Let \(f: X \to Y\) be a morphism of schemes, and suppose that \(Y\) can be covered by open subsets \(U_i\) such that for each \(i\) the induced map \(\inverseof{f}(U_i) \to U_i\) is an isomorphism. Then \(f\) is an isomorphism.
A scheme \(X\) is affine if and only if there is a finite set of elements \(f_1, \ldots, f_r \in A = \Gamma(X, \OO_X)\) such that the open subsets \(X_{f_i}\) are affine and \(f_1, \ldots, f_r\) generate the unit ideal in \(A\).
Part (b) uses Hartshorne II.2.4 and II.2.16(d).
Solution
1. Under the hypotheses of part (a), the local inverses\[ g_i:U_i\longrightarrow f^{-1}(U_i) \]glue to a morphism\[ g:Y\longrightarrow X. \]
For each \(i\), the restriction \[ f_i=f|_{f^{-1}(U_i)}:f^{-1}(U_i)\longrightarrow U_i \] is an isomorphism, so let \(g_i=f_i^{-1}\).
On \(U_i\cap U_j\), both \(g_i\) and \(g_j\) are inverse to the restriction \[ f:f^{-1}(U_i\cap U_j)\longrightarrow U_i\cap U_j. \] Hence they agree. Morphisms of schemes glue uniquely on an open cover of the source, so the \(g_i\) determine a morphism \(g:Y\to X\).
2. The morphism \(g\) is inverse to \(f\); therefore \(f\) is an isomorphism.
On every \(U_i\), \[ (f\circ g)|_{U_i}=f_i\circ g_i=\id_{U_i}, \] so \(f\circ g=\id_Y\). Likewise the opens \(f^{-1}(U_i)\) cover \(X\), and \[ (g\circ f)|_{f^{-1}(U_i)}=g_i\circ f_i=\id_{f^{-1}(U_i)}. \] Thus \(g\circ f=\id_X\).
3. If \(X\) is affine, then the condition in part (b) holds.
Write \(X=\Spec A\) and take \(f_1=1\in A\). Then \(X_{f_1}=X\) is affine and \(f_1\) generates the unit ideal.
4. Conversely, suppose that\[ A=\Gamma(X,\mathcal O_X) \]contains elements \(f_1,\ldots,f_r\) such that each \(X_{f_i}\) is affine and the \(f_i\) generate the unit ideal. Then the \(X_{f_i}\) cover \(X\).
Choose \(a_i\in A\) with \[ \sum_i a_if_i=1. \] If a point \(x\in X\) lay outside every \(X_{f_i}\), then every germ \((f_i)_x\) would lie in the maximal ideal \(\mathfrak m_x\subseteq\mathcal O_{X,x}\). Passing the displayed identity to the stalk would give \[ 1=\sum_i(a_i)_x(f_i)_x\in\mathfrak m_x, \] which is impossible. Hence the \(X_{f_i}\) cover \(X\).
5. The finite affine cover \(\{X_{f_i}\}\) satisfies the hypothesis of Hartshorne II.2.16(c).
For all \(i,j\), \[ X_{f_i}\cap X_{f_j}=X_{f_if_j}. \] Inside the affine scheme \(X_{f_i}\) this is the distinguished open where \(f_j|_{X_{f_i}}\) is invertible. Hence it is affine, and in particular quasi-compact.
6. Let\[ \eta:X\longrightarrow\Spec A \]be the canonical morphism corresponding under Hartshorne II.2.4 to the identity homomorphism \(A\to A\). Then\[ \eta^{-1}(D(f_i))=X_{f_i} \]for every \(i\).
For \(x\in X\), the prime ideal corresponding to \(\eta(x)\) is \[ \{a\in A:a_x\in\mathfrak m_x\}. \] Therefore \[ \eta(x)\in D(f_i) \iff (f_i)_x\notin\mathfrak m_x \iff x\in X_{f_i}. \]
7. For every \(i\), the restriction\[ \eta_i:X_{f_i}\longrightarrow D(f_i) \]is an isomorphism.
By 5, Hartshorne II.2.16(d) applies to \(X\), so \[ \Gamma(X_{f_i},\mathcal O_X)\cong A_{f_i}. \] By II.2.1, \[ D(f_i)\cong\Spec A_{f_i}. \] Since \(X_{f_i}\) is affine, \[ X_{f_i} \cong \Spec\Gamma(X_{f_i},\mathcal O_X) \cong \Spec A_{f_i} \cong D(f_i). \] Under these identifications, \(\eta_i\) is induced by the localization map \(A_{f_i}\to\Gamma(X_{f_i},\mathcal O_X)\), which is the isomorphism above. Hence \(\eta_i\) is an isomorphism.
8. The distinguished opens \(D(f_1),\ldots,D(f_r)\) cover \(\Spec A\).
Their complement is \[ V(f_1,\ldots,f_r). \] Since the \(f_i\) generate the unit ideal, this is \[ V(1)=\varnothing. \]
9. The canonical morphism \(\eta:X\to\Spec A\) is an isomorphism. Consequently \(X\) is affine.
By 8, the \(D(f_i)\) cover \(\Spec A\). By 6 and 7, for every \(i\) the induced map \[ \eta^{-1}(D(f_i))=X_{f_i}\longrightarrow D(f_i) \] is an isomorphism. Part (a), proved in <1>1–<1>2, therefore shows that \(\eta\) is an isomorphism.
10. Hence\[ \boxed{ X\text{ is affine} \iff \begin{array}{c} \text{there exist finitely many }f_i\in\Gamma(X,\mathcal O_X)\\ \text{generating the unit ideal such that every }X_{f_i}\text{ is affine.} \end{array} } \]
The forward implication is 3 and the reverse implication is <1>4–<1>9.
11. Q.E.D.
Steps <1>1–<1>2 prove part (a), and steps <1>3–<1>10 prove part (b).