## 77.21 Functoriality of quotient stacks

A morphism of groupoids in algebraic spaces gives an associated morphism of quotient stacks.

Lemma 77.21.1. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $f : (U, R, s, t, c) \to (U', R', s', t', c')$ be a morphism of groupoids in algebraic spaces over $B$. Then $f$ induces a canonical $1$-morphism of quotient stacks

$[f] : [U/R] \longrightarrow [U'/R'].$

Proof. Denote $[U/_{\! p}R]$ and $[U'/_{\! p}R']$ the categories fibred in groupoids over the base site $(\mathit{Sch}/S)_{fppf}$ associated to the functors (77.20.0.1). It is clear that $f$ defines a $1$-morphism $[U/_{\! p}R] \to [U'/_{\! p}R']$ which we can compose with the stackyfication functor for $[U'/R']$ to get $[U/_{\! p}R] \to [U'/R']$. Then, by the universal property of the stackyfication functor $[U/_{\! p}R] \to [U/R]$, see Stacks, Lemma 8.9.2 we get $[U/R] \to [U'/R']$. $\square$

Let $B \to S$ and $f : (U, R, s, t, c) \to (U', R', s', t', c')$ be as in Lemma 77.21.1. In this situation, we define a third groupoid in algebraic spaces over $B$ as follows, using the language of $T$-valued points where $T$ is a (varying) scheme over $B$:

1. $U'' = U \times _{f, U', t'} R'$ so that a $T$-valued point is a pair $(u, r')$ with $f(u) = t'(r')$,

2. $R'' = R \times _{f \circ s, U', t'} R'$ so that a $T$-valued point is a pair $(r, r')$ with $f(s(r)) = t'(r')$,

3. $s'' : R'' \to U''$ is given by $s''(r, r') = (s(r), r')$,

4. $t'' : R'' \to U''$ is given by $t''(r, r') = (t(r), c'(f(r), r'))$,

5. $c'' : R'' \times _{s'', U'', t''} R'' \to R''$ is given by $c''((r_1, r'_1), (r_2, r'_2)) = (c(r_1, r_2), r'_2)$.

The formula for $c''$ makes sense as $s''(r_1, r'_1) = t''(r_2, r'_2)$. It is clear that $c''$ is associative. The identity $e''$ is given by $e''(u, r) = (e(u), r)$. The inverse of $(r, r')$ is given by $(i(r), c'(f(r), r'))$. Thus we do indeed get a groupoid in algebraic spaces over $B$.

Clearly the maps $U'' \to U$ and $R'' \to R$ define a morphism $g : (U'', R'', s'', t'', c'') \to (U, R, s, t, c)$ of groupoids in algebraic spaces over $B$. Moreover, the maps $U'' \to U'$, $(u, r') \mapsto s'(r')$ and $R'' \to U'$, $(r, r') \mapsto s'(r')$ show that in fact $(U'', R'', s'', t'', c'')$ is a groupoid in algebraic spaces over $U'$.

Lemma 77.21.2. Notation and assumption as in Lemma 77.21.1. Let $(U'', R'', s'', t'', c'')$ be the groupoid in algebraic spaces over $B$ constructed above. There is a $2$-commutative square

$\xymatrix{ [U''/R''] \ar[d] \ar[r]_{[g]} & [U/R] \ar[d]^{[f]} \\ \mathcal{S}_{U'} \ar[r] & [U'/R'] }$

which identifies $[U''/R'']$ with the $2$-fibre product.

Proof. The maps $[f]$ and $[g]$ come from an application of Lemma 77.21.1 and the other two maps come from Lemma 77.20.2 (and the fact that $(U'', R'', s'', t'', c'')$ lives over $U'$). To show the $2$-fibre product property, it suffices to prove the lemma for the diagram

$\xymatrix{ [U''/_{\! p}R''] \ar[d] \ar[r]_{[g]} & [U/_{\! p}R] \ar[d]^{[f]} \\ \mathcal{S}_{U'} \ar[r] & [U'/_{\! p}R'] }$

of categories fibred in groupoids, see Stacks, Lemma 8.9.3. In other words, it suffices to show that an object of the $2$-fibre product $\mathcal{S}_ U \times _{[U'/_{\! p}R']} [U/_{\! p}R]$ over $T$ corresponds to a $T$-valued point of $U''$ and similarly for morphisms. And of course this is exactly how we constructed $U''$ and $R''$ in the first place.

In detail, an object of $\mathcal{S}_ U \times _{[U'/_{\! p}R']} [U/_{\! p}R]$ over $T$ is a triple $(u', u, r')$ where $u'$ is a $T$-valued point of $U'$, $u$ is a $T$-valued point of $U$, and $r'$ is a morphism from $u'$ to $f(u)$ in $[U'/R']_ T$, i.e., $r'$ is a $T$-valued point of $R$ with $s'(r') = u'$ and $t'(r') = f(u)$. Clearly we can forget about $u'$ without losing information and we see that these objects are in one-to-one correspondence with $T$-valued points of $R''$.

Similarly for morphisms: Let $(u'_1, u_1, r'_1)$ and $(u'_2, u_2, r'_2)$ be two objects of the fibre product over $T$. Then a morphism from $(u'_2, u_2, r'_2)$ to $(u'_1, u_1, r'_1)$ is given by $(1, r)$ where $1 : u'_1 \to u'_2$ means simply $u'_1 = u'_2$ (this is so because $\mathcal{S}_ U$ is fibred in sets), and $r$ is a $T$-valued point of $R$ with $s(r) = u_2$, $t(r) = u_1$ and moreover $c'(f(r), r'_2) = r'_1$. Hence the arrow

$(1, r) : (u'_2, u_2, r'_2) \to (u'_1, u_1, r'_1)$

is completely determined by knowing the pair $(r, r'_2)$. Thus the functor of arrows is represented by $R''$, and moreover the morphisms $s''$, $t''$, and $c''$ clearly correspond to source, target and composition in the $2$-fibre product $\mathcal{S}_ U \times _{[U'/_{\! p}R']} [U/_{\! p}R]$. $\square$

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