# Existence and uniqueness of adjoints with respect to pairings

Let $$V,W,L$$ be $$R$$-modules over a commutative ring $$R$$. A pairing is an $$R$$-linear map $$V\otimes W\to L$$. An adjoint of an endomorphism $$f:V\to V$$ w.r.t a pairing $$V\otimes W\overset{g}{\to}L$$ is an endomorphism $$f^{\dagger ^g}:W\to W$$ such that $$g(f\otimes 1)=g(1\otimes f^{\dagger^g})$$.

When do such things exist and when are they unique? That is, what assumptions are needed on the modules involved, the pairing $$g$$, and $$f$$ itself? What if we suppose $$g$$ is a perfect pairing? (Perhaps this questions is as easily answerable for nice monoidal categories.)

For instance the adjugate of a linear endomorphism can be defined as an adjoint with respect to a canonical pairing. I would simply like to understand which "dualizability" assumptions are needed and where.

I'm going to denote adjoints by $$*$$ rather than $$\dagger_g$$, for notational simplicity.

Yes if $$g$$ is a perfect pairing, then adjoints always exist and are unique.

Let's exploit the tensor-hom adjunction and let $$g_V: W\to \newcommand\Hom{\operatorname{Hom}}\Hom(V,L)$$ be the obvious map. Now for any $$f$$, we can consider the map $$g_f : W\to \Hom(V,L)$$ defined by $$g_f(w) = g_V(w)\circ f$$. Then in order for an adjoint to exist, we must be able to solve the equation $$g_Vf^* = g_f.$$

Therefore if $$g_V$$ is an isomorphism (i.e. if $$g$$ is a perfect pairing), there is a unique $$f^*$$ satisfying the equation, $$f^*=g_V^{-1}g_f$$.

Let's generalize slightly. When can we solve $$g_Vf^*=g_f$$? Consider the following diagram $$\require{AMScd} \begin{CD} W @>g_f>> \Hom(V,L) \\ @Vf^*VV @| \\ W @>>g_V> \Hom(V,L) \end{CD}$$

Well, one answer to when we can find such a $$f^*$$ is if $$g_V$$ is surjective and $$W$$ is projective. In this case $$f^*$$ won't be unique. In fact, this is roughly the most general we can get, though.

To generalize this slightly, observe that the image of $$g_f$$ had better be a subset of the image of $$g_V$$, otherwise there is no way we can solve it. However if the image of $$g_f$$ is a subset of the image of $$g_V$$, then we can replace $$\Hom(V,L)$$ with $$\newcommand\im{\operatorname{im}}\im g_V$$, so that $$g_V$$ is now surjective to its image. Then as long as $$W$$ is projective, we can lift $$g_f$$ along $$g_V$$ to find $$f^*$$.

My final version of the second answer:

As long as $$W$$ is projective, and for every $$w$$, there exists $$w'$$ so that $$g_V(w)\circ f = g_V(w')$$, then there exists a (possibly not unique) "adjoint" $$f^*$$ solving $$g_Vf^* = g_f$$.

• So simple, so easy. Thanks! – Arrow Dec 24 '18 at 23:36
• @Arrow in the case of perfect pairings, yes, I'm currently editing, since I think I should be able to make this slightly more general. – jgon Dec 24 '18 at 23:37
• @Arrow, I generalized slightly. – jgon Dec 24 '18 at 23:49
• Very cool. Thank you! – Arrow Dec 24 '18 at 23:50