An adjunction is a triple $(F, U, \zeta)$, where

  • $F\colon C\to D$ and $U\colon D\to C$ are functors and
  • $\zeta$ is an isomorphism between the functors $\operatorname{Hom}(-, U(-))$ and $\operatorname{Hom}(F(-), -)$.

Can it happen that for functors $F\dashv U$ there are two different natural isomorphisms $\zeta$ and $\zeta'$ such that $(F, U, \zeta)$ and $(F, U, \zeta')$ are adjunctions?

How different can $\zeta$ and $\zeta'$ be? For instance, each adjunction $(F, U, \zeta)$ induces an equivalence between the subcategories

  • $C_{\zeta}:=\{A\in C\mid \eta_A\colon A\to U(F(A))\text{ is an isomorphism}\}\leq C$
  • $D_{\zeta}:=\{B\in D\mid \epsilon_B\colon F(U(B))\to A\text{ is an isomorphism}\}\leq D$,

where $\eta$ and $\epsilon$ are the unit and counit induced by $\zeta$, respectively.

Can it happen that $C_{\zeta}\neq C_{\zeta'}$ and $D_{\zeta}\neq D_{\zeta'}$?

  • 1
    $\begingroup$ You should be able to modify $\zeta$ by composing with automorphisms of $U$ or $V$. Conversely, it looks to me like such a pair of $\zeta$ and $\zeta'$ are going to give an automorphism of $U$ and/or $V$ by Yoneda. $\endgroup$ – jgon Aug 20 '20 at 21:06
  • 2
    $\begingroup$ Does this answer your question? Are adjunctions unique? $\endgroup$ – Arnaud D. Aug 21 '20 at 8:32

Given a functor $U:\mathcal D\to\mathcal C$, a left adjoint $F$ (plus the adjunction unit) can be defined much more locally: if for every $x\in\mathcal C$, we have an initial object $\eta_x:x\to U(F_x)$ in the comma category $(x\downarrow U)$, then we can fix a choice of such an initial object for every $x$ and compile them into a left adjoint $F:\mathcal C\to\mathcal D$ induced by sending $x\mapsto F_x$, where the adjunction unit will be $\eta_x$. This is discussed in proposition 1.9 here.

In particular, if for some $x\in\mathcal C$ one choice of $\eta_x:x\to U(F_x)$ is an isomorphism, then all possible choices of the unit $\eta'_x:x\to U(F_x')$ must be isomorphisms, because initial objects are unique up to unique isomorphism. In particular, for any two natural isomorphisms $\zeta,\zeta':\operatorname{Hom}(-,U(-))\to\operatorname{Hom}(F(-),-)$, we will have $\mathcal C_\zeta=\mathcal C_{\zeta'}$. Dually, we also have $\mathcal D_\zeta=\mathcal D_{\zeta'}$.

However, this also reveals that the choice of unit and counit are componentwise only unique up to unique isomorphism (in the appropriate comma category) and thus not strictly unique. Since the unit and counit are uniquely determined by the natural isomorphism $\zeta:\operatorname{Hom}(-,U(-))\to\operatorname{Hom}(F(-),-)$ (in particular, $\eta_x:x\to U(F(x))$ is the preimgae under $\zeta$ of $\operatorname{id}:F(x)\to F(x)$ and $\epsilon_y:F(U(y))\to y$ is the image under $\zeta$ of $\operatorname{id}:U(y)\to U(y)$), this shows that the natural isomorphism $\zeta$ will not be unique either.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy