Topologizing Borel space so that certain functions become continuous Let $X$ and $Y$ be compact metric spaces. Let $f:X \to Y$ be a Borel measurable map and suppose that $T:X \to X$ is a homeomorphism. Can one change the topology on $X$ such that


*

*$X$ is still a compact metrizable space, with the same Borel sets.

*The map $T$ remains continuous.

*The map $f:X \to Y$ is continuous.


?
Note: If $Y$ is not compact, we can find counterexamples by taking $f$ to be an unbounded function. The answer is YES if we just want complete metrizability of the new topology on $X$, instead of compactness.
 A: No - This follows from: If $\tau$ properly extends the usual topology on $[0, 1]$, then $\tau$ is not compact. The same is true of any compact Hausdorff space.
A: The answer is NO even if we forget $T$ (Or equivalently, let $T=Id_X$). I will later post a measure theoretic weakening of this question, which I am more interested in.
Counterexample: Let $X=[0,1]$ and $Y=[0,1] \cup \{-1\}$ and define $f:X \to Y$ by $f(x)=x$ for $x \neq 0$ and $f(0)=-1$. Thus we would require a topology on $X$ such that $\{0\}$ and $(x,1]$ for all $0<x<1$ are open. Thus we have an open cover of $X$ given by $$\{ (x,1] | x \in (0,1) \} \cup \{0\},$$ which has no finite subcover.  
A: (fubini has a nice answer, I'm just giving a concrete example I thought in the meantime while being offline.)
Take $X=Y=[0,1]$ and $f=\chi_{\{0\}}$, the indicator function of the singleton $\{0\}$. That is,
$$
f(x)= \begin{cases} 1 & x=0 \\ 0 & 0<x\leq 1.\end{cases}
$$
Then $f$ is Borel, and if you make it continuous, then $\{0\}$ must be open. Then 
$$
\{0\}, (\tfrac{1}{n},1]
$$
with $n\in\mathbb{N}$, is an open cover of $[0,1]$ without a finite subcover.
