Let $\mathbf{C}$ be a small category. For a presheaf $P \in \what{\mathbf{C}}$, let $\intord P$ denote the category of elements of $P$, whose elements are pairs $(X \in \mathbf{C}, x \in P(C))$ and whose morphisms $(X, x) \xrightarrow{f} (Y, y)$ are maps $f : X \rightarrow Y$ such that $P(f)(y) = x$. There is a natrual projection $\pi_P : \intord P \rightarrow \mathbf{C}$, and we denote the Yoneda embedding as $\mathbf{y} : \mathbf{C} \rightarrow \what{\mathbf{C}}$.

Theorem: For any presheaf $P$, we have:

$$P \cong \colim \left( \textstyle\intord P \xrightarrow{\pi_P} \mathbf{C} \xrightarrow{\mathbf{y}} \what{\mathbf{C}} \right) = \colim(\mathbf{y} \circ \pi_P)$$ $$\newcommand{\diagram}{\mathbf{y} \circ \pi_P}$$

Proof: We will construct a limiting cocone over the diagram $\diagram$ with vertex $P$. The cone will be

$$\left( h_A \xrightarrow{\alpha^{A,a}} P \right)_{(A, a) \in \intord P}$$

where $\alpha^{A,a}$ is defined to be the natural transformation that $a \in P(A)$ corresponds to under the Yoneda correspondence. Precisely, $\alpha^{A,a}_X(g) = P(g)(a)$. To confirm that this actually forms a cone, we must show that for a map $(A, P(f)(b)) \xrightarrow{f} (B, b)$ that $\beta^{B,b} \circ h_f = \beta^{A,a}$. Indeed, for any $X$, we have:

Where the values are equal since $P$ is a contravariant functor. Now let

$$\left( h_A \xrightarrow{\beta^{A,a}} Q \right)_{(A, a) \in \intord P}$$

be another cocone. We must find a universal map $\gamma : P \rightarrow Q$ so that $\beta^{A,a} = \gamma \circ \alpha^{A,a}$. Define $\gamma_X : P(x) \rightarrow Q(x)$ by letting $\gamma_X(x)$ be precisely the element of $Q(x)$ that $\beta^{X,x}$ corresponds to under the Yoneda correspondence. Explicitly, $\gamma_X(x) = \beta^{X,x}_X(1_X)$. We confirm that this is actually a natural transformation, by considering the naturality square for a map $f : A \rightarrow B$:

Now, since $f$ can also be seen as a map $(A, P(f)(b)) \rightarrow (B, b)$ in $\intord P$, we have

$$\beta^{A, P(f)(b)}_A(1_A) = \left( \beta^{B,b}_A \circ (f \circ \dash) \right)(1_A) = \beta^{B,b}_A(f)$$

since the $\beta$'s form a cocone. On the other hand, since $\beta^{B,b}$ is a natural transformation itself, the diagram

commutes, and we have

$$Q(f)(\beta^{B,b}_B(1_B)) = \beta^{B,b}_A(f \circ 1_B) = \beta^{B,b}_A(f)$$

and the two expressions are equal.

Next, we must check that $\beta^{A,a} = \gamma \circ \alpha^{A,a}$. Consider:

The two expressions are equal due to the fact that $(X, P(g)(a)) \xrightarrow{g} (A, a)$ is a map in $\intord P$ and the $\beta$'s form a cone:

$$\beta^{X,P(g)(a)}_X(1_X) = \left( \beta^{A,a}_X \circ (g \circ \dash) \right)(1_X) = \beta^{A,a}_X(g)$$

Finally, for uniqueness, suppose that $\delta : P \rightarrow Q$ is another map with $\beta^{A,a} = \delta \circ \alpha^{A,a}$. Recall that for any $x \in P(X)$, we have $x = \alpha^{X,x}_X(1_X)$ (the Yoneda correspondence). As such, for any $x$:

$$\delta_X(x) = \delta_X(\alpha^{X,x}_X(1_X)) = \beta^{X,x}_X(1_X) = \gamma_X(x)$$

where the last step is by definition. Thus, $\delta = \gamma$.

Therefore, $P$ is the vertex of a limiting cocone over $\diagram$, showing the claim.