Lemma 89.24.1. Let $(U, R, s, t, c)$ be a smooth groupoid in functors on $\mathcal{C}_\Lambda$. Assume $U$ and $R$ satisfy (RS). Then $[U/R]$ satisfies (RS).

Proof. Let

$\xymatrix{ & (A_2, x_2) \ar[d]^{(f_2, a_2)} \\ (A_1, x_1) \ar[r]^{(f_1, a_1)} & (A, x) }$

be a diagram in $[U/R]$ such that $f_2: A_2 \to A$ is surjective. The notation is as in Remarks 89.5.2 (9). Hence $f_1: A_1 \to A, f_2: A_2 \to A$ are maps in $\mathcal{C}_\Lambda$, $x \in U(A)$, $x_1 \in U(A_1)$, $x_2 \in U(A_2)$, and $a_1, a_2 \in R(A)$ with $s(a_1) = U(f_1)(x_1)$, $t(a_1) = x$ and $s(a_2) = U(f_2)(x_2)$, $t(a_2) = x$. We construct a fiber product lying over $A_1 \times _ A A_2$ for this diagram in $[U/R]$ as follows.

Let $a = c(i(a_1), a_2)$, where $i: R \to R$ is the inverse morphism. Then $a \in R(A)$, $x_2 \in U(A_2)$ and $s(a) = U(f_2)(x_2)$. Hence an element $(a, x_2) \in R(A) \times _{s, U(A), U(f_2)} U(A_2)$. By smoothness of $s : R \to U$ there is an element $\widetilde{a} \in R(A_2)$ with $R(f_2)(\widetilde{a}) = a$ and $s(\widetilde{a}) = x_2$. In particular $U(f_2)(t(\widetilde{a})) = t(a) = U(f_1)(x_1)$. Thus $x_1$ and $t(\widetilde{a})$ define an element

$(x_1, t(\widetilde{a})) \in U(A_1) \times _{U(A)} U(A_2).$

By the assumption that $U$ satisfies (RS), we have an identification $U(A_1) \times _{U(A)} U(A_2) = U(A_1 \times _ A A_2)$. Let us denote $x_1 \times t(\widetilde{a}) \in U(A_1 \times _ A A_2)$ the element corresponding to $(x_1, t(\widetilde{a})) \in U(A_1) \times _{U(A)} U(A_2)$. Let $p_1, p_2$ be the projections of $A_1 \times _ A A_2$. We claim

$\xymatrix{ (A_1 \times _ A A_2, x_1 \times t(\widetilde{a})) \ar[d]_{(p_1, e(x_1))} \ar[rr]_-{(p_2, i(\widetilde{a}))} & & (A_2, x_2) \ar[d]^{(f_2, a_2)} \\ (A_1, x_1) \ar[rr]^{(f_1, a_1)} & & (A, x) }$

is a fiber square in $[U/R]$. (Note $e: U \to R$ denotes the identity.)

The diagram is commutative because $c(a_2, R(f_2)(i(\widetilde{a}))) = c(a_2, i(a)) = a_1$. To check it is a fiber square, let

$\xymatrix{ (B, z) \ar[d]_{(g_1, b_1)} \ar[rr]_{(g_2, b_2)} & & (A_2, x_2) \ar[d]^{(f_2, a_2)} \\ (A_1, x_1) \ar[rr]^{(f_1, a_1)} & & (A, x) }$

be a commutative diagram in $[U/R]$. We will show there is a unique morphism $(g, b) : (B, z) \to (A_1 \times _ A A_2, x_1 \times t(\widetilde{a}))$ compatible with the morphisms to $(A_1, x_1)$ and $(A_2, x_2)$. We must take $g = (g_1, g_2) : B \to A_1 \times _ A A_2$. Since by assumption $R$ satisfies (RS), we have an identification $R(A_1 \times _ A A_2) = R(A_1) \times _{R(A)} R(A_2)$. Hence we can write $b = (b'_1, b'_2)$ for some $b'_1 \in R(A_1)$, $b'_2 \in R(A_2)$ which agree in $R(A)$. Then $((g_1, g_2), (b'_1, b'_2)) : (B, z) \to (A_1 \times _ A A_2, x_1 \times t(\widetilde{a}))$ will commute with the projections if and only if $b'_1 = b_1$ and $b'_2 = c(\widetilde{a}, b_2)$ proving unicity and existence. $\square$

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