I am trying to answer to this question: What can be said about adjunctions between groups (regarded as one-object categories)?

Here is what I have arrived at till now, but I can't conclude in some way. I mean, it seams to me that one last step is missing.

Let's consider $G_1$ and $G_2$ groups and let's view them as one-object categories $\mathcal{G}_1$ and $\mathcal{G}_2$, whose only objects are $G_1$ and $G_2$ respectively.\ Let's then take $F:\mathcal{G}_1\longrightarrow\mathcal{G}_2$, $H:\mathcal{G}_2\longrightarrow\mathcal{G}_1$ functors and suppose $F\dashv H$. This is equivalent to state that the map $\eta_{G_1}:G_1\longrightarrow HF(G_1)$ is initial in $(G_1\Rightarrow H)$ (it suffices to state it for this only map $\eta_{G_1}$ since $G_1$ is the only object of $\mathcal{G}_1$). However, being $\mathcal{G}_1$ and $\mathcal{G}_2$ one-object categories, we already know how the functors $F$ and $H$ are defined on objects, namely $F(G_1)=G_2$ and $H(G_2)=G_1$. So the map $\eta_{G_1}$ is nothing but a function from $G_1$ to itself.\ Let's now look at the comma category $(G_1\Rightarrow H)$. Its objects are the maps of the form $g_1:G_1\longrightarrow H(G_2)$, so they are simply the functions from $G_1$ to itself. A map between $g_1$ and $g_1'$ in $(G_1\Rightarrow H)$ is a function $g_2:G_2\longrightarrow G_2$ such that $g_1'=H(g_2)\circ g_1$, so we can write it as ``$g_1\rightarrow H(g_2)$''.\ Now, we want to impose that $\eta_{G_1}$ is an initial object in $(G_1\Rightarrow H)$. So let $g_1$ be an arbitrary object in $Ob(\mathcal{G}_1)$, and we force the map $g_2\in (G_1\Rightarrow H)(\eta_{G_1},g_1)$ to be unique. Thus we must have $g_2$ to be the unique map such that $g_1=H(g_2)\circ \eta_{G_1}$.

So now, what can we say really interesting about this $g_2$???

Thanks in advance for any help!

  • $\begingroup$ $\eta$ is not a function from $G_1$ to itself. It's a morphism in the category $\mathcal G_1$ from the unique object to itself. What are the morphisms in $\mathcal G_1$, again? $\endgroup$ – Kevin Carlson Jan 13 '17 at 19:14
  • $\begingroup$ You're right! I don't really know what they are in reality, indeed I have always thought of as the elements of the group $G_1$ . What are they effectively? $\endgroup$ – any_one Jan 13 '17 at 19:18
  • $\begingroup$ No, they are indeed elements of the group-not functions. $\endgroup$ – Kevin Carlson Jan 13 '17 at 22:33
  • $\begingroup$ Mmm ok...thank you :) $\endgroup$ – any_one Jan 14 '17 at 13:04

Since groups as one-object categories are a special case of groupoids, that is, categories in which every morphism is an isomorphism, for any adjunction $F\dashv H\colon\mathcal G_1\to\mathcal G_2$ between two group(oid)s, the unit and counit natural transformations $\mathrm{id}_{\mathcal G_1}\overset{\eta}\Rightarrow HF$ and $FH\overset\epsilon\Rightarrow\mathrm{id}_{\mathcal G_2}$ will be natural isomorphisms. In particular, all adjoint pairs of functors between two group(oid)s are adjoint equivalences.

In the special case of the adjunction $F\dashv H\colon\mathcal G_1\to\mathcal G_2$ of one-object categories (i.e. monoids), we can identify $\eta$ and $\epsilon$ with elements of $\mathcal G_1$ and $\mathcal G_2$ respectively, which must satisfy the naturality conditions

  1. $HF(g_1)\eta=\eta g_1$ for all $g_1\in\mathcal G_1$
  2. $g_2\epsilon=\epsilon FH(g_2)$ for all $g_2\in\mathcal G_2$

Furthermore, the zig-zag identities which makes the unit and counit into an adjunction between the two functors (i.e. monoid homomorphisms) are

  1. $\epsilon F(\eta)=\mathrm{id}_{\mathcal G_2}$
  2. $H(\epsilon)\eta=\mathrm{id}_{\mathcal G_1}$

By multiplying the naturality conditions with the inverses given by the zig-zag identity, we end up reducing all four conditions to the following two

  1. $H(\epsilon F(g_1))\eta=g_1$ for all $g_1\in\mathcal G_1$
  2. $g_2=\epsilon F(H(g_2)\eta))$ for all $g_2\in\mathcal G_2$

Indeed, the zig-zag identities are the special case where $g_1$ and $g_2$ are the identity elements of the groups. In particular, in the case of a group (where left and right inverses coincide), we have $HF(g_1)=\eta^{-1}g_1\eta$ and $FH(g_2)=\epsilon^{-1}g_2\epsilon$ where $\eta^{-1}=H(\epsilon)$ and $\epsilon^{-1}=F(\eta)$. Thus an adjunction between two groups consists of a pair of group homomorphisms $F\colon\mathcal G_1\rightleftarrows\mathcal G_2\colon H$ whose composites are inner automorphisms so that $F$ sends the inner automorphism of $\mathcal G_1$ to the inverse inner automorphism of $\mathcal G_2$ and reversly for $H$.


Every natural transformation between functors between groups is an equivalence, so every adjunction is an equivalence.

  • $\begingroup$ And equivalences between groups viewed as categories are isomorphisms of the groups, for they induce isomorphisms between the endomorphism monoid of the objects. $\endgroup$ – Mariano Suárez-Álvarez Jan 13 '17 at 19:58

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