Product of Composite Morphisms

Theorem
Let $\mathbf C$ be a metacategory.

Let $f \times f': A \times A' \to B \times B'$ and $g \times g': B \times B' \to C \times C'$ be two composable products of morphisms in $\mathbf C$.

Then:


 * $\left({g \circ f}\right) \times \left({g' \circ f'}\right) = \left({g \times g'}\right) \circ \left({f \times f'}\right)$

where $\times$ signifies product of morphisms.

Proof
The situation is efficiently captured in the following commutative diagram:


 * $\begin{xy}

<-5em,0em>*+{A}           = "A", <0em,0em>*+{A \times A'}  = "P", <5em,0em>*+{A'}           = "A2", <-5em,-5em>*+{B}          = "B", <0em,-5em>*+{B \times B'} = "Q", <5em,-5em>*+{B'}          = "B2", <-5em,-10em>*+{C}         = "C", <0em,-10em>*+{C \times C'} = "R", <5em,-10em>*+{C'}         = "C2",

"P";"A" **@{-} ?>*@{>} ?*!/^.8em/{\operatorname{pr}_1}, "P";"A2" **@{-} ?>*@{>} ?*!/_.8em/{\operatorname{pr}_2}, "Q";"B" **@{-} ?>*@{>} ?*!/^.8em/{\operatorname{pr}_1}, "Q";"B2" **@{-} ?>*@{>} ?*!/_.8em/{\operatorname{pr}_2}, "R";"C" **@{-} ?>*@{>} ?*!/_.8em/{\operatorname{pr}_1}, "R";"C2" **@{-} ?>*@{>} ?*!/^.8em/{\operatorname{pr}_2},

"A";"B"  **@{-} ?>*@{>}  ?*!/^.8em/{f}, "A2";"B2" **@{-} ?>*@{>} ?*!/_.8em/{f'}, "P";"Q"  **@{--} ?>*@{>} ?*!/_1.4em/{f \times f'},

"B";"C"  **@{-} ?>*@{>}  ?*!/^.8em/{g}, "B2";"C2" **@{-} ?>*@{>} ?*!/_.8em/{g'}, "Q";"R"  **@{--} ?>*@{>} ?*!/_1.4em/{g \times g'}, \end{xy}$

The result is immediate from deleting the $B$s:


 * $\begin{xy}

<-5em,0em>*+{A}           = "A", <0em,0em>*+{A \times A'}  = "P", <5em,0em>*+{A'}           = "A2", <-5em,-10em>*+{C}         = "C", <0em,-10em>*+{C \times C'} = "R", <5em,-10em>*+{C'}         = "C2",

"P";"A" **@{-} ?>*@{>} ?*!/^.8em/{\operatorname{pr}_1}, "P";"A2" **@{-} ?>*@{>} ?*!/_.8em/{\operatorname{pr}_2}, "R";"C" **@{-} ?>*@{>} ?*!/_.8em/{\operatorname{pr}_1}, "R";"C2" **@{-} ?>*@{>} ?*!/^.8em/{\operatorname{pr}_2},

"A";"C"  **@{-} ?>*@{>}  ?*!/^1.1em/{g \circ f}, "A2";"C2" **@{-} ?>*@{>} ?*!/_1.3em/{g' \circ f'}, "P";"R"  **@{--} ?>*@{>} ?*!/_2em/{\left({g \circ f}\right) \times \\ \left({g' \circ f'}\right)}, \end{xy}$

The uniqueness of $\left({g \circ f}\right) \times \left({g' \circ f'}\right)$ implies it equals $\left({g \times g'}\right) \circ \left({f \times f'}\right)$.