Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

Your Answer


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

Browse other questions tagged or ask your own question.