Proof.
We will prove the first assertion; the second follows by a similar argument. Let $\mathscr {F}$ and $\mathscr {G}$ be functors from $\operatorname{\mathcal{C}}$ to the category of sets, and let $T: (\int _{\operatorname{\mathcal{C}}} \mathscr {F}) \rightarrow (\int _{\operatorname{\mathcal{C}}} \mathscr {G})$ be a functor for which the diagram
\[ \xymatrix@R =50pt@C=50pt{ \int _{\operatorname{\mathcal{C}}} \mathscr {F} \ar [rr]^-{T} \ar [dr] & & \int _{\operatorname{\mathcal{C}}} \mathscr {G} \ar [dl] \\ & \operatorname{\mathcal{C}}& } \]
is strictly commutative, where the vertical maps are the forgetful functors. We wish to show that there is a unique natural transformation of functors
\[ f: \mathscr {F} \rightarrow \mathscr {G} \quad \quad \{ f_ C: \mathscr {F}(C) \rightarrow \mathscr {G}(C) \} _{C \in \operatorname{\mathcal{C}}} \]
for which the functor $T$ is given on objects by the construction $T(C,x) = (C, f_ C(x) )$. Note that this requirement uniquely determines the function $f_ C: \mathscr {F}(C) \rightarrow \mathscr {G}(C)$ for each object $C \in \operatorname{\mathcal{C}}$. We must show that the resulting collection $\{ f_ C \} _{C \in \operatorname{\mathcal{C}}}$ is a natural transformation: that is, for every morphism $u: C \rightarrow D$ in the category $\operatorname{\mathcal{C}}$, the diagram of sets
\[ \xymatrix@R =50pt@C=50pt{ \mathscr {F}(C) \ar [r]^-{f_ C} \ar [d]^-{\mathscr {F}(u)} & \mathscr {G}(C) \ar [d]^-{ \mathscr {G}(u)} \\ \mathscr {F}(D) \ar [r]^-{ f_ D) } & \mathscr {G}(D) } \]
is commutative. Fix an element $x \in \mathscr {F}(C)$, so that $u$ can be regarded as a morphism from $(C, x)$ to $(D,\mathscr {F}(u)(x) )$ in the category $\int _{\operatorname{\mathcal{C}}} \mathscr {F}$. Applying the functor $T$, we deduce that $u$ can also be regarded as a morphism from $(C, f_ C(x))$ to $(D, f_ D( \mathscr {F}(u)(x) ) )$ in the category $\int _{\operatorname{\mathcal{C}}} \mathscr {G}$. It follows that $\mathscr {G}(u)( f_ C(x) ) = f_ D( \mathscr {F}(u)(x) )$, as desired.
$\square$