## 88.23 Smooth groupoids in functors on the base category

The notion of smoothness for groupoids in functors on $\mathcal{C}_\Lambda$ is defined as follows.

Definition 88.23.1. Let $(U, R, s, t, c)$ be a groupoid in functors on $\mathcal{C}_\Lambda$. We say $(U, R, s, t, c)$ is smooth if $s, t: R \to U$ are smooth.

Remark 88.23.2. We note that this terminology is potentially confusing: if $(U, R, s, t, c)$ is a smooth groupoid in functors, then the quotient $[U/R]$ need not be a smooth category cofibred in groupoids as defined in Definition 88.9.1. However smoothness of $(U, R, s, t, c)$ does imply (in fact is equivalent to) smoothness of the quotient morphism $U \to [U/R]$ as we shall see in Lemma 88.23.4.

Remark 88.23.3. Let $(\underline{R_0}, \underline{R_1}, s, t, c)|_{\mathcal{C}_\Lambda }$ be a prorepresentable groupoid in functors on $\mathcal{C}_\Lambda$. Then $(\underline{R_0}, \underline{R_1}, s, t, c)|_{\mathcal{C}_\Lambda }$ is smooth if and only if $R_1$ is a power series over $R_0$ via both $s$ and $t$. This follows from Lemma 88.8.6.

Lemma 88.23.4. Let $(U, R, s, t, c)$ be a groupoid in functors on $\mathcal{C}_\Lambda$. The following are equivalent:

1. The groupoid in functors $(U, R, s, t, c)$ is smooth.

2. The morphism $s : R \to U$ is smooth.

3. The morphism $t : R \to U$ is smooth.

4. The quotient morphism $U \to [U/R]$ is smooth.

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 88.5.2 (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). $\square$

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