in $X_i$ and such that $U=\cup_{\alpha}\prod_i U_{i,\alpha}$ and such that $p_i\in U_{i,\alpha}$ for every $\alpha$ and $i\neq j$ (otherwise, we'd have some summand of our intersection which was empty). Then, however,

$$ U\cap s_j(X_j)=\cup_{\alpha}((\prod_i U_{i,\alpha})\cap s_j(X_j))=\cup_{\alpha}( U_{j,\alpha}\times \prod_{i\neq j} \{p_i\}), $$

and thus, $\pi_j(U\cap s_j(X_j))= \cup_{\alpha} U_{j,\alpha},$ which is open in $X_j$ by assumption. So yes, the restriction of the projection is an open map in this case.