Explicit Proof that the Category of Groups is Total

The definition of a total category is very abstract; furthermore, it is not immediately clear how it is possible for any category which is not essentially small to satisfy the definition, much less a wide variety of the algebraic and topological categories which are considered in practice. Thus, to illustrate the definition, we give an explicit construction of the functor $$L : [\Grp^{\op},\Set] \to \Grp$$ that is left adjoint to the Yoneda embedding $y : \Grp \hookrightarrow [\Grp^{\op},\Set]$.

Fix a functor $T : \Grp^{\op} \to \Set$. To construct the group $L(T)$, we will make use of the usual cogroup structure on $\IZ$ in $\Grp$, which includes

• the comultiplication homomorphism $\mu : \IZ \to \IZ * \IZ'$, $1 \mapsto 1 \cdot 1'$ (where $\IZ'$ denotes a copy of $\IZ$),
• the coidentity homomorphism $\varepsilon : \IZ \to 0$,
• the coinverse homomorphism $\iota : \IZ \to \IZ$.

Also, let $i_1,i_2 : \IZ \rightrightarrows \IZ * \IZ'$ denote the coprojections. We define the group $L(T)$ as the group generated by elements $e(x)$, one for each element $x \in T(\IZ)$, subject to the following relations:

• $e(T\mu(x)) = e(Ti_1(x)) \cdot e(Ti_2(x))$ for each $x \in T(\IZ * \IZ')$,
• $e(T\varepsilon(x)) = 1$ for each $x \in T0$,
• $e(T\iota(x)) = e(x)^{-1}$ for each $x \in T\IZ$,

We first need to define a natural transformation $\eta_T : T \to \Hom({-}, L(T))$. For each group $H$ we define the function $\eta_T(H) : TH \to \Hom(H, L(T))$ by sending $x \in TH$ to $h \mapsto e(Th(x))$, where we abuse notation to identify $h \in H$ with the corresponding morphism $\IZ \to H$ mapping $1 \mapsto h$, so that $Th : TH \to T\IZ$. To see that this defines a group homomorphism from $H$ to $L(T)$, note that for $h, h' \in H$ we have three commutative diagrams of the form

$$\begin{CD} T(H) @> = >> T(H)\\ @V T(hh') VV @VVV\\ T(\IZ * \IZ') @>>> T(\IZ) \end{CD}$$

where on the bottom we use $T\mu, Ti_1, Ti_2$, and on the right we use $h h', h, h'$. Applying this to $x\in TH$, we get $Th(x)$, $Th'(x)$, and $T(h h')(x)$, respectively. Thus, the relation $e(T\mu(y)) = e(Ti_1(y)) \cdot e(Ti_2(y))$ with $y \coloneqq T(h h')(x)$ implies $$e(T(hh')(x)) = e(Th(x)) \cdot e(Th'(x)),$$ as required. Similar proofs show that the map $H \to L(T)$ respects inverses and the identity. We leave it as an exercise for the reader to show this is natural in $H$.

We now need to show that for each group $G$ and natural transformation $\alpha : T \to y_G$, there exists a unique group homomorphism $\varphi : L(T) \to G$ such that $$\alpha = y_{\varphi} \circ \eta_T : T \to \Hom({-}, L(T)) \to \Hom({-}, G).$$ We start with uniqueness: suppose $x \in T\IZ$. Then by hypothesis, $$\alpha_{\IZ} = (y_{\varphi})_{\IZ} \circ (\eta_T)_{\IZ} : T\IZ \to \Hom(\IZ, L(T)) \to \Hom(\IZ, G).$$ For each $x \in T\IZ$, the first step on the right hand side maps $x \mapsto (1 \mapsto e(x))$, and the second step then maps this to $1 \mapsto \varphi(e(x))$. Therefore, $$\varphi(e(x)) = \alpha_{\IZ}(x)(1)$$ for each $x$, which establishes the uniqueness of $\varphi$.

For the existence part, the first step is to show there is a group homomorphism $L(T) \to G$ with the images of $e(x)$ required by the previous part, i.e. $e(x) \mapsto \alpha_{\IZ}(x)(1)$. To prove this, we need to check that the relations in $L(T)$ are satisfied in $G$. Now, for each $x \in T(\IZ * \IZ')$, we have three commutative diagrams of the form

$$\begin{CD} T(\IZ * \IZ') @> \alpha_{\IZ * \IZ'} >> \Hom(\IZ * \IZ', G) @> \simeq >> UG \times UG\\ @VVV @VVV @VVV\\ T(\IZ) @> \alpha_{\IZ} >> \Hom(\IZ, G) @> \simeq >> UG \end{CD}$$

applying naturality to $\mu, i_1, i_2 : \IZ \to \IZ * \IZ'$. On the right hand side, we get multiplication, first projection, and second projection respectively. From this, we conclude that the images of $e(T\mu(x))$ and $e(Ti_1(x)) \cdot e(Ti_2(x))$ in $UG$ agree for any element $x \in T(\IZ * \IZ')$. Similar proofs show that the other relations are also satisfied.

Finally, we need to show $\alpha = y_{\varphi} \circ \eta_T$, i.e. $\alpha_H = (y_{\varphi})_H \circ (\eta_T)_H$ for each group $H$. By definition, for each $x \in TH$, the first step gives the homomorphism $h \mapsto e(Th(x))$; then the second step is formed by composition with $\varphi$. By the specification of $\varphi$, this gives the homomorphism $h \mapsto \alpha_{\IZ}(Th(x))(1)$. However, by the assumption that $\alpha$ is a natural transformation, for each $h \in H$ we have a commutative diagram

$$\begin{CD} TH @> \alpha_H >> \Hom(H, G) \\ @V Th VV @VV {-} \circ h V \\ T\IZ @> \alpha_{\IZ} >> \Hom(\IZ, G). \end{CD}$$

Applying this to $x \in TH$ gives exactly that $\alpha_{\IZ}(Th(x))(1) = \alpha_H(x)(h)$. $\square$