Suppose $X$ is a scheme, and $T$ its closet subset. I want to ask that is there a canonical way to obtain a quasi-coherent $\mathscr O_X$-ideal $I$ such that $T=supp \mathscr O_X/I$? This is the inverse question of the construction of closed subschemes. Could you give a counterexample or a proof?
Recently I'm reading algebraic geometry but I don't know why the inverse image of a closed subscheme is also a closed subscheme. So I thought that I may need to solve the above question. Or could you give another way to show "the inverse image of a closed subscheme is also a closed subscheme?" Thank you.
The canonical way to obtain a quasi-coherent sheaf of ideals $I$ with $V(I)=T$ is to put the reduced induced subscheme structure on $T$ and then let $I$ be the kernel of $\mathcal{O}_X\to\mathcal{O}_T$. In general, though, there may be many different subscheme structures on $T$, so the collection of quasi-coherent sheaves of ideals $I$ which have $V(I)=T$ as sets can be large.
This is not really relevant to your second paragraph, though - the way we define the preimage of a closed subscheme $T\subset X$ under a map $Y\to X$ is to take the fiber product $Y\times_X T$ (this is the definition as subschemes - if you were only interested in the definition as subsets, you can just use the fact that the map on topological spaces associated to a morphism of schemes is continuous by definition). The proof that the morphism of schemes $Y\times_X T\to Y$ is a closed immersion boils down to checking the affine case where you can see that if $X=\operatorname{Spec} A$, $Y=\operatorname{Spec} B$, and $T=\operatorname{Spec} A/I$, then $Y\times_X T=\operatorname{Spec} B\otimes_A A/I \cong \operatorname{Spec} B/IB$.