# Algebra operations as natural transformations

Apologies in advance if the following makes little to no sense, but here goes ..

Denote $$m_G : G\times G\to G$$ the multiplication of a group $$G$$. Does it make sense to think of the map $$m_G$$ as some kind of morphism in a (at the moment unspecified) category?

Even more, could we think of the family $$m_G$$ (indexed by the class of groups) as a natural transformation of some or other functors?

Let's call our mystery categories $$\mathcal A$$ and $$\mathcal B$$ with mystery functors $$X,Y :\mathcal A\to\mathcal B$$ such that the natural transformation condition holds: $$m_HX(f) = Y(f)m_G$$ where $$f$$ is a morphism in $$\mathcal A$$ and $$G,H$$ are groups.

For this to make sense, we need a category with objects $$G\times G$$ and $$G$$. So, we extend (?) the category of groups by $$\mathcal B_0 := \mbox{Grp}_0 \cup \{G\times G \mid G\in\mbox{Grp}_0\}$$. Morphisms in 'separate components' remain as they are in $$\mbox{Grp}$$ and $$\mbox{Grp}\times\mbox{Grp}$$ respectively. There would be no morphisms of the form $$G\to H\times H$$ and $$\mathcal B(G\times G,H) := \{\varphi m_G \mid \varphi \in \mbox{Hom}(G,H)\}$$ where for every $$x,y\in G$$ $$\varphi m_G (x,y) := \varphi (xy) = \varphi (x)\varphi (y) =: m_H(\varphi,\varphi)(x,y)$$. The identities are $$1_G$$ or $$(1_G,1_G)$$ depending on the component and composition of the morphisms would happen naturally.

1. Is it guaranteed $$(A,B) \neq (A',B') \implies \mathcal B(A,B)\cap\mathcal B(A',B') =\emptyset, A,A',B,B'\in\mathcal B_0$$?
2. Taking $$\mathcal A = \mbox{Grp}$$ with $$X :\mathcal A\to\mathcal B$$ given such that $$X(G) = G\times G$$ and for every morphism $$f:G\to H$$, $$X(f) = (f,f)$$. Put $$Y:\mathcal A\to\mathcal B$$ as the embedding, then we would have $$m_G$$ as a natural transformation $$X\Rightarrow Y$$.

I omit the routine checks here, for they aren't important for this discussion. I am interested in whether this idea of regarding families of operations as natural transformations is, call it, well-founded

Questionnaire.

1. Would such an approach be the only one? How else (if at all) would we regard the family $$m_G$$ as a natural transformation?
2. Is this a more general thing in universal algebra? Given a class of algebras with certain operations of various arities, could we regard every family of operations as a natural transformation? (For instance, inverse operation or unit element operation of groups)
• @H.H. the components must also be morphisms themselves in the destination category so if it were to work, then that must involve some 'weird' categories. I'm concerned about the way I defined $\mathcal B$ yet I can't find any rule that says a category can't contain objects of different 'type'. – Alvin Lepik Mar 18 at 11:27

Let $$\DeclareMathOperator\Set{Set}\Set$$ denote the category of sets and $$\DeclareMathOperator\Grp{Grp}\Grp$$ denote the category of groups. Let $$\Upsilon:\Grp\to\Set$$ denote the forgetful functor and $$\Delta:\Set\to\Set$$ be the diagonal functor $$\Delta(X)=X\times X$$.
Then you are looking for a natural transformation $$\mu:\Delta\circ\Upsilon\to\Upsilon$$. If $$X$$ is a group, then $$\Upsilon(X)$$ is its underlying set. For each group $$X$$ let $$\mu_X:(\Delta\circ\Upsilon)(X)\to\Upsilon(X)$$ be its composition law. If $$f:X\to Y$$ is a group homomorphism, then we have a commutative diagram$$\require{AMScd}$$: $$\begin{CD} \Upsilon(X)\times\Upsilon(X)@=(\Delta\circ\Upsilon)(X)@>\mu_X>>\Upsilon(X)\\ @V\Upsilon(f)\times\Upsilon(f)VV@V(\Delta\circ\Upsilon)(f)VV@VV\Upsilon(f)V\\ \Upsilon(Y)\times\Upsilon(Y)@=(\Delta\circ\Upsilon)(Y)@>>\mu_Y >\Upsilon(Y) \end{CD}$$ which proves naturalness of $$\mu_X$$.