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I am going through the adjuntions chapter of MacLane's CWM. I will follow his notation; it would be very difficult to try to describe all of the notation here. So excuse me for not explaining everything. Plus, I think everyone has a copy of CWM somewhere handy :)

In the proof of Corollary 1 (proof that left adjoints are unique up to natural isomorphism), he says "It is easy to verify that $\theta:F\to F'$ is natural". I am not sure if what I did is the proof he had in mind. Because I did not find this proof obvious (or at least as easy as one would expect re. proving naturality); maybe I am doing something wrong. So if someone could have a look at this, I would be grateful.

Let $f:x\to y$. We must prove that $\theta_y\circ Ff=F'f\circ \theta_x:Fx\to F'y$. Taking the "transpose" of the first morphism, we have \begin{eqnarray} \phi(\theta_y\circ Ff)&=& G(\theta_y \circ Ff)\circ \eta_y\\ &=&G\theta_y\circ GFf\circ\eta_x\\ &=&G\theta_y\circ\eta_y\circ f\\ &=&\eta_y'\circ f\\ &=&GF'f\circ \eta_x'\\ &=&GF'f\circ G\theta_x\circ \eta_x\\ &=&G(F'f\circ \theta_x)\circ \eta_x\\ &=&\phi(F'f\circ \theta_x). \end{eqnarray}

(where the third and fifth equalities hold by the definition of $F$ and $F'$ respectively on arrows -see Theorem 2.(ii) for example- and the fourth and sixth equalities hold by the definition of the isomorphism $\theta$ -see Corollary 1.)

Since the adjunct is unique, we must have $\theta_y\circ Ff=F'f\circ \theta_x$, as required.

(I know that this Corollary is immediately deduced from the Yoneda Lemma, if you consider the bijections on the hom-sets in the definitions of the adjunctions, but here I am only trying to prove naturality "manually", as asked in MacLane's proof.) Thanks for any help.

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"I think everyone has a copy of CWM somewhere handy" I wish, that thing is expensive! – Najib Idrissi Apr 19 '14 at 7:02
up vote 1 down vote accepted

Written equalities are just confusing when dealing with categorical properties. Prefer diagrams to it.

Suppose $F,F' \colon A \to B$ admits both $G \colon B \to A$ as right adjoint. By definition, there are bijections $$ \iota(x,y) \colon \hom_A(x,Gy) \simeq \hom_B(Fx,y) \\ \iota'(x,y) \colon \hom_A(x,Gy) \simeq \hom_B(F'x,y)$$ natural in $x$ and $y$ (i.e. $\iota,\iota'$ is a natural transformation between the functors $\hom_A(-,G(?)),\, \hom_B(F(-),?) \colon A^\mathrm{op} \times B \to \mathsf{Set}$).

Define $\vartheta(x) \colon Fx \to F'x$ as the image by $\iota(x,F'x)$ of the unit component at $x$ of $F' \dashv G$ : $$\vartheta(x) = \iota(x,F'x)(\eta'_x \colon x \to GF'x) = \iota(x,F'x)\circ {\iota'}^{-1}(x,F'x) (\operatorname{id}_{F'x} \colon F'x \to F'x).$$ Then naturality comes from naturality of $\iota$ : if $f \colon x_0 \to x_1$ is an arrow in $A$, then commutes $$ \require{AMScd} \begin{CD} \hom_B(F'x_0,F'x_0) @>\sim>\iota'^{-1}(x_0,F'x_0)> \hom_A(x_0,GF'x_0) @>\sim>\iota(x_0,F'x_0)> \hom_B(Fx_0,F'x_0) \\ @V F'(f) \,\circ\, -VV @V GF'(f) \,\circ\, - VV @VV F'(f) \,\circ\, -V \\ \hom_B(F'x_0,F'x_1) @>\sim>\iota'^{-1}(x_0,F'x_1)> \hom_A(x_0,GF'x_1) @>\sim>\iota(x_0,F'x_1)>\hom_B(Fx_0,F'x_1) \\ @A -\,\circ\, F'(f) AA @A -\,\circ\, f AA @AA -\,\circ\,F(f) A \\ \hom_B(F'x_1,F'x_1) @>\sim>\iota'^{-1}(x_1,F'x_1)> \hom_A(x_1,GF'x_1) @>\sim>\iota(x_1,F'x_1)> \hom_B(Fx_1,F'x_1). \end{CD} $$ By definition of the composition : $F'(f) \circ \operatorname{id_{F'x_0}} = \operatorname{id_{F'x_1}} \circ F'(f)$, which concludes.

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Thanks! So, is my attempt wrong? – Niels.Remb05 Apr 20 '14 at 6:53
@Niels.Remb05 No, it seems pretty much the same as my diagram (I don't have CWM at hand though, so I can't check for sure). But I think it is much more readable as a diagram than as a sequence of equalities (maybe it is just me). One thing however : at the end "Since the adjunct is unique" should be "Since $\phi$ is bijective". – Pece Apr 20 '14 at 8:45

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