[proofplan] The proof is a direct comparison of the two definitions. We first unpack the morphisms in the category of elements: a morphism from $(A,u)$ to $(X,x)$ is exactly a morphism $f:A \to X$ in $\mathcal C$ satisfying $F(f)(u)=x$. The universal-element condition says that for every such target pair $(X,x)$ there is exactly one such $f$, while the initial-object condition says exactly that there is exactly one morphism from $(A,u)$ to every object of $\int_{\mathcal C}F$. [/proofplan] [step:Unpack morphisms out of $(A,u)$ in the category of elements] Let $(X,x)$ be an object of $\int_{\mathcal C}F$, so $X$ is an object of $\mathcal C$ and $x \in F(X)$. By definition of the category of elements, a morphism \begin{align*} (A,u) \longrightarrow (X,x) \end{align*} in $\int_{\mathcal C}F$ is a morphism $f:A \to X$ in $\mathcal C$ such that \begin{align*} F(f)(u)=x. \end{align*} Thus the hom-set from $(A,u)$ to $(X,x)$ in $\int_{\mathcal C}F$ is precisely \begin{align*} \operatorname{Hom}_{\int_{\mathcal C}F}((A,u),(X,x)) = \{f \in \operatorname{Hom}_{\mathcal C}(A,X) : F(f)(u)=x\}. \end{align*} [/step] [step:Translate the universal-element condition into the same hom-set condition] By definition, $(A,u)$ is a universal element of $F$ exactly when, for every object $X$ of $\mathcal C$ and every element $x \in F(X)$, there exists a unique morphism $f:A \to X$ in $\mathcal C$ satisfying \begin{align*} F(f)(u)=x. \end{align*} Equivalently, for every object $(X,x)$ of $\int_{\mathcal C}F$, the set \begin{align*} \{f \in \operatorname{Hom}_{\mathcal C}(A,X) : F(f)(u)=x\} \end{align*} has exactly one element. [/step] [step:Translate the initial-object condition into the same uniqueness condition] By definition, $(A,u)$ is an initial object of $\int_{\mathcal C}F$ exactly when, for every object $(X,x)$ of $\int_{\mathcal C}F$, there exists a unique morphism \begin{align*} (A,u) \longrightarrow (X,x) \end{align*} in $\int_{\mathcal C}F$. Using the description of morphisms from the first step, this is exactly the assertion that, for every object $X$ of $\mathcal C$ and every element $x \in F(X)$, there exists a unique morphism $f:A \to X$ in $\mathcal C$ satisfying \begin{align*} F(f)(u)=x. \end{align*} This is precisely the universal-element condition for $(A,u)$. [/step] [step:Conclude the equivalence] The universal-element condition and the initial-object condition have both been shown to be the same statement: for every object $X$ of $\mathcal C$ and every $x \in F(X)$, there is a unique morphism $f:A \to X$ such that $F(f)(u)=x$. Therefore $(A,u)$ is a universal element of $F$ if and only if $(A,u)$ is an initial object of $\int_{\mathcal C}F$. [/step]