Lemma 68.6.4. With notation and assumptions as in Situation 68.6.1. If

1. $f$ is surjective,

2. $f_0$ is locally of finite presentation,

then $f_ i$ is surjective for some $i \geq 0$.

Proof. Choose an affine scheme $V_0$ and a surjective étale morphism $V_0 \to Y_0$. Choose an affine scheme $U_0$ and a surjective étale morphism $U_0 \to V_0 \times _{Y_0} X_0$. Diagram

$\xymatrix{ U_0 \ar[d] \ar[r] & V_0 \ar[d] \\ X_0 \ar[r] & Y_0 }$

The vertical arrows are surjective and étale by construction. We can base change this diagram to $B_ i$ or $B$ to get

$\vcenter { \xymatrix{ U_ i \ar[d] \ar[r] & V_ i \ar[d] \\ X_ i \ar[r] & Y_ i } } \quad \text{and}\quad \vcenter { \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y } }$

Note that $U_ i, V_ i, U, V$ are affine schemes, the vertical morphisms are surjective étale, the limit of the morphisms $U_ i \to V_ i$ is $U \to V$, and the morphisms $U_ i \to X_ i \times _{Y_ i} V_ i$ and $U \to X \times _ Y V$ are surjective (as base changes of $U_0 \to X_0 \times _{Y_0} V_0$). In particular, we see that $X_ i \to Y_ i$ is surjective if and only if $U_ i \to V_ i$ is surjective and similarly $X \to Y$ is surjective if and only if $U \to V$ is surjective. Since $f_0$ is locally of finite presentation, so is the morphism $U_0 \to V_0$. Hence the lemma follows from the case of schemes (Limits, Lemma 32.8.14). $\square$

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