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Let $C$ be a site and $p : F \mapsto C$ $p' : F' \mapsto C$ two categories fibred in groupoids over $C$ which are stacks (see, e.g., the definition here https://stacks.math.columbia.edu/tag/0268). For an object $X$, I will wright $F(X)$ the category whose objects $Y$ verify $p(Y) = X$ and whose arrows $f$ verify $p(f) = Id_{X}$. By definition of a fibred categories, if $f : X \mapsto X'$ is a morphism in $C$, then I can choose for each $x \in F(X)$ a pullback $f^{*}x$ by $f$ (and two such pullbacks are isomorph).


The problem : Let consider a morphism $g : F \mapsto F'$ of fibred categories. I suppose that for each object $X$ in $C$, the induced functor $F(X) \mapsto F'(X)$ is fully faithful. I also suppose for each $x $ in $F'(X)$, there exists a cover $\{f_{i} : X_{i} \mapsto X\}$ such that $f_{i}^{*}x \in F'(X_{i})$ is in the essential image of $g$.

I want to show $g$ is an isomorphism. I try a few things which give nothing. Do you have ideas?

Many thanks.

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    $\begingroup$ This is Tag 046N, isn't it? $\endgroup$ Commented Jun 2 at 13:25
  • $\begingroup$ It works! Many thanks! $\endgroup$
    – Analyse300
    Commented Jun 2 at 16:38
  • $\begingroup$ You are using the wrong arrow $\mapsto$ where you should write $\to.$ The $\mapsto$ arrow is used to describe a mapping on terms, as in $f:\mathbb R\to \mathbb R, x\mapsto x^2.$ $\endgroup$ Commented Jun 3 at 6:43
  • $\begingroup$ Indeed! You are right all morphism are not mappings! Thanks. $\endgroup$
    – Analyse300
    Commented Jun 3 at 11:19
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    $\begingroup$ What @Kevin Carlson is trying to say is that when you write $f: X \rightarrow Y$, it means that $X$ (that is, $Y$) is the domain (that is, the codomain) of the morphism $f$. This has nothing to do with whether or not $f$ is a function. But, if $f$ is a function, then we write $f: x \mapsto y_x$ to mean that $f(x)=y_x$ for all $x \in X$. It is misleading to use $\mapsto$ for $\rightarrow$. $\endgroup$ Commented Jul 31 at 17:53

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