If product space of two spaces is normal, then are the two spaces normal?

Question

We know that product of normal spaces need not be normal,for e.g. ${R_{l}}^2$. My question is regarding the converse. If $X$ & $Y$ are two spaces such that $X \times Y$ is normal, then can we claim that $X$ & $Y$ are normal spaces? In case of regular, hausdorff or T-1 spaces, this appear to be true and can be proven.

Answer 1

If $A,A'$ are disjoint closed subsets of $X$, then $A\times Y$, $A'\times Y$ are disjoint closed of $X\times Y$. Accordingly, let $O$, $O'$ be open neighbourhoods. Write $O=\bigcup _i U_i\times V_i$, $O'=\bigcup _j U'_j\times V'_j$ where each $U_i, U'_j$ is open in $X$ and $V_i, V'_j$ is open in $Y$. Pick $y\in Y$. Then $$ \bigcup \{\, U_i\mid y\in V_i\,\},\quad \bigcup \{\, U'_j\mid y\in V'_j\,\}$$ are disjoint open neighbourhoods of $A$ and $A'$.

Answer 2

We will only consider the case that $X$ and $Y$ are non-empty; otherwise $X \times Y = \emptyset$ is normal, but we can make no statement about $X$ (if $Y = \emptyset$) or $Y$ (if $X = \emptyset$). If we assume that $X, Y \neq \emptyset$, then the statement is true:

Recall that a topological space $X$ is normal if and only if the Tietze extension theorem holds for $X$, i.e. if for every closed subset $A \subseteq X$, every continuous function $f \colon X \to \mathbb{R}$ can be extended to a continuous function $F \colon X \to \mathbb{R}$.

Suppose that $X \times Y$ is normal, let $A \subseteq X$ be a closed set and let $f \colon A \to \mathbb{R}$ be a continuous function. Let $p \colon X \times Y \to X$ be the projection onto the first coordinate. Then $$ B := p^{-1}(A) = A \times Y $$ is a closed subset of $X \times Y$. Define $g \colon B \to \mathbb{R}$ as $g := f \circ p|_B$, i.e. by $g(a,y) = f(a)$ for all $(a,y) \in B$. Then $g$ is continuous, and can thus be extended to a continuous function $G \colon X \times Y \to \mathbb{R}$. Let $y_0 \in Y$ and consider the map $$ F \colon X \to \mathbb{R}, \quad x \mapsto G(x,y_0). $$ Then $F$ is a continuous extension of $f$.

This shows that $X$ is normal, if $Y$ is non-empty. If both $X$ and $Y$ are non-empty, then it follows that they are both normal.