Lemma 8.4.7. Let $\mathcal{C}$ be a site. Let $\mathcal{S}_1$, $\mathcal{S}_2$ be stacks over $\mathcal{C}$. Let $F : \mathcal{S}_1 \to \mathcal{S}_2$ be a $1$-morphism. Then the following are equivalent

1. $F$ is fully faithful,

2. for every $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and for every $x, y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_{1, U})$ the map

$F : \mathit{Mor}_{\mathcal{S}_1}(x, y) \longrightarrow \mathit{Mor}_{\mathcal{S}_2}(F(x), F(y))$

is an isomorphism of sheaves on $\mathcal{C}/U$.

Proof. Assume (1). For $U, x, y$ as in (2) the displayed map $F$ evaluates to the map $F : \mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_{1, V}}(x|_ V, y|_ V) \to \mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_{2, V}}(F(x|_ V), F(y|_ V))$ on an object $V$ of $\mathcal{C}$ lying over $U$. Now, since $F$ is fully faithful, the corresponding map $\mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_1}(x|_ V, y|_ V) \to \mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_2}(F(x|_ V), F(y|_ V))$ is a bijection. Morphisms in the fibre category $\mathcal{S}_{1, V}$ are exactly those morphisms between $x|_ V$ and $y|_ V$ in $\mathcal{S}_1$ lying over $\text{id}_ V$. Similarly, morphisms in the fibre category $\mathcal{S}_{2, V}$ are exactly those morphisms between $F(x|_ V)$ and $F(y|_ V)$ in $\mathcal{S}_2$ lying over $\text{id}_ V$. Thus we find that $F$ induces a bijection between these also. Hence (2) holds.

Assume (2). Suppose given objects $U$, $V$ of $\mathcal{C}$ and $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_{1, U})$ and $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_{1, V})$. To show that $F$ is fully faithful, it suffices to prove it induces a bijection on morphisms lying over a fixed $f : U \to V$. Choose a strongly Cartesian $f^*y \to y$ in $\mathcal{S}_1$ lying above $f$. This results in a bijection between the set of morphisms $x \to y$ in $\mathcal{S}_1$ lying over $f$ and $\mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_{1, U}}(x, f^*y)$. Since $F$ preserves strongly Cartesian morphisms as a $1$-morphism in the $2$-category of stacks over $\mathcal{C}$, we also get a bijection between the set of morphisms $F(x) \to F(y)$ in $\mathcal{S}_2$ lying over $f$ and $\mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_{2, U}}(F(x), F(f^*y))$. Since $F$ induces a bijection $\mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_{1, U}}(x, f^*y) \to \mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_{2, U}}(F(x), F(f^*y))$ we conclude (1) holds. $\square$

Comment #471 by Kestutis Cesnavicius on

Typo: should be $\Mor_{\cS_2}(F(x), F(y))$. Also, I think, it's not entirely clear what is meant by $\Mor_{\cS_1}(x, y)$, etc.: all morphisms in $\cS_1$ between the indicated objects or only morphisms in the fiber category? (I guess the later.) Perhaps it would be better to require that $\Mor_{(\cS_1)_U}(x, y) ...$ induced by $F$ be bijective?

Proof: The implication $(1) \Rightarrow (2)$ is clear. For the converse $(2) \Rightarrow (1)$, given $x \in \Ob(\cS_{1, U})$ and $y \in \Ob(\cS_{1, V})$, it suffices to restrict to morphisms lying over a fixed $f\colon U \ra V$ when checking the claimed bijectivity. Choose a strongly Cartesian $f^*y \ra y$ in $\cS_1$ lying above $f$. This results in an identification $\Mor_{\cS_1}(x, y)_{/f} \cong \Mor_{(\cS_1)_U}(x, f^*y)$. Since $F$ preserves strongly Cartesian morphisms, likewise $\Mor_{\cS_2}(F(x), F(y))_{/f} \cong \Mor_{(\cS_2)_U}(F(x), F(f^*y))$. Since $F\colon \Mor_{(\cS_1)_U}(x, f^*y) \isomto \Mor_{(\cS_2)_U}(F(x), F(f^*y))$ by assumption, the claim follows. Q.E.D.

P.S. Capitalization of Cartesian' is not consistent throughout Stacks project. Likewise forNoetherian' and 'Artinian' (I would capitalize all of these in all instances). Similarly for 'fiber' vs. 'fibre'.

Comment #472 by Kestutis Cesnavicius on

Typo: should be $\mathrm{Mor}_{\mathcal{S}_2}(F(x), F(y))$. Also, I think, it's not entirely clear what is meant by $\mathrm{Mor}_{\mathcal{S}_1}(x, y)$, etc.: all morphisms in $\mathcal{S}_1$ between the indicated objects or only morphisms in the fiber category? (I guess the later.) Perhaps it would be better to require that $\mathrm{Mor}_{(\mathcal{S}_1)_U}(x, y) ...$ induced by $F$ be bijective?

Proof: The implication $(1) \Rightarrow (2)$ is clear. For the converse $(2) \Rightarrow (1)$, given $x \in \mathrm{Ob}(\mathcal{S}_{1, U})$ and $y \in \mathrm{Ob}(\mathcal{S}_{1, V})$, it suffices to restrict to morphisms lying over a fixed $f\colon U \rightarrow V$ when checking the claimed bijectivity. Choose a strongly Cartesian $f^*y \rightarrow y$ in $\mathcal{S}_1$ lying above $f$. This results in an identification $\mathrm{Mor}_{\mathcal{S}_1}(x, y)_{/f} \cong \mathrm{Mor}_{(\mathcal{S}_1)_U}(x, f^*y)$. Since $F$ preserves strongly Cartesian morphisms, likewise $\mathrm{Mor}_{\mathcal{S}_2}(F(x), F(y))_{/f} \cong \mathrm{Mor}_{(\mathcal{S}_2)_U}(F(x), F(f^*y))$. Since $F\colon \mathrm{Mor}_{(\mathcal{S}_1)_U}(x, f^*y) \xrightarrow{\sim} \mathrm{Mor}_{(\mathcal{S}_2)_U}(F(x), F(f^*y))$ by assumption, the claim follows. Q.E.D.

P.S. Capitalization of Cartesian' is not consistent throughout Stacks project. Likewise forNoetherian' and 'Artinian' (I would capitalize all of these in all instances). Similarly for 'fiber' vs. 'fibre'.

Comment #486 by on

If $\mathcal{S}$ is a category fibred over a category $\mathcal{C}$, and if $x, y$ are objects lying over $U$, then the symbol $\mathit{Mor}_\mathcal{S}(x, y)$ is a presheaf of sets on $\mathcal{C}/U$. This construction is discussed in Definition 8.2.2.

Thanks for the proof and the typo! Added here. I slightly retyped your proof to make is very clear (I hope) what is really going on here. Also, I think there should be a lemma in the chapter on categories where we talk about when functors between fibred categories are fully faithful (i.e., the trivial thing of comparing morphisms in the whole category with morphisms in the fibre categories).

Yes, slowly over time we try to improve notation. I think cartesian should be lower case. I think Noetherian should be upper case. I am not sure about artinian. Fibre versus fibre. Glueing versus gluing. Etc, etc. Too much to fix all of these and people have different opinions. What is more important is to get the mathematics right!

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