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be a square of right adjoints which is commutative up-to a natural isomorphism $\varphi\colon v_*f_* \to g_*u_*$ (one can suppose it is the identity), where the left adjoint of $f_*$ is denoted by $f^*$, the unit by $\eta_f \colon 1_B \to f_*f^*$, the counit by $\epsilon_f \colon f^* f_* \to 1_A$ and similarly for $g_*$, $u_*$ and $v_*$.

There are two "mate" squares:

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given by precomposition and postcomposition of the appropriate units and counits, namely $$\alpha = (\epsilon_g\cdot u^*\cdot f_!)\circ (g_!\cdot \varphi \cdot f_!)\circ (g_!\cdot v^* \cdot \eta_F)\ :\ g^*v_* \to u_*f^*$$ and $$\beta = (\epsilon_v\cdot f_*\cdot u^*)\circ (v^* \cdot \varphi^{-1}\cdot u^*)\circ (v^*\cdot g_* \cdot \eta_u)\ :\ v^*g_* \to f_* u^*\,.$$ Finally, consider the mate square of $\beta$

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that is

$$\gamma = (\epsilon_f\cdot u^*\cdot g^*)\circ (f^*\cdot \beta \cdot g^*)\circ (f^* \cdot v^* \eta_g)\ :\ f^*v^* \to u^*g^*\,.$$

We can form the following triangle of natural transformations

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I should have managed to prove that it is commutative by graphical calculus and in the case I am interested in I can easily check that it commutes.

Question. Where can I find a reference for the commutativity of this triangle?

I imagine there are some very general results stating that between two compositions of functors of a certain type among the adjoint pairs described above, there exists a unique natural transformation which is a pasting of $\varphi$, $\alpha$, $\beta$, $\gamma$ and the units and counits.

I was addressed to this nice article of Ryan Reich, but skimming quickly through the user guide it seemed to me that I was not in the right condition to use his general results; although most likely his techniques allow to prove this particular case.

  • $\begingroup$ It seems unlikely to me that you'll find a reference for your particular commutative quadrilateral; it doesn't look important enough in general to have been written down explicitly in any abstract theory. There are lots of similar diagrams one could write down; the theory of mates and pasting gives us a way to check that they commute, not an exhaustive list of the ones that commute. (But I could easily be wrong!) $\endgroup$ Jan 19 '18 at 18:04

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