Which satisfy the triangle identities (a.k.a zig-zag identities)

$(\epsilon F)(F \eta) = 1_F$

$(G \epsilon)(\eta G) = 1_G$

Properties

Lemma 9.3.2

If $A$ s a category and $B$ is a precategory then the type “$F$ is a left adjoint” is a mere proposition?.

Proof. Suppose we are given $(G, \eta, \epsilon)$ with the triangle identities and also $(G', \eta', \epsilon')$. Define $\gamma: G \to G'$ to be $(G' \epsilon )(\eta G')$. Then

$\begin{aligned}
\delta \gamma &= (G \epsilon')(\eta G')(G' \epsilon) (\eta' G)\\
&= (G \epsilon')(G F G' \epsilon_))\eta G' F G)(\eta' G)\\
&= (G \epsilon ')(G \epsilon' F G)(G F \eta' G)(\eta G)\\
&= (G \epsilon)(\eta G)\\
&= 1_G
\end{aligned}$

Now we need to know that when $\eta$ and $\epsilon$ are [transported]] along this identity, they become equal to $\eta'$ and $\epsilon '$. By Lemma 9.1.9,

Lemma 9.1.9 needs to be included. For now as transports are not yet written up I didn’t bother including a reference to the page category. -Ali

this transport is given by composing with $\gamma$ or $\delta$ as appropriate. For $\eta$, this yields

$(G' \epsilon F)(\eta' G F)\eta = (G' \epsilon F)(G' F \eta)\eta'=\eta'$

using Lemma 9.2.8 (see natural transformation) and the traingle identity. The case of $\epsilon$ is similar. FInally, the triangle identities transport correctly automatically, since hom-sets are sets. $\square$