# Proving that the counit of adjunction is a natural transformation

Let $$F:\mathcal A\to\mathcal B, G:\mathcal B\to \mathcal A$$ be functors such that $$F$$ is left adjoint to $$G$$. I'm trying to prove that the counit of adjunction $$\epsilon: FG\to 1_\mathscr{B}$$ is a natural transformation.

So the task is to prove that the following square commutes:

i.e. $$\overline {1_{G(B')}}\circ FG(g)=1_\mathscr{B}(g)\circ\overline {1_{G(B)}}$$.

I tried writing down the conditions of $$F$$ and $$G$$ being adjoint. The first one is the commutativity of this square:

The LHS of the last equation is equal to the RHS of the equation we need to establish.

Then I tried to use the other requirement on the functors being adjoint, namely, the commutativity of this square:

(it's assumed that there's an arrow $$f:A'\to A$$ in the background).

I put $$A=G(B'), A'=G(B)$$ and got the highlighted equality from the commutativity of the diagram:

If I could set $$f=id$$, that would tell me $$FG(g)=\overline {G(g)}$$, and the proof would be finished. But I can't do that (I can't set $$B'=B$$, because I wouldn't be able to use the map $$G(g)$$ in the diagram).

So how should I proceed?

• After the second diagram, I should have applied Lemma 2.2.4 from Leinster to finish the proof. The third and fourth diagrams aren't needed. – user634426 Jun 20 '20 at 17:32

There are lots of ways to show this, because there's actually a lot going on here. For example, $$\epsilon_B$$ is actually a universal morphism; that's one way you can show this. But a simple way is to use the relations that are given by the definition of an adjunction. To be clear, I’ll write $$\varphi: \text{Hom}_{\mathcal{B}}(F(A), B) \to \text{Hom}_{\mathcal{A}}(A, G(B))$$ so that $$\epsilon_B = \varphi^{-1}(1_{G(B)})$$
Now recall that for $$k: B \to B’$$ and $$h: A’ \to A$$, the diagrams below commute by naturality of $$\varphi$$. Now for any $$f: F(A) \to B$$, the above diagrams give us the relations $$\varphi(k \circ f) = G(k) \circ \varphi(f) \qquad \varphi(f \circ F(h)) = \varphi(f) \circ h.$$ To show that $$\epsilon_B$$ is natural, you’ll need to show that the diagram below commutes.
Observe that we have the relation $$\text{Hom}_{B}(F(G(B), B’) \cong \text{Hom}_{A}(G(B), G(B’))$$ By just substituting $$A = G(B)$$. But, by our formulas given by the definition of an adjunction, we get that $$\varphi(\epsilon_{B’} \circ F(G(g))) = 1_{G(B’)} \circ G(g) = G(g)$$ while $$\varphi(g \circ \epsilon_B) = G(g) \circ \varphi(\epsilon_B) = G(g) \circ 1_{G(B’)} = G(g).$$ So we see that $$\epsilon_{B’} \circ F(G(g))$$ and $$g \circ \epsilon_B$$ are sent to the same element. But since $$\phi$$ is an isomorphism, this implies that $$\epsilon_{B’} \circ F(G(g)) = g \circ \epsilon_B$$. Hence the diagram must commutes, so you have naturality.