We give a proof of Yoneda's Lemma. Thus, we have to show that $\mathcal{C}\to\hat{\mathcal{C}}$ is a faithful functor. Let $X$ and $Y$ be two objects belonging to $\mathcal{C}$ we want to show that

$\psi : {\rm Hom}(X,Y)\to {\rm Hom}(X(.),Y(.))$ $f\mapsto (f_K:X(K)\to Y(K))_K$ is bijective.

Let's start with injectivity. Let $f$ and $g$ be two morphisms from $X$ to $Y$ which are having the same mappings for the points $f_K=g_K$ for all K. Let's show that $f=g$ What happens for the $X$ points? For the $X$ points, we have $f=g$ and the range of $f_X$ and of $g_X$ of the $X$ point of X which is $Id_X$ is exactly the $X$ points of $Y$ which are $f$ and $g$ Hence $f=g$

Now for surjectivity: let $\psi: X(.)\to Y(.)$ a morphism of functors. We need to show that this morphism comes from an arrow $f$ which should be the range of $Id_X$ by the map $\phi_X$ Thus, let $f=\psi_X(Id_X)$ Let's verify that $f_K=\psi_K$ for all $K$ Let $p:K\to X$ be a $K$ point of $X$ $p$ is a morphism between the two types of points $K$ and $X$ and in this case we have the following commutative diagram:

$I don't know how to do diagrams with latex, it's too hard$ If you make $Id(X)$ turn in the diagram one verifies that $\psi_K(p)=f\circ p=f_K(p)$ which proves the surjectivity.