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Let $\mathcal C$ be a category. Grothendieck[1] defines the category of arrows in $\mathcal C$ to be the category of functors $$\Delta^1 \to \mathcal C,$$ where $\Delta^1$ is the category consisting of two objects $0, 1$ where we have the identity arrows, and one additional arrow $0 \to 1$. So in effect, an object in the category of arrows is is just an arrow $$X \to Y$$ between two objects $X, Y \in \mathcal C$, and a morphism of two arrows is a natural transformation of functors, i.e. a commutative diagram $\require{AMScd}$ \begin{CD} X' @>>>X\\ @VVV @VVV\\ Y' @>>> Y .\end{CD} Let's denote this category by $\DeclareMathOperator{\Arr}{Arr}\Arr(\mathcal C)$. Grothendieck then claims that the fiber $\Arr(\mathcal C)_S$ over some $S \in \mathcal C$ of the forgetful functor $$\Arr(\mathcal C) \ni(X \to Y) \mapsto Y \in \mathcal C$$ is (canonically isomorphic to) the category $\mathcal C/S$ of objects of $\mathcal C$ over $S$. I don't understand this. Clearly, both categories have as objects arrows $$X \to S.$$ But I think that the morphisms differ. In the fiber $\Arr(\mathcal C)_S$ a morphism is a commutative square \begin{CD} X @>\beta>>Y\\ @VpVV @VVqV\\ S @>\alpha>> S ,\end{CD} whereas in $\mathcal C/S$ a morphism is a triangle \begin{align*} X \rightarrow Y\\ \searrow \swarrow\,\, \\ S .\,\,\,\,\, \end{align*} To illustrate my point, consider the following schemes over $\mathbb A^2$ (everything over $\mathbb C$): $$\mathbb P^1 \times \mathbb A^2 \xrightarrow{\operatorname{pr}_2} \mathbb A^2, \quad \mathbb P^1 \xrightarrow{\text{const.}} \mathbb A^2.$$ Then in $(\text{Sch}/\mathbb A^2)$ there will is no morphism $\mathbb P^1 \times \mathbb A^2 \to \mathbb P^1$. However in $\Arr(\text{Sch})_{\mathbb A^2}$, the projection $\operatorname{pr}_1: \mathbb P^1 \times \mathbb A^2 \to \mathbb P^1$ can be made into a morphism, if complemented by the constant map $\mathbb A^2 \to \mathbb A^2$ with the same image as $\mathbb P^1 \to \mathbb A^2$. Did I make any mistake here?

[1] SGA 1, Exposé VI, Example 11 (a)

red_trumpet
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1 Answers1

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You seem to be missing a small but important detail in the definition of the fiber category. Given a functor $F : \mathcal{C}\to\mathcal{D}$ and an object $X\in\mathcal{D},$ the fiber of $F$ over $X$ is defined to be the category of $\widetilde{X}\in\mathcal{C}$ such that $F(\widetilde{X}) = X$ with morphisms $u : \widetilde{X}\to\widetilde{X}'$ such that $F(u) = \operatorname{id}_X.$ I've quoted the definition from SGA below, found on page 158 at the start of section 4.

Soit $\mathcal{F}$ une catégorie sur $\mathcal{E}$, et soit $S\in\operatorname{Ob}(\mathcal{E}).$ On appelle catégorie-fibre de $\mathcal{F}$ en $S$ le sous-catégorie $\mathcal{F}_S$ de $\mathcal{F}$ image réciproque de la sous-catégorie ponctuelle de $\mathcal{E}$ définie par $S.$ Donc les objets de $\mathcal{F}_S$ sont les objets $\xi$ de $\mathcal{F}$ tels que $p(\xi) = S,$ ses morphismes sont les morphismes $u$ de $\mathcal{F}$ tels que $p(u) = \operatorname{id}_S,$ i.e. les $S$-morphismes dans $\mathcal{F}.$

Stahl
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