Lemma 8.12.6. With notation and assumptions as in Lemma 8.12.5. Set $u_ p\mathcal{S} = R^{-1}u_{pp}\mathcal{S}$, see Categories, Section 4.27. Then $u_ p\mathcal{S}$ is a fibred category over $\mathcal{D}$.

Proof. We use the description of $u_ p\mathcal{S}$ given just above Categories, Lemma 4.27.11. Note that the functor $p_{pp} : u_{pp}\mathcal{S} \to \mathcal{D}$ transforms every element of $R$ to an identity morphism. Hence by Categories, Lemma 4.27.16 we obtain a canonical functor $p_ p : u_ p\mathcal{S} \to \mathcal{D}$ extending the given functor. This is how we think of $u_ p\mathcal{S}$ as a category over $\mathcal{D}$.

First we want to characterize the $\mathcal{D}$-strongly cartesian morphisms in $u_ p\mathcal{S}$. A morphism $f : X \to Y$ of $u_ p\mathcal{S}$ is the equivalence class of a pair $(f' : X' \to Y, r : X' \to X)$ with $r \in R$. In fact, in $u_ p\mathcal{S}$ we have $f = (f', 1) \circ (r, 1)^{-1}$ with obvious notation. Note that an isomorphism is always strongly cartesian, as are compositions of strongly cartesian morphisms, see Categories, Lemma 4.33.2. Hence $f$ is strongly cartesian if and only if $(f', 1)$ is so. Thus the following claim completely characterizes strongly cartesian morphisms. Claim: A morphism

$(a, b, \alpha ) : X_1 = (U_1, \phi _1 : V_1 \to u(U_1), x_1) \longrightarrow (U_2, \phi _2 : V_2 \to u(U_2), x_2) = X_2$

of $u_{pp}\mathcal{S}$ has image $f = ((a, b, \alpha ), 1)$ strongly cartesian in $u_ p\mathcal{S}$ if and only if $\alpha$ is a strongly cartesian morphism of $\mathcal{S}$.

Assume $\alpha$ strongly cartesian. Let $X = (U, \phi : V \to u(U), x)$ be another object, and let $f_2 : X \to X_2$ be a morphism of $u_ p\mathcal{S}$ such that $p_ p(f_2) = b \circ b_1$ for some $b_1 : U \to U_1$. To show that $f$ is strongly cartesian we have to show that there exists a unique morphism $f_1 : X \to X_1$ in $u_ p\mathcal{S}$ such that $p_ p(f_1) = b_1$ and $f_2 = f \circ f_1$ in $u_ p\mathcal{S}$. Write $f_2 = (f'_2 : X' \to X_2, r : X' \to X)$. Again we can write $f_2 = (f'_2, 1) \circ (r, 1)^{-1}$ in $u_ p\mathcal{S}$. Since $(r, 1)$ is an isomorphism whose image in $\mathcal{D}$ is an identity we see that finding a morphism $f_1 : X \to X_1$ with the required properties is the same thing as finding a morphism $f'_1 : X' \to X_1$ in $u_ p\mathcal{S}$ with $p(f'_1) = b_1$ and $f'_2 = f \circ f'_1$. Hence we may assume that $f_2$ is of the form $f_2 = ((a_2, b_2, \alpha _2), 1)$ with $b_2 = b \circ b_1$. Here is a picture

$\xymatrix{ & & (U_1, V_1 \to u(U_1), x_1) \ar[d]^{(a, b, \alpha )} \\ (U, V \to u(U), x) \ar[rr]^{(a_2, b_2, \alpha _2)} & & (U_2, V_2 \to u(U_2), x_2) }$

Now it is clear how to construct the morphism $f_1$. Namely, set $U' = U \times _{U_2} U_1$ with projections $c : U' \to U$ and $a_1 : U' \to U_1$. Pick a strongly cartesian morphism $\gamma : x' \to x$ lifting the morphism $c$. Since $b_2 = b \circ b_1$, and since $u(U') = u(U) \times _{u(U_2)} u(U_1)$ we see that $\phi ' = (\phi , \phi _1 \circ b_1) : V \to u(U')$. Since $\alpha$ is strongly cartesian, and $a \circ a_1 = a_2 \circ c = p(\alpha _2 \circ \gamma )$ there exists a morphism $\alpha _1 : x' \to x_1$ lifting $a_1$ such that $\alpha \circ \alpha _1 = \alpha _2 \circ \gamma$. Set $X' = (U', \phi ' : V \to u(U'), x')$. Thus we see that

$f_1 = ((a_1, b_1, \alpha _1) : X' \to X_1, (c, \text{id}_ V, \gamma ) : X' \to X) : X \longrightarrow X_1$

works, in fact the diagram

$\xymatrix{ (U', \phi ' : V \to u(U'), x') \ar[d]_{(c, \text{id}_ V, \gamma )} \ar[rr]_{(a_1, b_1, \alpha _1)} & & (U_1, V_1 \to u(U_1), x_1) \ar[d]^{(a, b, \alpha )} \\ (U, V \to u(U), x) \ar[rr]^{(a_2, b_2, \alpha _2)} & & (U_2, V_2 \to u(U_2), x_2) }$

is commutative by construction. This proves existence.

Next we prove uniqueness, still in the special case $f = ((a, b, \alpha ), 1)$ and $f_2 = ((a_2, b_2, \alpha _2), 1)$. We strongly advise the reader to skip this part. Suppose that $g_1, g'_1 : X \to X_1$ are two morphisms of $u_ p\mathcal{S}$ such that $p_ p(g_1) = p_ p(g'_1) = b_1$ and $f_2 = f \circ g_1 = f \circ g'_1$. Our goal is to show that $g_1 = g'_1$. By Categories, Lemma 4.27.13 we may represent $g_1$ and $g'_1$ as the equivalence classes of $(f_1 : X' \to X_1, r : X' \to X)$ and $(f'_1 : X' \to X_1, r : X' \to X)$ for some $r \in R$. By Categories, Lemma 4.27.14 we see that $f_2 = f \circ g_1 = f \circ g'_1$ means that there exists a morphism $r' : X'' \to X'$ in $u_{pp}\mathcal{S}$ such that $r' \circ r \in R$ and

$(a, b, \alpha ) \circ f_1 \circ r' = (a, b, \alpha ) \circ f'_1 \circ r' = (a_2, b_2, \alpha _2) \circ r'$

in $u_{pp}\mathcal{S}$. Note that now $g_1$ is represented by $(f_1 \circ r', r \circ r')$ and similarly for $g'_1$. Hence we may assume that

$(a, b, \alpha ) \circ f_1 = (a, b, \alpha ) \circ f'_1 = (a_2, b_2, \alpha _2).$

Write $r = (c, \text{id}_ V, \gamma ) : (U', \phi ' : V \to u(U'), x')$, $f_1 = (a_1, b_1, \alpha _1)$, and $f'_1 = (a'_1, b_1, \alpha '_1)$. Here we have used the condition that $p_ p(g_1) = p_ p(g'_1)$. The equalities above are now equivalent to $a \circ a_1 = a \circ a'_1 = a_2 \circ c$ and $\alpha \circ \alpha _1 = \alpha \circ \alpha '_1 = \alpha _2 \circ \gamma$. It need not be the case that $a_1 = a'_1$ in this situation. Thus we have to precompose by one more morphism from $R$. Namely, let $U'' = \text{Eq}(a_1, a'_1)$ be the equalizer of $a_1$ and $a'_1$ which is a subobject of $U'$. Denote $c' : U'' \to U'$ the canonical monomorphism. Because of the relations among the morphisms above we see that $V \to u(U')$ maps into $u(U'') = u(\text{Eq}(a_1, a'_1)) = \text{Eq}(u(a_1), u(a'_1))$. Hence we get a new object $(U'', \phi '' : V \to u(U''), x'')$, where $\gamma ' : x'' \to x'$ is a strongly cartesian morphism lifting $\gamma$. Then we see that we may precompose $f_1$ and $f'_1$ with the element $(c', \text{id}_ V, \gamma ')$ of $R$. After doing this, i.e., replacing $(U', \phi ' : V \to u(U'), x')$ with $(U'', \phi '' : V \to u(U''), x'')$, we get back to the previous situation where in addition we now have that $a_1 = a'_1$. In this case it follows formally from the fact that $\alpha$ is strongly cartesian (!) that $\alpha _1 = \alpha '_1$. This shows that $g_1 = g'_1$ as desired.

We omit the proof of the fact that for any strongly cartesian morphism of $u_ p\mathcal{S}$ of the form $((a, b, \alpha ), 1)$ the morphism $\alpha$ is strongly cartesian in $\mathcal{S}$. (We do not need the characterization of strongly cartesian morphisms in the rest of the proof, although we do use it later in this section.)

Let $(U, \phi : V \to u(U), x)$ be an object of $u_ p\mathcal{S}$. Let $b : V' \to V$ be a morphism of $\mathcal{D}$. Then the morphism

$(\text{id}_ U, b, \text{id}_ x) : (U, \phi \circ b : V' \to u(U), x) \longrightarrow (U, \phi : V \to u(U), x)$

is strongly cartesian by the result of the preceding paragraphs and we win. $\square$

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