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my question is regarding topology.

Given two topological spaces $X, Y$ and we know that $Y$ is $T_4$.

Let $f:X \rightarrow Y$ be a function such that for any continuous function $\phi:Y \rightarrow \mathbb{R}$, the composition $\phi \circ f$ is continuous.

Show that $f$ it self is continuous.

My attempt:

we want to show that for any open set $V$, $f^{-1}[V]$ is open. If we could find a function $g$ such that $g:Y \rightarrow \mathbb{R}$ would be zero in exactly $Y \setminus V$ and positive else where, then we could take the preimage of the open set $(0,\infty)$ which would be open in $X$ and be exactly $f^{-1}[V]$.

As I looked around, it seems we can build such a function when knowing that a set is $G_{\delta}$, and the space is normal but here we don't know that the set is $G_{\delta}$.

In the exam from which I took the question, the first question was to quote Urysohn's lemma, so the solution probably uses the lemma.

Any suggestions?

Nadav Kalma
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1 Answers1

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We prove that $f$ is continuous at each point, i.e for every $x\in X$ and $V$ an open neighborhood of $f(x)$, $f^{-1}(V)\subset X$ contains an open neighborhood of $x$.

Let $x\in X$ and let $f(x)\in V\subset Y$ be open. The sets $\{f(x)\}$ and $Y\setminus V$ are disjoint and closed. Therefore by Urysohn's lemma (as $Y$ is normal) there exists a continuous function $\phi:Y\to [0,1]$ such that: $$\phi(x)=0,\,\phi(Y\setminus V)=\{1\}$$ Thinking of $\phi$ as a function from $Y$ to $\mathbb{R}$ (via the continuous inclusion $\iota:[0,1]\to\mathbb{R}$), we have: $$\phi^{-1}(\mathbb{R}\setminus\{1\})\subset V$$ Hence: $$x\in f^{-1}(\phi^{-1}(\mathbb{R}\setminus\{1\}))\subset f^{-1}(V)$$ And $f^{-1}(\phi^{-1}(\mathbb{R}\setminus\{1\}))$ is an open subset that contains $x$. This concludes the proof.

Math101
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  • Why is it enough for $f^{-1}(V)$ to contains an open neighborhood of $x$? The definition of continuous function states that $f^-1(V)$ is an open neighborhood of $x$. – Amit Jul 03 '23 at 13:56
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    @Amit There is a definition of continuity and point-wise continuity. What I proved is that $f$ is pointwise continuous at each point, and hence is continuous. – Math101 Jul 03 '23 at 13:59
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    @Amit, to show that a set is open you can show that it can be presented as a union of open sets, and that is what GBA showed. Repeating the process for each $ x \in f^{-1}[V}$ will show the set is open. – Nadav Kalma Jul 03 '23 at 14:06
  • Got it, Thanks! – Amit Jul 03 '23 at 14:40