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 $y : \Grp \hookrightarrow [\Grp^{\op},\Set]$ and therefore witnesses that $\Grp$ is total.

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 $\sigma : \IZ \to \IZ$.

Also, let $\iota_1, \iota_2 : \IZ \rightrightarrows \IZ * \IZ'$ 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:

• $e(T(\mu)(x)) = e(T(\iota_1)(x)) \cdot e(T(\iota_2)(x))$ for each $x \in T(\IZ * \IZ')$,
• $e(T(\varepsilon)(x)) = 1$ for each $x \in T(0)$,
• $e(T(\sigma)(x)) = e(x)^{-1}$ 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 $\eta_T(H) : T(H) \to \Hom(H, L(T))$ by sending $x \in T(H)$ to the map $$H \to L(T), \quad h \mapsto e(T(\widetilde{h})(x)),$$ where $\widetilde{h} : \IZ \to H$ denotes the homomorphism mapping $1 \mapsto h$, so that $T(\widetilde{h}) : T(H) \to T(\IZ)$. To see that this defines a group homomorphism from $H$ to $L(T)$, for $h,h' \in H$ we define the homomorphism $[h,h'] : \IZ * \IZ' \to H$ by $[h,h'] \circ \iota_1 = \widetilde{h}$ and $[h,h'] \circ \iota_2 = \widetilde{h'}$. Note that $[h,h'] \circ \mu = \widetilde{hh'}$. Hence,

$$\begin{align*} e(T(\widetilde{hh'})(x)) &= e(T(\mu)(T([h,h'])(x))) \\ & = e(T(\iota_1)(T([h,h'])(x))) \cdot e(T(\iota_2)(T([h,h'])(x))) \\ & = e(T(\widetilde{h})(x)) \cdot e(T(\widetilde{h'})(x)), \end{align*}$$

as required. Similar proofs show that the map $H \to L(T)$ respects inverses and the identity. We leave it as an exercise to verify that $\eta_T(H)$ 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 that $\varphi$ is such a homomorphism. Then, in particular, $$\alpha_{\IZ} = (y_{\varphi})_{\IZ} \circ (\eta_T)_{\IZ} : T(\IZ) \to \Hom(\IZ, L(T)) \to \Hom(\IZ, G).$$ 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

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

by applying naturality to $\mu, \iota_1, \iota_2 : \IZ \to \IZ * \IZ'$, where $U : \Grp \to \Set$ denotes the forgetful functor. 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 $e(T(\iota_1)(x)) \cdot e(T(\iota_2)(x))$ in $U(G)$ 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 $x \in T(H)$, the first step gives the homomorphism $h \mapsto e(T(\widetilde{h})(x))$; then the second step is formed by composition with $\varphi$. By the specification of $\varphi$, this gives the homomorphism $h \mapsto \alpha_{\IZ}(T(\widetilde{h})(x))(1)$. However, by the assumption that $\alpha$ is a natural transformation, for each $h \in H$ we have a commutative diagram

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

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