A morphism of groupoids in algebraic spaces gives an associated morphism of quotient stacks.
Lemma 78.21.1. 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 (78.20.0.1). 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 78.21.1. 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 $B$:
$U'' = U \times _{f, U', t'} R'$ so that a $T$-valued point is a pair $(u, r')$ with $f(u) = t'(r')$,
$R'' = R \times _{f \circ s, U', t'} R'$ so that a $T$-valued point is a pair $(r, r')$ with $f(s(r)) = t'(r')$,
$s'' : R'' \to U''$ is given by $s''(r, r') = (s(r), r')$,
$t'' : R'' \to U''$ is given by $t''(r, r') = (t(r), c'(f(r), r'))$,
$c'' : R'' \times _{s'', U'', t''} R'' \to R''$ 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 78.21.2. Notation and assumption as in Lemma 78.21.1. Let $(U'', R'', s'', t'', c'')$ be the groupoid in algebraic spaces over $B$ constructed above. There is a $2$-commutative square which identifies $[U''/R'']$ with the $2$-fibre product.
\[ \xymatrix{ [U''/R''] \ar[d] \ar[r]_{[g]} & [U/R] \ar[d]^{[f]} \\ \mathcal{S}_{U'} \ar[r] & [U'/R'] } \]
Proof. The maps $[f]$ and $[g]$ come from an application of Lemma 78.21.1 and the other two maps come from Lemma 78.20.2 (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
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
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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