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This may be a stupid question, but I just can't find out the solution.

I'm confusion on how to deal with adjoint pairs, which means a pair of $1$-morphisms $(L,R)$ with two $2$-morphisms $\eta\colon 1\to R\circ L$ and $\epsilon\colon L\circ R\to 1$ satisfies the triangle identities, see here.

In the case $L, R$ are functors, things are wonderful since $(L,R)$ is an adjoint pair iff they are adjoint functors to each other, meaning there is a natural bijection \begin{equation} Hom(L-,-)\cong Hom(-,R-). \end{equation} Then, one can deal with the Hom-sets and everything is fine.

However, I have no idea how to deal adjoint pairs in an arbitrary $2$-category.

For example, when one wants to show the adjoint functor is unique up to unique isomorphism, he/she only need to use the Hom-set statement and follows Yoneda lemma, while I haven't seen any way to prove this simple property for the general adjoint pairs.

So, the question is how to deal with the adjoint pairs in a $2$-category? In particular, are there something like the Hom-set statement for adjoint pairs and propositions like Yoneda lemma?

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    $\begingroup$ Hi GaXiv! This is probably going to get closed here, since it’s not really research level. But if you re-ask it at math.stackexchange, it should get some good answers. The short answer is: yes, you can make a hom-set/Yoneda-lemma like argument, by looking at maps into the 0-cels C and D (or out of them), and the natural functors between hom-cats induced by composition with L, R. Alternatively, look (in the 1-cat setting) at how(e.g.) the natural iso you obtain between two adjoints can be explicitly described in terms of the unit/co-unit (once you know how the transposition iso can be). $\endgroup$
    – Peter LeFanu Lumsdaine
    Oct 31, 2015 at 15:46
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    $\begingroup$ The argument that adjoints are unique up to unique isomorphism in a 2-category can be done purely equationally; it is a categorification of the argument that two-sided inverses are unique in a monoid. It's cleanest to describe using string diagram notation, though. $\endgroup$
    – Qiaochu Yuan
    Oct 31, 2015 at 17:40
  • $\begingroup$ Suppose $(L',R)$ together with $\eta': 1\to RL'$ and $\epsilon':L'R\to 1$ are also an adjoint pair. Then one obtains $(\epsilon \circ 1_{L'}) \cdot (1_{L} \circ \eta') : L \to L'$ and $(\epsilon \circ 1_L) \cdot (1_{L'} \circ \eta) : L' \to L$ which are inverse to each other (using the triangle identities). $\endgroup$
    – Nex
    Nov 1, 2015 at 7:18

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