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Background

I am reviewing some category theory, which I did not learn too well the first time around. One text I am using is Mac Lane's. Near the beginning of the chapter on adjunctions (pg 80), he gives the basic definition of an adjunction $<F,G,\phi>$ ($G$ right-adjoint to $F$) in terms of the natural isomorphism $\phi$: $$ \phi_{x,a}:A(Fx,a)\cong X(x,Ga) $$ Of course, this is not explicitly defined in general, since it depends on the particular adjunction.

Anyway, just on the next page, he begins to motivate and establish the definition of the unit, $\eta_x$. Taking $a=Fx$ above, we have $$ \phi_{x,Fx}:A(Fx,Fx)\cong X(x,GFx). $$ So, Mac Lane defines $$ \eta_x=\phi(1_{Fx}):x\rightarrow GFx $$ and claims that by "Yoneda's Proposition III.2.1, this $\eta_x$ is a universal arrow".

My confusion begins here. In terms of the variables we have used above, it states that, given a functor $G:A\rightarrow X$, an arrow $\eta_x:x\rightarrow GFx$ is universal from $x$ to $GF$ if and only if the function specified by the mapping $$ f':Fx\rightarrow a \quad\longmapsto \quad Gf'\circ \eta_x $$ is a natural bijection (in $a$): $$ A(Fx,a)\cong X(x,Ga) $$

Note that I abridged slightly the proposition, taking the relevant part.

Question

Our adjunction isomorphism $\phi$ does not necessarily involve the assignment above. (Or am I missing something?) So how can Mac Lane use this proposition to establish the universality of $\eta_x$?

Thanks, sorry if it is something trivial.

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    $\begingroup$ I don't understand, what you mean by "of course, this is not explicitly defined in general, since it depends on the particular adjunction"? $\endgroup$ Jul 2, 2014 at 17:10
  • $\begingroup$ Thanks for the responses and my apologies for the inappropriate posting. $\endgroup$
    – user48413
    Jul 3, 2014 at 15:26

2 Answers 2

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By Yoneda for every natural transformation $\tau \colon \mathbf C(c,-) \to F$, where $F \colon \mathbf C \to \mathbf{Set}$ and $\mathbf C(c,-)$ is the covariant $\hom$-functor, is a family of functions of the form

$$\tau_x \colon \mathbf C(c,x) \to F(x)$$ such that $\tau_x(f)=F(f)\circ \eta_x$ for every $f \in \mathbf C(c,x)$ (where $\eta_x=\tau_x(1_x)$).

By properties of natural transformations for $\mathbf {Set}$-valued functors a $\tau$ as above is a natural isomorphism if and only if all the $\tau_x$ are bijection. This amounts to say that for every $x \in \mathbf C$ and every $\alpha \in F(x)$ there exists a unique $f \colon c \to x$ in $\mathbf C$ such that $$\alpha=\tau_x(f)=F(f)(\eta_x)$$.

In the present case where $\mathbf C= A$ and $$F=\mathbf X(x,G(-))\colon \mathbf A \to \mathbf {Set}$$ by what previously said the natural morphism $\varphi_{x,-} \colon \mathbf A(F(x),-) \cong X(x,G(-))$ is such that for every $a \in A$ and $f \in A(F(x),a)$ we have $$\varphi_{x,a}(f)=X(x,G(f))(\eta_x) = G(f) \circ \eta_x\ .$$ This is an isomorphism such if and only if for every $g \in X(x,G(a))$ there's a unique $f \in A(F(x),a)$ such that $G(f)\circ \eta_x = g$, and this is exactly the universal property.

Hope this helps.

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Our adjunction isomorphism $\phi$ does not necessarily involve the assignment above.

Yes, it does, by Yoneda's lemma. The bijection $\phi$ is natural, hence its value $\eta_x$ at the identity determines its value at any $f'$ by exactly the formula given.

(A question like this would be better asked on math.SE, I think.)

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