Adjoint group homomorphisms

Adjoint functors are a core topic in category theory. As group homomorphisms are essentially functors, when do they form an adjunction?

Consider a group \(G\). Its delooping is a one-object category, \(\mathbf{B}G\), such that it has a single object \(\bullet_G\) and each \(g\in G\) is treated as a morphism \(\bullet_G \to \bullet_G\). The morphism composition is then the usual group composition. A group homomorphism \(f\colon G\to H\) is then essentially the same as a functor \(\mathbf{B}f\colon \mathbf{B}G\to \mathbf{B}G\) (with the difference that the functor \(\mathbf{B}f\) additionally maps the object \(\bullet_G\) to the object \(\bullet_H\)).

Consider now two group homomorphisms, \(f\colon G\to H\) and \(u\colon H\to G\). If we treat them as functors, \(\mathbf{B}f\) and \(\mathbf{B}u\), under which circumstances are they adjoint? In other words, what are the properties of homomorphisms of \(f\) and \(u\), such that \(\mathbf{B}f \dashv \mathbf{B}u\)?

For an adjunction we need a natural isomorphism \(\psi_{\bullet_G\bullet_H}\colon \mathrm{Hom}_{\mathbf{B}H}( \mathbf{B}f(\bullet_G), \bullet_H) \to \mathrm{Hom}_{\mathbf{B}G}(\bullet_G, \mathbf{B}g(\bullet_H) )\).This is just a convoluted way of writing that we need a bijection \(\psi\colon H\to G\) with additional properties, making it the natural isomorphism.

What properties? Let’s introduce some notation. For an arrow \(h\in H\) we write \(h^\flat = \psi(h)\). If \(\phi\colon G\to H\) be the inverse, \(\phi = \psi^{-1}\), and \(g\in G\), we write \(g^\sharp = \phi(g)\). Of course, we have \((h^\flat)^\sharp = h\) and \((g^\sharp)^\flat = g\) for all \(h\) and \(g\).

The first property (naturality in \(\mathbf{B}H\)) reads that for every \(h_1\) and \(h_2\) in \(H\), we have \[ u(h_1) h_2^\flat = (h_1h_2)^\flat, \] or equivalently, \[ u(h_1) \psi(h_2) = \psi(h_1h_2). \]

The second property (naturality in \(\mathbf{B}G\)) reads that for every \(g\in G\) and \(h \in H\) it holds that \[ (h \circ f(g))^\flat = h^\flat g. \]

This looks asymmetric, but we can fix it. As \(G\) and \(H\) have to be bijections, we can write \(h = g_1^\sharp\). Rename \(g\) to \(g_2\). Then, \[ (g_1^\sharp f(g_2))^\flat = g_1 g_2, \]

or equivalently \[ g_1^\sharp f(g_2) = (g_1 g_2)^\sharp, \]

or equivalently \[ \phi(g_1) f(g_2) = \phi(g_1g_2). \]

To summarise, we require \[ \phi(g_1)f(g_2) = \phi(g_1 g_2), \quad u(h_1) \psi(h_2) = \psi(h_1 h_2). \]

for all \(g_1, g_2\in G\) and \(h_1, h_2 \in H\). Note that these conditions differ from the definition of a homomorphism, having \(f(g_2)\) instead of \(\phi(g_2)\) and \(u(h_1)\) instead of \(\psi(h_1)\). They are also slightly asymmetric, which is related to the fact that adjoint functors also have this asymmetry: we want \(f\) to be the left adjoint and \(g\) to be the right adjoint.

These conditions are, actually, very restrictive: if we take $g_1 to be the identity element, then we have \[ \phi(1_G) f(g) = \phi(g) \]

for all \(g\in G\). We can therefore write \(f\) as a composition of bijective functions, i.e., \(f\) is also bijective. And as \(f\) is a bijective group homomorphism, it has to be a group isomorphism.

Analogous argument proves that any right adjoint \(u\) has to be a group isomorphism as well. Let’s now think how \(f\) and \(u\) can be related. A natural candidate for a right adjoint \(u\) for \(f\) would be its inverse.

Proposition: If \(f\colon G\to H\) is a group isomorphism, then we have an adjunction \(f \dashv f^{-1}\).

Proof: Define \(\phi(g) = f(g)\), so that \(\psi(h) = f^{-1}(h)\). The naturality equations are then obvious.

If \(u\) is a group isomorphism, then also \(u^{-1}\) is a group isomorphism, and we obtain \(u^{-1} \dashv u\) adjunction. The above adjunction (i.e., use the inverse) is almost the only one existing: adjoint functors are unique up to a natural isomorphism.

Let’s therefore think what it means for two group homomorphisms \(f_1\colon G\to H\) and \(f_2\colon G\to H\) to be naturally isomorphic. As usual, we have only a single object \(\bullet_G\) in the category \(\mathbf{B}G\), indexing a single arrow \(a\colon \bullet_G \to \bullet_G\), i.e., \(a\in G\). As it is a natural transformation, we have to have \[ f_1(g) a = a f_2(g) \]

for every \(g\in G\) and \(f_2(g) = a^{-1} f_1(g) a\). By writing \(\sigma_a\colon h\to h\) to be the inner automorphism \(h\mapsto a^{-1} h a\), we see that \(f_2 = \sigma_a f_1\) for some \(a\).

This allows us to formulate the theorem characterising adjunctions:

Theorem: Every group isomorphism \(f\colon G\to H\) has both left and right adjoints, given by \(\sigma_{\hat g} f^{-1}\) for arbitrary elements \({\hat g}\in G\). Conversely, if \(f\dashv u\) is any adjunction, then \(f\) and \(u\) are group isomorphisms and \(u = \sigma_{\hat g} f^{-1}\) for some \(\hat g \in G\).

At this point we see that the only examples of adjoint group homomorphisms arise from group isomorphisms and (appropriately conjugated) inverses.

Where are the monads?

Finally, let’s take a look at a monad arising from an adjunction. Before we start, let’s fix some notation. Take \(f \dashv u\) for \(u = \sigma_{\hat g} f^{-1}\) and \(\hat{g} \in G\). Hence, we have \(uf = \sigma_{\hat g}\colon G\to G\). The other composition can be obtained as \[ fu(h) = f\left( \hat g^{-1} f^{-1}(h) \hat g \right) = f(\hat g)^{-1} h f(\hat g) = \sigma_{\hat h} h \]

for \(\hat h = f(\hat g)\).

We need one more thing: the \(\phi = \psi^{-1}\) bijection. Let’s take \(\phi(g) = \hat h f(g)\) and \(\psi(h) = \hat g^{-1} f^{-1}(h)\). We have \[ \psi( \phi(g) ) = \hat g^{-1} f^{-1} ( \hat h f(g) ) = \hat g^{-1} f^{-1}(\hat h) g = \hat g^{-1} \hat g g = g \]

and

\[ \phi( \psi(h) ) = \hat h f( \hat g^{-1} f^{-1}(h) ) = \hat h f(\hat g)^{-1} h = h, \]

so that these are indeed inverse to each other. We can also verify that

\[ \phi(g_1)f(g_2) = \hat h f(g_1)f(g_2) = \hat h f(g_1 g_2) = \phi(g_1 g_2) \] and \[ u(h_1) \psi(h_2) = \left(\hat g^{-1} f^{-1}(h_1) \hat g\right) \left( \hat g^{-1} f^{-1}(h_2) \right) = \hat g^{-1} f^{-1}(h_1h_2) = \psi(h_1 h_2). \]

The unit of the adjunction is given by \[ \eta = (1_H)^{\flat} = \psi(1_H) = \hat g^{-1} \]

and is a natural transformation from the identity functor \(1_{\mathbf{B}G}\) to the functor given by \(uf = \sigma_{\hat g}\), which can be verified as \[ \hat g^{-1} (-) = \left( \hat g^{-1} ( - ) \hat g\right) \hat g^{-1}. \]

What is the counit? Analogously, it is \(\epsilon = (1_G)^\sharp = \hat h\). Let’s quickly verify the triangle identities:

\(\epsilon f(\eta) = \hat h f( \hat g^{-1} ) = 1_H\) and \(u(\epsilon) \eta = (\hat g^{-1} f^{-1}(\epsilon) \hat g) \hat g^{-1} = \hat g^{-1} f^{-1}(\hat h) = 1_G\).

The monad now is given by a functor \(t = uf = \sigma_{\hat g}\colon G\to G\), the unit \(\eta = \hat g^{-1}\), and the multiplication \(\mu = u\epsilon f\), which is a natural transformation from \(t^2\) to \(t\). More explicitly, multiplication is given by \(\mu = u(\epsilon) = \sigma_{\hat g}f^{-1}(\epsilon) = \sigma_{\hat g}( \hat g ) = \hat g\). It is a natural transformation as \[ \mu \, t^2(g) = t(g)\, \mu \]

for every \(g\in G\): both sides evaluate to \(\hat g^{-1}\, g\, \hat g^2\).

If \(\hat g\) is taken to be the identity element, then this monad is just the identity monad. What is the meaning of this one? I am not sure…