I’m studying differentiable stacks from this paper (Gregory Ginot), and I’m trying to understand why the “Yoneda embedding” from the category of differentiable manifolds $\textbf{Man}$ to the category of “categories fibered in groupoids” (which for simplicity I’ll call “prestacks”) $\mathbf{Prestacks}$ is a fully faithful functor.
(If you don’t know about prestacks you can simply substitute $\mathbf{Prestacks} $ with the category of functors $\text{Fun}(-,\textbf{Man})$, where an arrow between $F:\mathbf{C}\to \mathbf{Man}$ and $G:\mathbf{D}\to \mathbf{Man}$ is a functor $H:\mathbf{C}\to\mathbf{D}$ such that $G\circ H=F$. This is possible because the former category is a full subcategory of the latter.)
The Yoneda embedding maps a manifold to the prestack $\pi_M:\text{Hom}(-,M)\to \textbf{Man}$. This prestack maps an object $f$ to $\text{dom}(f)$ and an it basically doesn’t modify the arrows.
Regarding the arrows, the Yoneda embedding maps an arrow $F$ to the standard pullback of smooth maps $F^\ast$.
To prove that the Yoneda embedding is a bijection I need to prove that for any manifolds $M,N$ the map induced by the Yoneda embedding
$$C^\infty(M,N)\to \text{Hom}_{\textbf{Prestacks}}(\pi_M,\pi_N),\ \ \ F\mapsto F^*$$
is bijective. The only possible candidate for the inverse map is $\alpha \mapsto \alpha(\text{id}_M)$. Obviously $F^\ast (\text{id}_M)=F$, so this direction is easy.
I’m having some trouble in the other direction. I can’t prove that $(\alpha (\text{id}_M))^*=\alpha$. Could you help me?
The word "prestacks" already has a well established meaning (the functors to descent data need to be fully faithful) so you probably shouldn't use the word prestack to refer to Grothendieck fibrations above $Man$.
The trick is to see that not any functor $yM\to YN$ is allowed as a map of Grothendieck fibrations, but only those which preserve cartesian arrows. An object in the $L$th fiber of $yM$ is a map $u:L\to M$ of smooth manifolds. There is additionally a horizontal cartesian arrow $u\leadsto id_M$ in the total category of $yM$ which indicates that $u$ as an object in $yM$ is the reindexing of $id_M$ along the morphism $u:L\to M$ in the base. Any map $\alpha:yM\to yN$ must send that cartesian arrow $u\leadsto id_M$ to a cartesian arrow $\alpha_L(u)\leadsto \alpha_M(id_M)$ also lying above $u:L\to M$ in the base. You can check that this implies that $\alpha_L(u) = \alpha_M(id_M)\circ u$. This means that the action of $\alpha$ is completely determined once you know what $\alpha_M(id_M)$ is.
The argument above is a special case of the argument which you do when you prove the fibered Yoneda lemma. The fibered Yoneda lemma says that \begin{align} Fib_{Man}(yM,C) \to C_M, A\mapsto A(id_M) \end{align} is an equivalence for every Grothendieck fibration $C$ above $Man$. The fact that $C$ is a Grothendieck fibration is very important, since that means that the action of some 1-cell $A:yM\to C$ or 2-cell out of $yM$ is completely determined by its action on the generic object $id_M$ of the discrete fibration $yM$.
Edit: To see that $\alpha(f) = \alpha(id_M)\circ f$ first note that there must be a cartesian arrow $\alpha(f) \leadsto \alpha(id_M)$ lying in above $f$ in the total category of $yN$. This cartesian arrow is the image of the cartesian arrow $f\leadsto id_M$ under $\alpha$. There is a canonical cartesian arrow $\alpha(id_M)\circ f \leadsto \alpha(id_M)$ above $f$ in the total category of $yN$. By the universal property of cartesian arrows we have that $\alpha(id_M)\circ f$ must be isomorphic to $\alpha(f)$ via a vertical isomorphism in the total category of $yN$. But the only vertical isomorphisms in the total category of $yN$ are identities, since $yN$ is a discrete fibered category. Hence we see that $\alpha(f) = \alpha(id_M)\circ f$.