Cooperator in a Unital Category

Let π’ž π’ž \mathcal{C} be a Unital Category, and f : ( X , Z ) : 𝑓 𝑋 𝑍 f:{{}(X,Z)} , g : ( Y , Z ) : 𝑔 π‘Œ 𝑍 g:{{}(Y,Z)} be a pair of morphisms with common codomain. A cooperator of the pair ( f , g ) 𝑓 𝑔 (f,g) is a map Ο† : X Γ— Y β†’ Z : πœ‘ β†’ 𝑋 π‘Œ 𝑍 \varphi:X\times Y\to Z that makes the following diagram commute:

X𝑋{X}XΓ—Yπ‘‹π‘Œ{{X\times Y}}Yπ‘Œ{Y}Z𝑍{Z}⟨mathrmidifx..β‹„lsei,0⟩\scriptstyle{\langle mathrm{id}{ifx..lse\par i},0\rangle}f𝑓\scriptstyle{f}Ο†πœ‘\scriptstyle{\varphi}⟨0,mathrmidifx..β‹„lsei⟩\scriptstyle{\langle 0,mathrm{id}{ifx..lse\par i}\rangle}g𝑔\scriptstyle{g}

Note that Ο† πœ‘ \varphi must be unique, as ⟨ m a t h r m i d i f x . . β‹„ l s e i , 0 ⟩ \langle mathrm{id}{ifx..lse\par i},0\rangle and ⟨ 0 , m a t h r m i d i f x . . β‹„ l s e i ⟩ \langle 0,mathrm{id}{ifx..lse\par i}\rangle must be Strongly Epimorphic Family.

Another way of viewing this condition is that ( f , g ) 𝑓 𝑔 (f,g) have a cooperator when X Γ— Y 𝑋 π‘Œ X\times Y acts like a Binary Biproduct with respect to f 𝑓 f and g 𝑔 g .

Note that we can extend this to a Sink f i : X i β†’ Y : subscript 𝑓 𝑖 β†’ subscript 𝑋 𝑖 π‘Œ f_{i}:X_{i}\to Y if π’ž π’ž \mathcal{C} has Indexed Cartesian Products; this is akin to asking that ∏ i : I X i subscript product : 𝑖 𝐼 subscript 𝑋 𝑖 \prod_{i:I}X_{i} act like an Indexed Biproduct. In the case of the empty family, this is akin to asking that the Terminal Object be a Zero Object, which is already the case in a Unital Category.

Intuition

In the Category of Groups, we can unfold the cooperator condition to get that

In fact, we can compute what the cooperator ought to be:

φ⁒(x,y)πœ‘π‘₯𝑦\displaystyle\varphi(x,y) =φ⁒(x+0,0+y)absentπœ‘π‘₯00𝑦\displaystyle=\varphi(x+0,0+y) (left and right identity)
=φ⁒((x,0)+(0,y))absentπœ‘π‘₯00𝑦\displaystyle=\varphi((x,0)+(0,y)) (defn of multiplication in product group)
=φ⁒(x,0)+φ⁒(0,y)absentπœ‘π‘₯0πœ‘0𝑦\displaystyle=\varphi(x,0)+\varphi(0,y) (Ο†πœ‘\varphi is a group homomorphism)
=f⁒(x)+g⁒(y)absent𝑓π‘₯𝑔𝑦\displaystyle=f(x)+g(y) (Ο†πœ‘\varphi is a cooperator)

However, we can run the same argument with Ο† ⁒ ( 0 + x , y + 0 ) πœ‘ 0 π‘₯ 𝑦 0 \varphi(0+x,y+0) to get that Ο† ⁒ ( x , y ) = g ⁒ ( y ) + f ⁒ ( x ) πœ‘ π‘₯ 𝑦 𝑔 𝑦 𝑓 π‘₯ \varphi(x,y)=g(y)+f(x) . In other words, the cooperator of f 𝑓 f and g 𝑔 g only exists if they commute.

Moreover, suppose let f , g : G β†’ H : 𝑓 𝑔 β†’ 𝐺 𝐻 f,g:G\to H be a parallel pair of group homomorphisms. If the cooperator Ο† πœ‘ \varphi of ( f , g ) 𝑓 𝑔 (f,g) exists, then the sum f + g 𝑓 𝑔 f+g is equal to Ο† ∘ Ξ΄ G : G β†’ H : πœ‘ subscript 𝛿 𝐺 β†’ 𝐺 𝐻 \varphi\circ\delta_{G}:G\to H : both of these are group homomorphisms, so the sum must be a group homomorphism as well!

References