# A Solvability Of The Couniversal Mapping Problem On Products

#### fresh_42

Mentor
2018 Award
Summary
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 $\mathcal{G}=\mathcal{K}\,?$

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#### Infrared

Gold Member
Sorry, I'm a little confused by the notation. You're asking whether the mapping problem is solvable for some examples of categories $\mathcal{G}$ and $\mathcal{K}$, but shouldn't you specify the functor $\mathscr{F}$ as well?

For example, if $\mathcal{G}=\text{R-Mod}$ and $\mathcal{K}=\text{R-Mod}$, you can recover tensor products by taking $\mathscr{F}(M_1\times M_2,M_3)$ to be the set of bilinear maps $M_1\times M_2\to M_3$, but I don't know what you want $\mathscr{F}$ to be in other examples.

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#### fresh_42

Mentor
2018 Award
Sorry, I'm a little confused by the notation. You're asking whether the mapping problem is solvable for some examples of categories $\mathcal{G}$ and $\mathcal{K}$, but shouldn't you specify the functor $\mathscr{F}$ as well?

For example, if $\mathcal{G}=\text{R-Mod}$ and $\mathcal{K}=\text{R-Mod}$, you can recover tensor products by taking $\mathscr{F}(M_1\times M_2,M_3)$ to be the set of bilinear maps $M_1\times M_2\to M_3$, but I don't know what you want $\mathscr{F}$ to be in other examples.
It is only given, that $\mathscr{F}$ is contravariant on the product of the objects of $\mathcal{G}$ and covariant on the objects of $\mathcal{K}$. That is for the tensor product: contravariant on $M_1 \times M_2$ and covariant on $M_3=M_1\otimes M_2$. The tensor product is only a solution, it's not meant to start with. In $R-Mod$ we start with bilinear morphisms which define $\mathscr{F}$. I am interested in any situation, where morphisms aren't given by this restriction. One could also ask:

Is there a first place contra- and second place covariant functor $\mathscr{F}\, : \,\left(Grp \times Grp\right) \times Grp \longrightarrow Set$ such that for any groups $P:=G_1 \times G_2$ there is a group $U_P$ and an $\mathscr{F}-$morphism $u_P\, : \,P\longrightarrow U_P$ with $f=\overline{f}u_P$ if given any group homomorphism $f\, : \,P \longrightarrow K$ uniquely extended to $\overline{f}\, : \,U_P\longrightarrow K\;$?

Do we have such an example without a given bilinear structure for $\mathscr{F}$ as in $R-Mod$ or when we identify Abelian groups with $\mathbb{Z}-$modules. Hence: Does it make sense in any reasonable way to define $G_1 \otimes G_2$ by other means than bilinearity?

#### WWGD

Gold Member
I prefer to understand it by just using the isomorphism between $B((V \times W),Z)$ and $L( (V \otimes W),Z)$ for vector spaces and its generalization and the algebra showing maps factor through.

#### fresh_42

Mentor
2018 Award
I prefer to understand it by just using the isomorphism between $B((V \times W),Z)$ and $L( (V \otimes W),Z)$ for vector spaces and its generalization and the algebra showing maps factor through.
Yes, but this already puts a structure on $\mathscr{F}$, namely bilinearity. I want to know, whether it can be done without such a structure given. Or is $u_G$ automatically such a structure?

#### Infrared

Gold Member
Is there a first place contra- and second place covariant functor $\mathscr{F}\, : \,\left(Grp \times Grp\right) \times Grp \longrightarrow Set$ such that for any groups $P:=G_1 \times G_2$ there is a group $U_P$ and an $\mathscr{F}-$morphism $u_P\, : \,P\longrightarrow U_P$ with $f=\overline{f}u_P$ if given any group homomorphism $f\, : \,P \longrightarrow K$ uniquely extended to $\overline{f}\, : \,U_P\longrightarrow K\;$?
I don't think you mean for $f$ to be a group homomorphism, but rather it should be a $\mathscr{F}$-morphism (since if you're trying to generalize tensor product, you should include the case that $f$ is a bilinear map, which are usually not morphisms for $\text{R-Mod}$, e.g. the map $\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$ given by $(a,b)\mapsto ab$ is bilinear, but not a group homomorphism).

Anyway, there are certainly some functors $\mathscr{F}$ that let your problem be solved, e.g. let $\mathscr{F}:(\text{Grp}\times\text{Grp})\times\text{Grp}\to\text{Set}$ given by $\mathscr{F}((G_1,G_2),H)=\text{Hom}_{\text{Grp}}(G_1\times G_2,H)$. Then $\mathscr{F}((G_1,G_2),\cdot):\text{Grp}\to\text{Set}$ is clearly represented by the group $G_1\times G_2$.

#### fresh_42

Mentor
2018 Award
The $Hom$ functor solves the universal mapping problem where the result is a contravariant functor, not a covariant one. It is the dual object/problem to the tensor product.

#### Infrared

Gold Member
$\text{Hom}$ is contravariant in the first argument and covariant in the second argument like you demanded. I don't see which conditions it fails.

#### fresh_42

Mentor
2018 Award
$\text{Hom}$ is contravariant in the first argument and covariant in the second argument like you demanded. I don't see which conditions it fails.
I do not see where the third factor comes in.

The tensor product of say vector spaces would be $\otimes \, : \,G_1\times G_2 \longrightarrow U_{G_1\times G_2}=G_1\otimes G_2$ and covariant in both arguments. It's a solution since all bilinear mappings $f\, : \,G_1\times G_2 \longrightarrow H$ extend to $\overline{f}\, : \,G_1\otimes G_2 \longrightarrow H$.

Now what does $Hom(G_1,G_2)\longrightarrow H$ mean? If we have a morphism $G_1\times G_2 \longrightarrow H$, how does it extend to $Hom(G_1,G_2) \longrightarrow H$? And if we consider $Hom(G_1\times G_2, ?)$ then what is the universal object $U_{G_1\times G_2}$?

#### Infrared

Gold Member
I am not claiming that Hom is the solution to the mapping problem, rather it is just my choice of $\mathscr{F}$. The solution to the mapping problem in this case is just the usual product of groups.

Let me re-phrase your question as I understand it. Please let me know if I'm misunderstanding something: You are looking for a functor $\mathscr{F}:(\text{Grp}\times\text{Grp})\times\text{Grp}$ that is contravariant in $\text{Grp}\times\text{Grp}$ and that is covariant in $\text{Grp}$ such that for all $(G_1,G_2)\in\text{Grp}\times\text{Grp}$, the functor $\mathscr{F}^{(G_1,G_2)}$ is representable, that is, naturally isomorphic to the functor $\text{Hom}(G,\cdot)$ for some fixed group $G$ (this should be equivalent to your language of a representing pair).

In my case, I choose the functor $\mathscr{F}((G_1,G_2),H)=\text{Hom}_{Grp}(G_1\times G_2,H)$. Restricting to the second argument, the (covariant) functor $\mathscr{F}^{(G_1,G_2)}$ is by definition represented by the group $G_1\times G_2$.

Note that if we switch from groups to abelian groups, and I re-define $\mathscr{F}((G_1,G_2),H)=\text{Bilin}(G_1\times G_2,H)$, the functor $\mathscr{F}^{(G_1,G_2)}$ is represented by the abelian group $G_1\otimes G_2$ (that is, we recover the tensor product).

To answer your specific questions, in my case $U_{(G_1,G_2)}=G_1\times G_2$ and $u_{(G_1,G_2)}\in\mathscr{F}((G_1,G_2),G_1\times G_2)=\text{Hom}(G_1\times G_2,G_1\times G_2)$ is the identity morphism.

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#### fresh_42

Mentor
2018 Award
Hmm, this means in words: the free product of groups is the correspondence in this case. I had hoped for a bit more structure for $u_{(G_1,G_2)}$. Seems the same can be done in $Top$. That means that bilinearity is intrinsic for tensor products. Makes sense, as realizations of e.g. algebras can often be constructed as quotients of tensor algebras. In case of groups we have to fall back on free products.

#### Infrared

Gold Member
My products are direct, not free. Anyway, this example was vacuous, the point was just to show that you could find a functor $\mathscr{F}$ for which the mapping problem could be solved. I think the real problem is to find examples of $\mathscr{F}$ for which this problem is interesting. I don't have a good example in mind for groups, like the bilinear maps functor is for R-modules.

• fresh_42

#### fresh_42

Mentor
2018 Award
My products are direct, not free. Anyway, this example was vacuous, the point was just to show that you could find a functor $\mathscr{F}$ for which the mapping problem could be solved. I think the real problem is to find examples of $\mathscr{F}$ for which this problem is interesting. I don't think there's a good choice for groups, like the bilinear maps functor is for R-modules.
With only two factors and no additional relations, where's the difference between free and direct?

#### Infrared

Gold Member
For example $\mathbb{Z}\times\mathbb{Z}$ is abelian, but $\mathbb{Z}*\mathbb{Z}$ is not.

Direct product is the product in the category of groups, and free product is the co-product

• WWGD