Theorem: Diagonal Functor on Product Category has Left Adjoint Iff Category has Binary Coproducts/Necessary Condition

Let $\mathbf C$ be a locally small category.
Let $\mathbf C \times \mathbf C$ be the product category of $\mathbf C$ with itself.
Let $\Delta: \mathbf C \to \mathbf C \times \mathbf C$ denote the diagonal functor.
Let $\Delta$ have a left adjoint.
Then $\mathbf C$ has all binary coproducts
such that the left adjoint is the coproduct functor.

Proof:
Let $\oplus:\mathbf {C \times C} \to \mathbf C$ be a left adjoint of $\Delta$.
==== $\mathbf C$ has all Binary Products ====
By definition of left adjoint there exists an adjunction $\tuple{\oplus, \Delta, \alpha}$.
Let $\iota : \operatorname{id}_{\mathbf {C \times C} } \to \Delta \oplus$ denote the unit of adjunction $\tuple{\oplus, \Delta, \alpha}$.
From Morphism of Unit of Adjunction is Universal:
:for each object $\tuple{C1, C_2} \in \mathbf {C \times C}$:
::the morphism $\iota_{\tuple{C1, C_2} }: \tuple{C1, C_2} \to \map {\paren{\Delta \oplus} } {C1, C_2}$ is a universal morphism from $\map \oplus {C_1, C_2}$ to $\Delta$
By definition of diagonal functor:
:$(1):\quad$for each object $C \in \mathbf C$:
::$\map {\paren{\Delta \oplus} } {C_1, C_2} = \tuple{\map \oplus {C_1, C_2}, \map \oplus {C_1, C_2} }$
and:
:$(2):\quad$for each morphism $f \in \mathbf C$:
::$\map {\paren{\Delta \oplus} } {f_1, f_2} = \tuple{\map \oplus {f_1, f_2}, \map \oplus {f_1, f_2} }$
By definition of a morphism of product cateogry:
:$(3):\quad$there exists morphisms $i_1: C_1 \to \oplus \tuple{C_1, C_2}$ and $i_2: C_2 \to \oplus \tuple{C_1, C_2}$ such that:
::$\iota_{\tuple{C_1, C_2}} = \tuple{i_1, i_2}$
By definition of universal morphism:
:$(4):\quad$for each object $\tuple{C_1, C_2} \in \mathbf {C \times C}$ and for each morphism $f: \tuple{C_1, C_2} \to \Delta X \in \mathbf{C \times C}$:
::$\exists f': \oplus \tuple{C_1, C_2} \to X  : \Delta f' \circ \iota_{\tuple{C_1, C_2} } = f$
It will be shown that for objects $C_1, C_2 \in \mathbf C$:
:$\tuple{\oplus \tuple{C_1, C_2}, i_1, i_2}$ is a binary coproduct.
Let $f_1: C_1 \to X, f_2: C_2 \to X$ be morphisms of $\mathbf C$.
Consider the morphism $\tuple{f_1, f_2} : \tuple{C_1, C_2} \to \tuple{X, X}$ of the product category $\mathbf{C \times C}$.
By definition of diagonal functor:
:$\tuple{f_1, f_2} : \tuple{C_1, C_2} \to \Delta X \in \mathbf{C \times C}$
By $(4)$:
:$\exists f': \oplus \tuple{C_1, C_2} \to X: \Delta f' \circ \iota_{\tuple{C_1, C_2} } = \tuple{f_1, f_2}$
By $(3)$:
:$\exists f': \oplus \tuple{C_1, C_2} \to X : \tuple{f', f'} \circ \tuple{i_1, i_2} = \tuple{f_1, f_2}$
By definition of composition in product category:
:$\exists f': \oplus \tuple{C_1, C_2} \to X : \tuple{f' \circ i_1, f' \circ i_2} = \tuple{f_1, f_2}$
That is:
:$\exists f': X \to \oplus \tuple{C_1, C_2} : f' \circ i_1 = f_1, f' \circ i_2 = f_2$
Hence $\tuple {\map \oplus {C_1,  C_2}, i_1, i_2 }$ is a binary coproduct by definition. 
It follows that for all objects $\tuple{C_1, C_2} \in \mathbf {C \times C}$:
:$(5):\quad$there exists morphisms $i_1: C_1 \to \map \oplus {C_1, C_2}$ and $i_2: C_2 \to \map \otimes {C_1, C_2}$ such that:
::$\tuple {\map \oplus {C_1,  C_2}, i_1, i_2 }$ is a binary coproduct
:and:
::$\iota_{\tuple{C_1, C_2}} = \tuple{i_1, i_2}$
It follows that $\mathbf C$ has all binary coproducts.
{{qed|lemma}}
==== Left Adjoint $\oplus$ is Coproduct functor ====
Let $\sqcup$ denote the product functor.
For all objects $\tuple{C_1, C_2} \in \mathbf {C \times C}$ we have:
{{begin-eqn}}
{{eqn | n = 6
      | l = \sqcup \tuple {C_1, C_2}
      | r = C_1 \sqcup C_2
      | c = {{Defof|Coproduct Functor}}
}}
{{eqn | r = \oplus \tuple{C_1, C_2}
      | c = By $(5)$
}}
{{end-eqn}}
By definition of coproduct of morphisms 
:$\sqcup \tuple {f_1, f_2}$ denotes the unique morphism guaranteed by the binary coproduct UMP that makes the following diagram commute:
::<nowiki>$\begin{xy}
<-8em,0em>*+{C_1} = "TL",
<0em,0em>*+{C_1 \sqcup C_2} = "TM",
<8em,0em>*+{C_2} = "TR",
<-8em,-6em>*+{ {C_1}'} = "BL",
<0em,-6em>*+{ {C_1}' \sqcup {C_2}'} = "BM",
<8em,-6em>*+{ {C_2}'} = "BR",
"TL";"BL" **@{-} ?>*@{>} ?<>(.5)*!/^1em/{f_1},
"TL";"TM" **@{-} ?>*@{>} ?<>(.5)*!/_1em/{i_1},
"TR";"TM" **@{-} ?>*@{>} ?<>(.5)*!/^1em/{i_2},
"BL";"BM" **@{-} ?>*@{>} ?<>(.5)*!/^1em/{j_1},
"BR";"BM" **@{-} ?>*@{>} ?<>(.5)*!/_1em/{j_2},
"TM";"BM" **@{--} ?>*@{>} ?<>(.5)*!/_2.7em/{f_1 \sqcup f_2},
"TR";"BR" **@{-} ?>*@{>} ?<>(.5)*!/_1em/{f_2},
\end{xy}$</nowiki>
where $\tuple{C_1 \sqcup C_2, i_1, i_2}$ and $\tuple{ {C_1}' \sqcup {C_2}', j_1, j_2}$ are the binary coproducts of $C_1, C_2$ and ${C_1}', {C_2}'$ respectively.
Let $\tuple{f_1: C_1 \to {C_1}', f_2: C_2 \to {C_2}'} \in \mathbf {C \times C}$ be a morphism of the product category $\mathbf {C \times C}$.
Let $\tuple {\map \oplus {C_1,  C_2}, i_1, i_2 }$ and $\tuple {\map \oplus { {C_1},  '{C_2}'}, j_1, j_2 }$ denote the binary coproducts of $\tuple{C_1,  C_2}$ and $\tuple { {C_1},  '{C_2}'}$ respectively.
By the definition of unit of adjunction, $\iota : \operatorname{id}_{\mathbf {C \times C} } \to \oplus \Delta$ is a natural transformation.
By definition of natural transformation, the following diagram commutes:
::<nowiki>$\begin{xy}
<-6em,0em>*+{\tuple{ C_1, C_2} } = "TL",
<6em,0em>*+{\paren{\Delta \oplus} {\tuple{C_1, C_2} } } = "TR",
<-6em,-6em>*+{\tuple{ {C_1}', {C_2}'} } = "BL",
<6em,-6em>*+{\paren{\Delta \oplus} {\tuple{ {C_1}', {C_2}'} } } = "BR",
"TL";"TR" **@{-} ?>*@{>} ?<>(.5)*!/_1em/{\iota_{\tuple{C_1, C_2} } },
"BL";"BR" **@{-} ?>*@{>} ?<>(.5)*!/^1em/{\iota_{\tuple{ {C_1}', {C_2}'} } },
"TL";"BL" **@{-} ?>*@{>} ?<>(.5)*!/^2em/{\tuple{f_1, f_2} },
"TR";"BR" **@{-} ?>*@{>} ?<>(.5)*!/_3.2em/{\paren{\Delta \oplus} {\tuple{f_1, f_2} } },
\end{xy}$</nowiki>
Substituting $(2)$, $(5)$, and $(6)$ into the diagram above, the following diagram commutes:
::<nowiki>$\begin{xy}
<-6em,0em>*+{\tuple{ C_1, C_2} } = "TL",
<6em,0em>*+{\tuple{C_1 \sqcup C_2, C_1 \sqcup C_2} } = "TR",
<-6em,-6em>*+{\tuple{ {C_1}', {C_2}'} } = "BL",
<6em,-6em>*+{\tuple{ {C_1}' \sqcup {C_2}', {C_1}' \sqcup {C_2}' } } = "BR",
"TL";"TR" **@{-} ?>*@{>} ?<>(.5)*!/_1em/{\tuple{i_1, i_2} },
"BL";"BR" **@{-} ?>*@{>} ?<>(.5)*!/^1em/{\tuple{j_1, j_2} },
"TL";"BL" **@{-} ?>*@{>} ?<>(.5)*!/^2em/{\tuple{f_1, f_2} },
"TR";"BR" **@{-} ?>*@{>} ?<>(.5)*!/_5.2em/{\tuple{\oplus \tuple{f_1, f_2}, \oplus \tuple{f_1, f_2} }  },
\end{xy}$</nowiki>
By definition of composition in product category the above diagram commutes in each component of the product category.
Hence the following diagram commutes:
::<nowiki>$\begin{xy}
<-8em,0em>*+{C_1} = "TL",
<0em,0em>*+{C_1 \sqcup C_2} = "TM",
<8em,0em>*+{C_2} = "TR",
<-8em,-6em>*+{ {C_1}'} = "BL",
<0em,-6em>*+{ {C_1}' \sqcup {C_2}'} = "BM",
<8em,-6em>*+{ {C_2}'} = "BR",
"TL";"BL" **@{-} ?>*@{>} ?<>(.5)*!/^1em/{f_1},
"TL";"TM" **@{-} ?>*@{>} ?<>(.5)*!/_1em/{i_1},
"TR";"TM" **@{-} ?>*@{>} ?<>(.5)*!/^1em/{i_2},
"BL";"BM" **@{-} ?>*@{>} ?<>(.5)*!/^1em/{j_1},
"BR";"BM" **@{-} ?>*@{>} ?<>(.5)*!/_1em/{j_2},
"TM";"BM" **@{-} ?>*@{>} ?<>(.5)*!/_2.7em/{\oplus \tuple{f_1, f_2} },
"TR";"BR" **@{-} ?>*@{>} ?<>(.5)*!/_1em/{f_2},
\end{xy}$</nowiki>
By uniqueness of $\sqcup \tuple {f_1, f_2}$:
:$\oplus \tuple {f_1, f_2} = \sqcup \tuple {f_1, f_2}$
This completes the proof that $\oplus = \sqcup$.
Hence the right adjoint $\oplus:\mathbf {C \times C} \to \mathbf C$ of $\Delta$ is the coproduct functor.