# Given the adjunction morphisms $\epsilon, \eta$, then $(\eta \circ L) \circ(L\circ \epsilon) = \text{id}_L$ is easy to check.

Given the adjunction morphisms $\epsilon, \eta$, then $(\eta \circ L) \circ(L\circ \epsilon) = \text{id}_L$ is easy to check.

This is from page 29 of "categories and sheaves". I have a major confusion contusion regarding this and not sure where to even begin...

EDIT:

$\text{id}_L$ is the identity natural transformation for the functor $L : C \to C'$.

$\epsilon : \text{id}_{C'} \to R\circ L$ is constructed as $X \mapsto \psi(\text{id}_{L(X)})$ where $\psi : \text{Hom}_{C'}(L(X), L(X)) \simeq \text{Hom}_C(X, R \circ L(X))$.

$\eta : \text{id}_C \to R\circ L$ is constructed as?

• It would probably help if you write up precisely how each of these things are defined (both for the purpose of answerers but also because that will bring you much closer to seeing why it is true). – Tobias Kildetoft Aug 29 '17 at 9:33
• This might help : math.stackexchange.com/questions/999797/… – Arnaud D. Aug 29 '17 at 9:42
• Thanks for helping, ya'll. Not only did I figure it out, I got a cool diagram to draw. See below. – BananaCats Category Theory App Aug 29 '17 at 10:36
• Part of your confusion is probably due to using terrible notation, using $\circ$ both for horizontal and vertical composites. In the two most popular conventions, the left hand side of the equation would be written as $(\eta \circ L) \cdot (L \circ \epsilon)$ or $(\eta L) \circ (L \epsilon)$. – Hurkyl Aug 30 '17 at 1:50
• Aside: you should put all relevant information in the body of the question -- you should not make essential information (such as the actual question you mean to ask!) title-only. (I've made an edit to correct this) – Hurkyl Aug 30 '17 at 1:51

## 1 Answer

$$\require{AMScd} \begin{CD} \text{id}_C(X) @>{\text{id}_C(f)}>> \text{id}_C(Y)\\ @V\epsilon_XVV @VV\epsilon_YV \\ R\circ L(X) @>{R\circ L(f)}>> R \circ L(Y) \end{CD}$$

Then functor it over to $$C'$$ by composing the whole diagram with $$L$$:

$$\require{AMScd} \begin{CD} L(X) @>{L(f)}>> L(Y)\\ @VL(\epsilon_X)VV @VVL(\epsilon_Y)V \\ L\circ R\circ L(X) @>{L\circ R\circ L(f)}>> L\circ R \circ L(Y) \end{CD}$$

from which we can use $$\eta : L \circ R \to \text{id}_{C'}$$ to yield:

$$\require{AMScd} \begin{CD} L(X) @>{L(f)}>> L(Y)\\ @VL(\epsilon_X)VV @VVL(\epsilon_Y)V \\ L\circ R\circ L(X) @>{L\circ R\circ L(f)}>> L\circ R \circ L(Y) \\ @V\eta_{L(X)}VV @VV\eta_{L(Y)}V \\ L(X) @>{L(f)}>> L(Y) \end{CD}$$

The above visuallly exhibits that $$(\eta \circ L ) \circ (L\circ \epsilon)$$ is indeed a natural map of functors $$L \to L$$. Additionally it shows that for any $$L(f) \in \text{Hom}_{C'}(L(X), L(Y))$$ the natural map, so composed, takes us back to $$L(f)$$. The only map that can do that is unique, $$\text{id}_L$$.