The Stacks project

Proof. Statement (2) is equivalent to (3) since the inverse $i: R \to R$ of $(U, R, s, t, c)$ is an isomorphism and $t = s \circ i$. By definition (1) is equivalent to (2) and (3) together, hence it is equivalent to either of them individually.

Finally we prove (2) is equivalent to (4). Unwinding the definitions:

1. Smoothness of $s: R \to U$ amounts to the following condition: If $f: B \to A$ is a surjective ring map in $\mathcal{C}_\Lambda $, $a \in R(A)$, and $y \in U(B)$ such that $s(a) = U(f)(y)$, then there exists $a' \in R(B)$ such that $R(f)(a') = a$ and $s(a') = y$.

2. Smoothness of $U \to [U/R]$ amounts to the following condition: If $f: B \to A$ is a surjective ring map in $\mathcal{C}_\Lambda $ and $(f, a) : (B, y) \to (A, x)$ is a morphism of $[U/R]$, then there exists $x' \in U(B)$ and $b \in R(B)$ with $s(b) = x'$, $t(b) = y$ such that $c(a, R(f)(b)) = e(x)$. Here $e : U \to R$ denotes the identity and the notation $(f, a)$ is as in Remarks 90.5.2 (9); in particular $a \in R(A)$ with $s(a) = U(f)(y)$ and $t(a) = x$.

If (4) holds and $f, a, y$ as in (2) are given, let $x = t(a)$ so that we have a morphism $(f, a): (B, y) \to (A, x)$. Then (4) produces $x'$ and $b$, and $a' = i(b)$ satisfies the requirements of (2). Conversely, assume (2) holds and let $(f, a): (B, y) \to (A, x)$ as in (4) be given. Then (2) produces $a' \in R(B)$, and $x' = t(a')$ and $b = i(a')$ satisfy the requirements of (4). $\square$