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Let $X, Y$ be topological spaces and let $f:X \rightarrow Y$ be a function and let $g = f \times f : X \times X \rightarrow Y \times Y$.

I want to show that if:

1) $Y$ is normal and

2) for all open sets $U$ of $Y \times Y$ which contain the diagonal $\Delta(Y)$, the interior of the inverse image, $(g^{-1}(U))^{\circ}$, contains the diagonal $\Delta(X)$

Then $f$ is continuous.

Since the product of normal spaces need not be normal, my current line of thought is to show that $g$ is continuous by composing $g$ with a projection mapping $p$ and invoking the universal property of the product topology or something. However, this route would seem to bypass the use of the second condition.

user133398
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1 Answers1

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Let $V$ be open in $Y$, and fix $x \in f^{-1}(V)$. Then since $Y$ is Hausdorff, $\{f(x)\}$ is closed in $Y$, as is $Y-V$; these two sets are disjoint since $f(x) \in V$. By the normality of $Y$ (we really only need regularity), pick open disjoint sets $U,W \subset Y$ with $f(x) \in U$ and $(Y-V) \subset W$. Then $W \cup V = Y$, so $D:=(W \times W) \cup (V \times V)$ is open in $Y \times Y$ and contains the diagonal. By assumption, the interior of $f^{-1}(D)$ contains the diagonal of $X$. Let \begin{align*} B = \{y \in X : (x,y) \in (f^{-1}(D))^{\circ}\}. \end{align*} Then $B$ is open because each $(x,y)$ is a point in an interior, and thus there exists an open basis set containing $(x,y)$ contained in the interior, which gives rise to an open neighborhood of $y$ contained in $B$. Furthermore, $f(B) \subset V$, because $f(x)$ is disjoint from $W$ and so no point $(x,y)$ with $w \in W$ lies in $f^{-1}(D)$, so $f$ must take $(x,y)$ to $V \times V$. Thus f$^{-1}(V)$ is open, so $f$ is continuous, as desired.

Mark Perlman
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