**Proof.**
Assume $s$ and $t$ open and $W \subset U$ open. Since $s$ is open we see that $W' = s(t^{-1}(W))$ is an open subspace of $U$. Now it is quite easy to using the functorial point of view that this is an $R$-invariant open subset of $U$, but we are going to argue this directly by some diagrams, since we think it is instructive. Note that $t^{-1}(W')$ is the image of the morphism

\[ A := t^{-1}(W) \times _{s|_{t^{-1}(W)}, U, t} R \xrightarrow {\text{pr}_1} R \]

and that $s^{-1}(W')$ is the image of the morphism

\[ B := R \times _{s, U, s|_{t^{-1}(W)}} t^{-1}(W) \xrightarrow {\text{pr}_0} R. \]

The algebraic spaces $A$, $B$ on the left of the arrows above are open subspaces of $R \times _{s, U, t} R$ and $R \times _{s, U, s} R$ respectively. By Lemma 77.11.4 the diagram

\[ \xymatrix{ R \times _{s, U, t} R \ar[rd]_{\text{pr}_1} \ar[rr]_{(\text{pr}_1, c)} & & R \times _{s, U, s} R \ar[ld]^{\text{pr}_0} \\ & R & } \]

is commutative, and the horizontal arrow is an isomorphism. Moreover, it is clear that $(\text{pr}_1, c)(A) = B$. Hence we conclude $s^{-1}(W') = t^{-1}(W')$, and $W'$ is $R$-invariant. This proves (1).

Assume now that $s$, $t$ are both open and quasi-compact. Then, if $W \subset U$ is a quasi-compact open, then also $W' = s(t^{-1}(W))$ is a quasi-compact open, and invariant by the discussion above. Letting $W$ range over images of affines étale over $U$ we see (2).
$\square$

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