Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Assume we have a commutative diagram of functors $$ \array{ X&\stackrel{f^*}{\to}& Y\\ {\scriptstyle{g^*}}\downarrow\phantom{m} & & \phantom{m}\downarrow{\scriptstyle{k^*}}\\ Z&\stackrel{h^*}{\to}& T } $$ (where the natural isomorphism filling the 2-cell is not explicitly written), and assume that all the functors $f^*,g^*,k^*,h^*$ are parts of biadjoint pairs $(f^*,f_*)$, $(g^*,g_*)$ etc. such that also $f_*,g_*,h_*,k_*$ give a commutative diagram $$ \array{ X&\stackrel{f_*}{\leftarrow}& Y\\ {\scriptstyle{g_*}}\uparrow\phantom{m} & & \phantom{m}\uparrow{\scriptstyle{k_*}}\\ Z&\stackrel{h_*}{\leftarrow}& T } $$ (for instance $f^*$ can be the restriction functor for finite dimensional finite groups reprentations and $f_*$ the induced representation functor).

Then we can form the following diagram of natural transformations $$ \array{ f_*f^*&\stackrel{\epsilon_f}{\to}& Id &\stackrel{\eta_g}{\to} g_*g^*\\ {\scriptstyle{f_*}\eta_k{f^*}}\downarrow\phantom{mmm} & & &\phantom{mmmm}\uparrow{\scriptstyle{g_*}\epsilon_h{g^*}}\\ f_*k_*k^*f^*&&\stackrel{\sim}{\longrightarrow}& g_*h_*h^*g^* } $$ where $\eta, \epsilon$ are the unit and counit of the adjuctions and where the isomorphism on the bottom horizontal arrow is the one given by the commutative diagrams above.

And the question is: when does this diagram commutes?

Always? (I'm not confident in this, but I admit I got lost in my computations, so it is very well possible I've missed something elementary here)

If not always (as I think), are there known "natural" conditions ensuring its commutativity?

are there known classical examples of the above situation in which the latter diagram does commute?

The problem originating this question is that of giving an "as much functorial as possible" construction of the discrete quantization functor considered by Freed-Hopkins-Lurie-Teleman in Topological Quantum Field Theories from Compact Lie Groups, at least for the case of 1-categories.

share|improve this question
add comment

1 Answer

up vote 4 down vote accepted

The answer is always, provided you choose compatible 2-cells filling the first two diagrams. Probably, you didn't see it because you removed from notation these 2-cells. Let us give a name to the natural isomorphism for the first square:

$$\zeta\colon k^\ast f^\ast\cong h^\ast g^\ast$$

You need not demand the second square to be commutative, it automatically commutes by uniqueness of adjoints. The wise choice is:

$$f_\ast k_\ast \stackrel{f_\ast k_\ast\eta_h}\longrightarrow f_\ast k_\ast h^\ast h_\ast \stackrel{f_\ast k_\ast h^\ast \eta_g h_\ast}\longrightarrow f_\ast k_\ast h^\ast g^\ast g_\ast h_\ast \stackrel{f_\ast k_\ast \zeta^{-1} g_\ast h_\ast}\longrightarrow f_\ast k_\ast k^\ast f^\ast g_\ast h_\ast \stackrel{f_\ast \epsilon_k f^\ast g_\ast h_\ast}\longrightarrow f_\ast f^\ast g_\ast h_\ast \stackrel{\epsilon_f g_\ast h_\ast}\longrightarrow g_\ast h_\ast$$

This choice is precisely taken from the usual proof of uniqueness of adjoints.

With this choice, it's a straightforward exercise on 2-categorical composition to check that your third diagram commutes. You must also use the equations satisfied by (co)units in an adjunction.

share|improve this answer
2  
Thanks! Indeed I had forgotten the golden rule: never, by no reason, forget to keep track of the 2-cells :) –  domenico fiorenza Jun 14 '13 at 5:56
    
You 're welcome! –  Fernando Muro Jun 14 '13 at 6:43
    
Dear Fernando, working out the proof in detail, it seems to me that the "normalization" conditions $g_*g^*\stackrel{\epsilon_g}{\to}id\stackrel{\eta_g}{\to} g_*g^*$ being the identity of $g_*g^*$ and $id\stackrel{\eta_k}{\to}k_*k^*\stackrel{\epsilon_k}{\to} id$ being the identity of $id$ are also needed for the diagram to commute, is this correct? –  domenico fiorenza Jun 17 '13 at 19:39
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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