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?
Best Answer
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.