A morphism \(f: X \to Y\) of schemes is quasi-compact if there is a cover of \(Y\) by open affines \(V_i\) such that \(\inverseof{f}(V_i)\) is quasi-compact for each \(i\).
Show that \(f\) is quasi-compact if and only if for every open affine subset \(V \subseteq Y\) the preimage \(\inverseof{f}(V)\) is quasi-compact.
Solution
1. If \(f^{-1}(V)\) is quasi-compact for every affine open \(V\subseteq Y\), then \(f\) is quasi-compact.
Choose any affine open cover \[ Y=\bigcup_iV_i. \] The hypothesis says that every \(f^{-1}(V_i)\) is quasi-compact, which is exactly the definition given in the problem.
2. Conversely, suppose \(f\) is quasi-compact in the stated sense. Thus there is an affine open cover\[ Y=\bigcup_iV_i \]such that every \(f^{-1}(V_i)\) is quasi-compact. Fix an arbitrary affine open\[ V\subseteq Y. \]For every \(y\in V\) there is an index \(i\) and a distinguished open\[ W_y=D(g_y)\subseteq V_i \]such that\[ y\in W_y\subseteq V\cap V_i. \]
Choose \(i\) with \(y\in V_i\). Then \(V\cap V_i\) is an open neighborhood of \(y\) in the affine scheme \(V_i\). Distinguished opens form a basis of an affine scheme, so there is some \(g_y\in\Gamma(V_i,\mathcal O_Y)\) with the asserted property.
3. For each \(y\in V\), the inverse image\[ f^{-1}(W_y) \]is quasi-compact.
Fix \(y\) and write \(W_y=D(g_y)\subseteq V_i\). Since \(f^{-1}(V_i)\) is quasi-compact, choose a finite affine open cover \[ f^{-1}(V_i)=U_1\cup\cdots\cup U_n. \] On each affine \(U_j\), the pullback of \(g_y\) is a global function, and \[ U_j\cap f^{-1}(W_y) \] is its distinguished nonvanishing locus. Hence it is affine, in particular quasi-compact.
Therefore \[ f^{-1}(W_y) = \bigcup_{j=1}^n\bigl(U_j\cap f^{-1}(W_y)\bigr) \] is a finite union of quasi-compact open subsets and is quasi-compact.
4. Finitely many of the opens \(W_y\) cover \(V\).
The sets \(W_y\) form an open cover of \(V\). Since \(V\) is affine, its underlying topological space is quasi-compact. Hence there are points \[ y_1,\ldots,y_m\in V \] such that \[ V=W_{y_1}\cup\cdots\cup W_{y_m}. \]
5. The inverse image \(f^{-1}(V)\) is quasi-compact.
By 4, \[ f^{-1}(V) = f^{-1}(W_{y_1})\cup\cdots\cup f^{-1}(W_{y_m}). \] Each term is quasi-compact by 3, so their finite union is quasi-compact.
6. Hence the two formulations of quasi-compactness are equivalent.
Step 1 proves one implication and steps <1>2–<1>5 prove the converse.
7. Q.E.D.
Step 6 is the required equivalence.