Often we have adjoint pairs $(A, B)$ (meaning $A$ is left adjoint to $B$). Sometimes we have adjoint triplets $(A,B,C)$ (meaning $A$ is left adjoint to $B$ and $B$ is left adjoint to $C$. No adjoint relation between $A$ and $C$ obviously, since they have the same source and target).

So the first question is: can we have quadruplets $(A,B,C,D)$ ? (meaning that $C$ is also left adjoint to $D$).

Second question: It would be then possible that we have $(A,D)$ besides $(A,B)$. But I think not, because I think that it is not possible to have a functor $A$ with two different right adjoints, $B$ and $D$. Is this correct?

  • 4
    $\begingroup$ See this MO thread for some good examples for the first question. For the second question: prove that the right adjoint of a functor is unique up to a unique isomorphism. $\endgroup$
    – t.b.
    Dec 16, 2011 at 14:51

1 Answer 1


Actually, for every natural number $\ge 2$, there is a maximal string of adjoint functors of that length. Here's one construction: let $\mathbf{n}$ denote the poset category $\{ 0 < \ldots < n - 1 \}$. I claim there are functors (i.e. order-preserving maps) \begin{align} d_i & : \mathbf{n} \to \mathbf{n} + 1 & & 0 \le i \le n \newline s_i & : \mathbf{n} + 1 \to \mathbf{n} & & 0 \le i \le n - 1 \end{align} such that $$d_n \dashv s_{n-1} \dashv d_{n-1} \dashv \cdots \dashv s_0 \dashv d_0$$ is a maximal string of adjoint functors. Moreover, if $\mathbf{C}$ is a category with a terminal object but no initial object, there are functors \begin{align} \partial_i & : [\mathbf{n} + 1, \mathbf{C}] \to [\mathbf{n}, \mathbf{C}] & & 0 \le i \le n \newline \sigma_i & : [\mathbf{n}, \mathbf{C}] \to [\mathbf{n} + 1, \mathbf{C}] & & 0 \le i \le n \end{align} such that $$\partial_0 \dashv \sigma_0 \dashv \cdots \dashv \partial_n \dashv \sigma_n$$ is a maximal string of adjoint functors. (Here $[\mathbf{D}, \mathbf{C}]$ denotes the category of all functors $\mathbf{D} \to \mathbf{C}$.)

It is also not hard to find an infinite string of adjoint functors: for example, $$\cdots \dashv \textrm{id} \dashv \textrm{id} \dashv \textrm{id} \dashv \cdots$$ is such a string, though a little degenerate.

As for your second question, recall that adjoints are unique up to unique natural isomorphism, i.e. if $F \dashv G$ and $F \dashv G'$, then there is a unique natural isomorphism $G \cong G'$ which interacts nicely with the unit and counit of the adjunction.

  • $\begingroup$ There's something odd with the first example: you define $s_i$ with $0\leq i \leq n-1$ but then in the chain below you use $s_n$. Also, wouldn't such a chain prove only that there is a maximal string of adjoint functors of odd length? (As it stands, the chain has length $2n+1$) $\endgroup$ May 21, 2012 at 14:17
  • $\begingroup$ @Bruno: $\sigma_n$ is not defined using $s_n$ (since it doesn't exist) and is quite different from the rest of them. If $\mathbf{C}$ has an initial object then there's also a $\sigma_{-1}$. Unless you are deliberately excluding posets from your definition of "category", the very first string of adjoints is a string of even length. $\endgroup$
    – Zhen Lin
    May 21, 2012 at 17:58
  • $\begingroup$ I never said anything about $\sigma_n$, I'm only talking about the first example (before the "moreover"): the second functor in the chain is $s_n$. $\endgroup$ May 22, 2012 at 1:50
  • $\begingroup$ @Bruno: Sorry, that's a typo. It's $s_{n-1}$, as the pattern suggests. $\endgroup$
    – Zhen Lin
    May 22, 2012 at 7:21
  • $\begingroup$ I'm sorry to bother you again. Could you please elaborate a bit on the examples? I've found that the first maps are called coface and codegeneracy maps, and they are the obvious injections and surjections. Is the second example a variation on the face/degeneracy maps? Can't you tweak the coface/codegeneracy maps you have above? FWIW, I'm not familiar at all with simplicial theory. $\endgroup$ May 31, 2012 at 3:43

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.