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- Is there a representing pair in case the category doesn't have two binary operations?

Let us assume a functor ##\mathscr{F}\, : \,\mathcal{G} \times \mathcal{K}\longrightarrow \mathcal{Set}## which is contravariant on ##\mathcal{G}## and covariant on ##\mathcal{K}##. The question whether for any object ##G \in \mathcal{G}## the covariant functor ##\mathscr{F}^{G}:=\mathscr{F}(G,\cdot)\, : \,\mathcal{K}\longrightarrow \mathcal{Set}## can be represented is called the couniversal mapping problem of ##\mathscr{F}## on ##\mathcal{K}##.

A representing pair ##(U_G,u_G) \in (\operatorname{Obj}(\mathcal{K}),\operatorname{Mor}_\mathscr{F}\, : \,G\longrightarrow U_G)## is a couniversal object ##U_G## and a couniversal ##\mathscr{F}-##morphism ##u_G##. The couniversal mapping problem is solvable, if for any object ##G\in \mathcal{G}## there is a representing pair for ##\mathscr{F}^{G}##. The following theorem makes it easier to sort out:

The couniversal mapping problem of ##\mathscr{F}\, : \,\mathcal{G} \times \mathcal{K}\longrightarrow \mathcal{Set}## is solvable if and only if for any object ##G\in \mathcal{G}## there is an object ##U_G \in \mathcal{K}## and an ##\mathscr{F}-##morphism ##u_G\, : \,G\longrightarrow U_G## such that:

Every ##\mathscr{F}-##morphism ##f\, : \,G\longrightarrow K## determines a unique ##\mathcal{K}-##morphism ##\overline{f}\, : \,U_G\longrightarrow K## with ##f= \overline{f} \circ u_G##.

Now let's consider products, i.e. a functor ##\mathscr{F}\, : \,\left(\mathcal{G} \times \mathcal{G}\right) \times \mathcal{K}\longrightarrow \mathcal{Set}## on product objects. I assume that the couniversal mapping problem cannot always be solved, since otherwise the definition of solvability wouldn't make much sense. But can it be solved on products of certain categories? I know a couple of examples (representing pairs) for categories with two binary operations, such as ##\mathcal{Module}## or ##\mathcal{Ring}##.

Is the couniversal mapping problem of products solvable in the category of (arbitrary, i.e. not necessarily Abelian) groups: ##\mathcal{G}=\mathcal{Grp}\times \mathcal{Grp}\, , \,\mathcal{K}=\mathcal{Grp}##, or as another example, in the category of topological spaces ##\mathcal{G}=\mathcal{Top}\times \mathcal{Top}\, , \,\mathcal{K}=\mathcal{Top}\,?## How do free groups play a role in here? Is commutativity a necessary condition? Do we need additional properties for topological spaces, like e.g. a vector space structure? Or can we say, that ##u_G\, : \, G\times G \longrightarrow U_G\in \mathcal{K}## automatically means a second binary structure on ##G##, esp. in the cases where ##\mathcal{G}=\mathcal{K}\,?##

A representing pair ##(U_G,u_G) \in (\operatorname{Obj}(\mathcal{K}),\operatorname{Mor}_\mathscr{F}\, : \,G\longrightarrow U_G)## is a couniversal object ##U_G## and a couniversal ##\mathscr{F}-##morphism ##u_G##. The couniversal mapping problem is solvable, if for any object ##G\in \mathcal{G}## there is a representing pair for ##\mathscr{F}^{G}##. The following theorem makes it easier to sort out:

The couniversal mapping problem of ##\mathscr{F}\, : \,\mathcal{G} \times \mathcal{K}\longrightarrow \mathcal{Set}## is solvable if and only if for any object ##G\in \mathcal{G}## there is an object ##U_G \in \mathcal{K}## and an ##\mathscr{F}-##morphism ##u_G\, : \,G\longrightarrow U_G## such that:

Every ##\mathscr{F}-##morphism ##f\, : \,G\longrightarrow K## determines a unique ##\mathcal{K}-##morphism ##\overline{f}\, : \,U_G\longrightarrow K## with ##f= \overline{f} \circ u_G##.

Now let's consider products, i.e. a functor ##\mathscr{F}\, : \,\left(\mathcal{G} \times \mathcal{G}\right) \times \mathcal{K}\longrightarrow \mathcal{Set}## on product objects. I assume that the couniversal mapping problem cannot always be solved, since otherwise the definition of solvability wouldn't make much sense. But can it be solved on products of certain categories? I know a couple of examples (representing pairs) for categories with two binary operations, such as ##\mathcal{Module}## or ##\mathcal{Ring}##.

Is the couniversal mapping problem of products solvable in the category of (arbitrary, i.e. not necessarily Abelian) groups: ##\mathcal{G}=\mathcal{Grp}\times \mathcal{Grp}\, , \,\mathcal{K}=\mathcal{Grp}##, or as another example, in the category of topological spaces ##\mathcal{G}=\mathcal{Top}\times \mathcal{Top}\, , \,\mathcal{K}=\mathcal{Top}\,?## How do free groups play a role in here? Is commutativity a necessary condition? Do we need additional properties for topological spaces, like e.g. a vector space structure? Or can we say, that ##u_G\, : \, G\times G \longrightarrow U_G\in \mathcal{K}## automatically means a second binary structure on ##G##, esp. in the cases where ##\mathcal{G}=\mathcal{K}\,?##

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