If $f:X \to [0,1]$ is an onto continuous map and $\{f^{-1} (y)\}$ is compact for every $y \in [0,1]$, then is X necessarily compact?
Now continuous image of a compact set is compact. Again $X$ is the uncountable union of compact sets but to get any conclusion?
No, it doesn't have to be. Counterexample: endow $[0,1]$ with the discrete topology, $\mathcal T_{dis}$. Call the standard topology on $[0,1]$ $\mathcal T_{Euc}$. Now consider the identity function: $$\mathrm{id}:([0,1],\mathcal T_{dis})\longrightarrow ([0,1],\mathcal T_{Euc}) \atop \qquad x\mapsto x$$ Then $f^{-1}(\{y\})=\{y\}$ for all $y\in[0,1]$ yet $([0,1],\mathcal T_{dis})$ is not compact.
Here is a simple counter-example which proves that $X$ is not necessarily compact. Let $X=[0,1)$ which is not closed then not compact. and $f(x)=x(1-x)$ for $x$ in $X$. Then the range is exactly $[0,1]$ and $f$ is continuous. EDIT. Topologies are usual ones on $X$ and on the range.
