Probabily it's trivial but I've no idea for a proof.
Let $f: X \rightarrow Y $ a continuous map between Topological Spaces, with $Im(f)$ closed in $Y$. I know there exist a covering $\{Y_i\}$ of $Y$ such that $\{f^{-1}(Y_i)\}$ is a covering of $X$ and $f: f^{-1}(Y_i) \rightarrow Y_i $ is closed. Is $f$ a closed function?
I suppose this is false, but i know it is true in the context of the Theory of Schemes (if $f: X \rightarrow Y $ a morphism between the schemes $X$ and $Y$, with $Im(f)$ closed in $Y$ and there exist a covering $\{Y_i\}$ of $Y$ such that $\{f^{-1}(Y_i)\}$ is a covering of $X$ and $f: f^{-1}(Y_i) \rightarrow Y_i $ is closed, then $f$ is a closed morphism ).
It seems clear if $Y$ is quasi-compact. Some help for a general proof?
Thank you!
Since the question is purely topological, the fact that you have the supplementary structure of a scheme on $Y$ is irrelevant.
The answer to the question is "no" if you consider an arbitrary covering $(Y_i)$ of $Y$: by taking for the $Y_i$'s the points of $Y$ (seen as singleton subsets of $Y$) you would conclude that every surjective continuous map is closed, which is false (cf. first projection $\mathbb R^2\to \mathbb R$).
However if $(Y_i)$ is an open covering, the result is true, because $F\subset Y$ is closed if and only if all the $F\cap Y_i$ are closed in the relative topology of $Y_i$.