[proofplan]
The proof is a direct unpacking of the terminal-object definition of a limit. First, the projections of the cone $(L,(p_j))$ turn every morphism $u:N\to L$ into a cone from $N$ to $D$, so the displayed map is well-defined. If $(L,(p_j))$ is terminal among cones, the unique factorization property gives exactly existence and uniqueness of the preimage of every cone under $\Phi_N$. Conversely, if every $\Phi_N$ is bijective, applying this bijection to an arbitrary cone produces the unique comparison morphism required for terminality. Naturality follows because both sides act on a morphism $v:M\to N$ by precomposition.
[/proofplan]
[step:Verify that the displayed assignment sends morphisms into cones]
Fix an object $N\in \operatorname{Ob}(\mathcal C)$. Define
\begin{align*}
\Phi_N:\mathcal C(N,L)&\to \operatorname{Cone}(N,D)\\
u&\mapsto (p_j\circ u)_{j\in \operatorname{Ob}(J)}.
\end{align*}
To see that $\Phi_N$ is well-defined, let $u:N\to L$ be a morphism in $\mathcal C$. For every arrow $a:j\to k$ in $J$, the cone identity for $(L,(p_j))$ gives
\begin{align*}
D(a)\circ (p_j\circ u)=(D(a)\circ p_j)\circ u=p_k\circ u.
\end{align*}
Thus $(p_j\circ u)_{j\in \operatorname{Ob}(J)}$ is a cone from $N$ to $D$.
[/step]
[step:Derive bijectivity from the terminal cone property]
Assume that $(L,(p_j)_{j\in \operatorname{Ob}(J)})$ is a limit of $D$. By definition, this means that $(L,(p_j))$ is a terminal object in the category of cones over $D$.
Fix an object $N\in \operatorname{Ob}(\mathcal C)$ and a cone $(q_j)_{j\in \operatorname{Ob}(J)}\in \operatorname{Cone}(N,D)$. The terminal property applied to the cone $(N,(q_j))$ gives a unique morphism $u:N\to L$ such that, for every object $j\in \operatorname{Ob}(J)$,
\begin{align*}
p_j\circ u=q_j.
\end{align*}
This says exactly that $\Phi_N(u)=(q_j)_{j\in \operatorname{Ob}(J)}$, so $\Phi_N$ is surjective. If $u,u':N\to L$ satisfy $\Phi_N(u)=\Phi_N(u')$, then
\begin{align*}
p_j\circ u=p_j\circ u'
\end{align*}
for every $j\in \operatorname{Ob}(J)$. Both $u$ and $u'$ are morphisms of cones from $(N,(p_j\circ u))$ to $(L,(p_j))$, so uniqueness in the terminal property gives $u=u'$. Hence $\Phi_N$ is injective. Therefore $\Phi_N$ is bijective for every object $N\in \operatorname{Ob}(\mathcal C)$.
[/step]
[step:Recover the terminal cone property from bijectivity]
Assume conversely that $\Phi_N$ is a bijection for every object $N\in \operatorname{Ob}(\mathcal C)$. Let $(N,(q_j)_{j\in \operatorname{Ob}(J)})$ be an arbitrary cone over $D$. Since $\Phi_N$ is surjective, there exists a morphism $u:N\to L$ such that
\begin{align*}
\Phi_N(u)=(q_j)_{j\in \operatorname{Ob}(J)}.
\end{align*}
By the definition of $\Phi_N$, this means that, for every $j\in \operatorname{Ob}(J)$,
\begin{align*}
p_j\circ u=q_j.
\end{align*}
Thus $u$ is a morphism of cones from $(N,(q_j))$ to $(L,(p_j))$.
If $u':N\to L$ is another morphism of cones from $(N,(q_j))$ to $(L,(p_j))$, then $p_j\circ u'=q_j$ for every $j\in \operatorname{Ob}(J)$, so
\begin{align*}
\Phi_N(u')=(q_j)_{j\in \operatorname{Ob}(J)}=\Phi_N(u).
\end{align*}
Since $\Phi_N$ is injective, $u'=u$. Therefore every cone over $D$ admits a unique morphism to $(L,(p_j))$, so $(L,(p_j))$ is terminal in the category of cones over $D$. Hence $(L,(p_j))$ is a limit of $D$.
[/step]
[step:Check naturality by comparing the two precomposition operations]
Let $v:M\to N$ be a morphism in $\mathcal C$. The hom-functor sends $v$ to the precomposition map
\begin{align*}
\mathcal C(N,L)&\to \mathcal C(M,L)\\
u&\mapsto u\circ v,
\end{align*}
and the cone functor sends $v$ to the precomposition map
\begin{align*}
\operatorname{Cone}(N,D)&\to \operatorname{Cone}(M,D)\\
(q_j)_{j\in \operatorname{Ob}(J)}&\mapsto (q_j\circ v)_{j\in \operatorname{Ob}(J)}.
\end{align*}
For every morphism $u:N\to L$,
\begin{align*}
\Phi_M(u\circ v)
&=(p_j\circ u\circ v)_{j\in \operatorname{Ob}(J)}\\
&=((p_j\circ u)\circ v)_{j\in \operatorname{Ob}(J)}.
\end{align*}
This is precisely the image of $\Phi_N(u)$ under precomposition by $v$. Hence the square expressing naturality in $N$ commutes. This completes the proof.
[/step]
