Subsection 7.3.5. (0PHE) · ScalingStacks

7.3.5. Pullback

Let $f:Z\to Z^{\prime}$ be a morphism of curves. We define a non-multiplicative “functor” $f^{\#}:\operatorname{add}\nolimits({\mathcal{S}}(Z^{\prime},1))\to\operatorname{add}\nolimits({\mathcal{S}}(Z,1))$ . It commutes with coproducts but is not a functor, i.e., it is not compatible with composition for a general $f$ . We put $f^{\#}(z^{\prime})=\coprod_{z\in f^{-1}(z^{\prime})}z$ . Given $\zeta^{\prime}\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z^{\prime},1)}(z^{\prime}_{1},z^{\prime}_{2})$ non-zero, we define $f^{\#}(\zeta^{\prime})$ to be

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$\operatorname{id}\nolimits$ if $\zeta^{\prime}=\operatorname{id}\nolimits$
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$0$ if $\zeta^{\prime}$ does not lift to an admissible class of paths in $Z$

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the composition

\coprod_{z\in f^{-1}(z^{\prime}_{1})}z\xrightarrow{\text{projection}}z_{1}\xrightarrow{\zeta}z_{2}\xrightarrow{\text{inclusion}}\coprod_{z\in f^{-1}(z^{\prime}_{2})}z

where $\zeta:z_{1}\to z_{2}$ is the unique lift of $\zeta^{\prime}$ , otherwise (cf Lemma 7.3.14).

We denote by $f^{-1}(\zeta^{\prime})$ the set of admissible lifts of $\zeta^{\prime}$ . We have $f^{\#}(\zeta^{\prime})=\sum_{\zeta\in f^{-1}(\zeta^{\prime})}\zeta$ .

Given $\zeta^{\prime}_{1}\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z^{\prime},1)}(z^{\prime}_{1},z^{\prime}_{2})$ and $\zeta^{\prime}_{2}\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z^{\prime},1)}(z^{\prime}_{2},z^{\prime}_{3})$ such that $f^{\#}(\zeta^{\prime}_{1})\neq 0$ and $f^{\#}(\zeta^{\prime}_{2})\neq 0$ , we have $f^{\#}(\zeta^{\prime}_{2})f^{\#}(\zeta^{\prime}_{1})=f^{\#}(\zeta^{\prime}_{2}\zeta^{\prime}_{1})$ (cf Lemma 7.3.13).

Given $f^{\prime}:Z^{\prime}\to Z^{\prime\prime}$ a morphism of curves, we have $(f^{\prime}f)^{\#}=f^{\#}f^{\prime\#}$ .

0PA4

Lemma 7.3.16. Let $\gamma^{\prime}$ be a smooth path in $Z^{\prime}$ . Consider the following assertions:

(1)

$\gamma^{\prime}$ lifts to a smooth path in $Z$
(2)

$\gamma^{\prime}([0,1])\subset f(Z)$ .
(3)

$[\gamma^{\prime}]$ lifts to a smooth homotopy class in $Z$
(4)

$\operatorname{supp}\nolimits([\gamma^{\prime}])\subset f(Z)$ .

We have $(1)\Leftrightarrow(2)\Rightarrow(3)\Leftrightarrow(4)$ .

Assume $f$ is strict. Then $(3)\Rightarrow(2)$ . Furthermore, if $\gamma^{\prime}$ is admissible and it lifts to a smooth path in $Z$ , then that path is admissible.

0PA5

Proof. The implications $(1)\Rightarrow(2)$ , $(1)\Rightarrow(3)\Rightarrow(4)$ are clear. We can assume that $\gamma^{\prime}$ is not constant, for otherwise the other implications are trivial.

Assume $(2)$ . Let $\hat{f}:\hat{Z}\to\hat{Z}^{\prime}$ be the map between non-singular covers corresponding to $f$ . Since $\gamma^{\prime}$ is smooth, it lifts uniquely to a path $\hat{\gamma}^{\prime}$ on $\hat{Z}^{\prime}$ and $\hat{\gamma}^{\prime}([0,1])\subset\hat{f}(\hat{Z})$ . Since $\hat{f}$ is an open embedding, it follows that $\hat{\gamma}^{\prime}$ is the image of a path of $\hat{Z}$ . Its image in $Z$ is a smooth path that lifts $\gamma^{\prime}$ , hence $(1)$ holds.

Assume $(4)$ . Let $\gamma^{\prime}_{0}$ be a minimal smooth path homotopic to $\gamma^{\prime}$ (cf Properties 7.3.5(4)). We have $\gamma^{\prime}_{0}([0,1])=\operatorname{supp}\nolimits([\gamma^{\prime}])\subset f(Z)$ , hence $\gamma^{\prime}_{0}$ lifts to a smooth path in $Z$ . So $(3)$ holds.

Assume $(3)$ and $f$ is strict. Note that $\gamma^{\prime}([0,1])\cap Z^{\prime}_{o}=\operatorname{supp}\nolimits([\gamma^{\prime}])\cap Z^{\prime}_{o}$ and $\gamma^{\prime}([0,1])\cap Z^{\prime}_{u}$ is contained in the union of the connected components of $Z^{\prime}_{u}$ that have a non-empty intersection with $\operatorname{supp}\nolimits([\gamma^{\prime}])$ (Properties 7.3.2(1)). Since $f(Z_{u})$ is open and closed in $Z^{\prime}_{u}$ , it follows that $\gamma^{\prime}([0,1])\subset f(Z)$ , so $(2)$ holds.

Assume $\gamma^{\prime}$ is admissible and lifts to $Z$ . Since $f$ is strict, it follows that the lift is oriented. ∎

Since quotient maps are strict, we have the following consequence of Lemma 7.3.16.

0PA6

Lemma 7.3.17. Assume $f$ is the quotient map of $Z$ by a finite relation. Every non-constant admissible path in $Z^{\prime}$ lifts uniquely to a path in $Z$ and that lift is admissible.

0PA7

Proposition 7.3.18. If $f$ is strict, then $f^{\#}:\operatorname{add}\nolimits({\mathcal{S}}(Z^{\prime},1))\to\operatorname{add}\nolimits({\mathcal{S}}(Z,1))$ is a functor.

0PA8

Proof. We need to check that $f^{\#}$ is compatible with composition. This is clear if $Z$ and $Z^{\prime}$ are non-singular. In general, consider two maps $\zeta^{\prime}_{1}$ and $\zeta^{\prime}_{2}$ in ${\mathcal{S}}(Z^{\prime},1)$ such that $f^{\#}(\zeta^{\prime}_{2}\circ\zeta^{\prime}_{1})\neq 0$ . Let $\hat{f}:\hat{Z}\to\hat{Z}^{\prime}$ be the map corresponding to $f$ between non-singular covers $q:\hat{Z}\to Z$ and $q^{\prime}:\hat{Z}^{\prime}\to Z^{\prime}$ .

We have

	$\displaystyle q^{\#}f^{\#}(\zeta^{\prime}_{2}\circ\zeta^{\prime}_{1})$	$\displaystyle=\hat{f}^{\#}q^{\prime\#}(\zeta^{\prime}_{2}\circ\zeta^{\prime}_{1})=\hat{f}^{\#}\bigl(q^{\prime\#}(\zeta^{\prime}_{2})\circ q^{\prime\#}(\zeta^{\prime}_{1})\bigr)=\bigl(\hat{f}^{\#}q^{\prime\#}(\zeta^{\prime}_{2})\bigr)\circ\bigl(\hat{f}^{\#}q^{\prime\#}(\zeta^{\prime}_{1})\bigr)$
		$\displaystyle=q^{\#}f^{\#}(\zeta^{\prime}_{2})\circ q^{\#}f^{\#}(\zeta^{\prime}_{1}),$

hence $f^{\#}(\zeta^{\prime}_{1})\neq 0$ and $f^{\#}(\zeta^{\prime}_{2})\neq 0$ since $q^{\#}f^{\#}(\zeta^{\prime}_{2}\circ\zeta^{\prime}_{1})\neq 0$ by Lemma 7.3.17. It follows that $f^{\#}(\zeta^{\prime}_{2}\circ\zeta^{\prime}_{1})=f^{\#}(\zeta^{\prime}_{2})\circ f^{\#}(\zeta^{\prime}_{1})$ . ∎

The construction $Z\mapsto\operatorname{add}\nolimits({\mathcal{S}}(Z,1))$ and $f\mapsto f^{\#}$ defines a contravariant functor from the category of curves with strict morphisms to the category of ${\mathbf{F}}_{2}$ -linear categories.

Lemma 7.3.17 and Proposition 7.3.18 have the following consequence.

0PA9

Proposition 7.3.19. Let $Z$ be a curve with an admissible relation $\sim$ and let $q:Z\to Z/\!\!\sim$ be the quotient map. The functor $q^{\#}:\operatorname{add}\nolimits({\mathcal{S}}(Z/\!\!\sim,1))\to\operatorname{add}\nolimits({\mathcal{S}}(Z,1))$ is faithful.

Note that Proposition 7.3.19 provides an identification of ${\mathcal{S}}(Z/\!\!\sim,1)$ with a (non-full) subcategory of $\operatorname{add}\nolimits({\mathcal{S}}(Z,1))$ .

0PAA

Example 7.3.20. We describe the image by the map $f^{\#}$ of two paths, the first of which is the constant path at the singular point of $Z_{exc}$ (we draw the lifts in the non-singular cover).

Andrew Manion, Raphael Rouquier

Original source: arXiv:2009.09627v2