# Construct a triangulated adjunction from a right triangulated adjoint functor?

In the Lemma 40. of the note on triangulated categories by Daniel Murfet, one finds the construction of a triangulated adjunction from a left (triangulated) adjoint triangulated functor, whose proof confuses me. I am going to, however, adapt the notations from the wiki page, for that notation seems to me more convenient and natural. In particular, what is called a left adjoint in the note above is a right adjoint in the wikian notation, I think.

Notations:
Given a right adjoint functor $G:C\to D,$ and a unit $\eta:1_D\to GF,$ where $F:D\to C$ is another functor, we can construct a $\text{Hom-set}$ adjunction $\Phi:\text{Hom}_C(F-, -)\rightarrow\text{Hom}_D(-,G-),$ as $\Phi_{Y,X}(f:F(Y)\to X)=G(f)\circ \eta_Y,$ and then we define $\varepsilon:FG\to 1_C$ as $\varepsilon_X=\Phi^{-1}_{G(X),X}(1_{G(X)}).$

Question:
Let $G$ be a triangulated functor and let $\eta$ be a triangulated natural transformation. How to prove that $\varepsilon$ is a triangulated natural transformation as well?

Attempt:
The proof given in the linked note quite confused me: it tells us to apply $G,$ and then compose with the isomorphism $\varphi_{G,X},$ and use the universality of $\eta,$ which is encoded in the bijectivity and the naturality of $\Phi.$ But this hint seems to lead to nowhere: We are trying to show $$\varepsilon_{\Sigma X}=\Sigma\varepsilon_X\circ\varphi_{F,GX}\circ F(\varphi_{G,X}).$$
How does applying $F$ help? And the composition with $\varphi_{G,X}$ doesn't seem very helpful either.

I tried another plan: Show that $\Phi$ has some good properties with respect to $\Sigma,$ that we can deduce from $\eta.$ So I rewrote the target equation as $$1_{G\Sigma X}=\Phi_{G\Sigma X,\Sigma X}(\Sigma\varepsilon_X\circ\varphi_{F,GX}\circ F(\varphi_{G,X}))=G(\Sigma\varepsilon_X)\circ\Phi_{\Sigma GX,\Sigma FGX}(\varphi_{F,GX})\circ\varphi_{G,X}=G(\Sigma\varepsilon_X)\circ\varphi_{G,FGX}^{-1}\circ\Sigma\eta_{GX}\circ\varphi_{G,X},$$
the last equality because $\eta$ is a triangulated natural transformation. But this appears to get even more complicated and entangled.
Thus any hints or references are greatly welcomed, thanks in advance.

• It turned out that the notations in the note and the wikian notations are not different: the difference results from an error on my part. But that does not affect the question here. Apr 22, 2015 at 23:54

Indeed the hint: apply $G$, and then compose with the isomorphism $\varphi_{G,X}$, and use the universality of $\eta$ can be translated into the following plan (Anyone not interested could just skip this part):
Since the universality of $\eta$ implies a unique factorisation, we can try to examine if the morphism $f:=\Sigma\varepsilon_X\circ\varphi_{F,GX}\circ F(\varphi_{G,X})$ solves the problem $\varepsilon$ solves, i.e. is it true that $G(f)\circ\eta_{\Sigma X}=1_{G\Sigma X}.$ If this is true, then by the universality of $\eta$ the proof is complete. But that simply translates to $\Phi_{G\Sigma X, \Sigma X}(f)=1_{G\Sigma X},$ just my proposed plan. And the hint goes further: since $\varphi_{G,X}$ is an isomorphism, if $\varphi_{G, X}\circ\Phi_{G\Sigma X, \Sigma X}(f)=\varphi_{G, X}$ then we have shown the requred identity.
Our goal is to show that $\Phi_{G\Sigma X, \Sigma X}(f)=1_{G\Sigma X},$ where $f:=\Sigma\varepsilon_X\circ\varphi_{F,GX}\circ F(\varphi_{G,X}).$ Then notice that as $\varphi_{G,X}$ is an isomorphism, we only have to show that $\varphi_{G, X}\circ\Phi_{G\Sigma X, \Sigma X}(f)=\varphi_{G, X}.$
Now expand and obtain: $$g:=\varphi_{G, X}\circ\Phi_{G\Sigma X, \Sigma X}(f)=\varphi_{G, X}\circ G(\Sigma\varepsilon_X)\circ\varphi_{G,FGX}^{-1}\circ\Sigma\eta_{GX}\circ\varphi_{G,X},$$ as $\varphi_G$ is a natural transformation we have $$g=\Sigma G\varepsilon_X\circ\varphi_{G,FGX}\circ\varphi_{G,FGX}^{-1}\circ\Sigma\eta_{GX}\circ\varphi_{G,X}=\Sigma G\varepsilon_X\circ\Sigma\eta_{GX}\circ\varphi_{G,X}\\=\Sigma(G(\varepsilon_X)\eta_{GX})\circ\varphi_{G,X}=\Sigma(\Phi_{GX,X}(\varepsilon_X))\circ\varphi_{G,X}=\varphi_{G,X},$$ as required.