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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)
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    $\begingroup$ @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'. $\endgroup$ – Alvin Lepik Mar 18 at 11:27
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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 $\mu_X$.

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    $\begingroup$ This looks much neater and cleaner than my attempt. Thanks, I have a much better understanding of this idea now. $\endgroup$ – Alvin Lepik Mar 18 at 12:31
  • $\begingroup$ Also thanks for demonstrating the amscd package, looks much more convenient than tikz syntax-wise. Can't believe I didn't know about it! $\endgroup$ – Alvin Lepik Mar 18 at 12:44

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