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.

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

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

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

**Proof.**
The maps $[f]$ and $[g]$ come from an application of Lemma 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

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

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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