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Let $F: \mathcal{C} \rightarrow \mathcal{D}$ be left adjoint to $G: \mathcal{D} \rightarrow \mathcal{C}$. Consider the unit $\eta$ and counit $\varepsilon$ of this adjunction.

Is it true that $GFG \varepsilon_B \circ \eta_{GFGB} = 1_{GFGB}$?

There is a split co-equaliser diagram in my notes, but I suspect that a couple of the arrows might be the wrong way round - the above morphism is equal to $\eta_{GB} \circ G\varepsilon_B$, which is backwards from the $\Delta$ identities. Any help would be appreciated.

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Perhaps you could include the diagram from your notes in your question. It might be easier to guess from there. What comes to my mind is to apply $GF$ to the triangular equation $G\varepsilon \circ \eta G = 1_G$ and evaluate the resulting natural maps at the object $B$. This would give $GFG\varepsilon_B \circ GF\eta_{GB} = 1_{GFGB}$. –  Marc Olschok Apr 25 '12 at 17:26
    
Turns out I wrote my diagram the wrong way round in my notes. I would have drawn the diagram but im not sure how to do them on this site in LaTeX. –  Paul Slevin Apr 25 '12 at 21:13
    
See this answer regarding commutative diagrams on this site.... Also, what is to be done with this question, are you still looking for an answer? –  ˈjuː.zɚ79365 Jun 15 '13 at 7:02
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