Soit $f:X\rightarrow$ un morphisme affine. Pour tout ouvert $U\subset Y $, le morphisme $f^{-1}(U)\rightarrow U$, restriction de f, est affine.
Translation: Let $f:X\rightarrow$ an affine morphism. For all open $U\subset Y $, the morphism $f^{-1}(U)\rightarrow U$, restriction of f, is affine.
Don't we need here $U$ to be affine also in $Y$?
This is from EGA-I Springer Verlag 1971 Edition
Let $\{V_\alpha\}_\alpha$ be an affine cover of $Y$ with affine preimages, and denote $\varphi_\alpha\colon \mathcal{O}_Y(V_\alpha) \to \mathcal{O}_X(f^{-1}(V_\alpha))$ to be the corresponding map on the coordinate rings of $V_\alpha$ and $f^{-1}(V_\alpha)$.
Then, $U$ is covered by $\{U \cap V_\alpha\}_\alpha$, and each $U \cap V_\alpha$ can be covered by open sets $\{D_{V_\alpha}(h_\beta)\}_{\beta}$ that are distinguished in $V_\alpha$. These open sets $\{D_{V_\alpha}(h_\beta)\}_{\alpha,\beta}$ form an affine open cover of $U$. The preimages of the $D_{V_\alpha}(h_\beta)$ are distinguished in $f^{-1}(V_\alpha)$; more precisely, they are of the form $D_{f^{-1}(V_\alpha)}(\varphi_\alpha(h_\beta))$, hence are affine.