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Suppose we are given functors $F:\mathcal{C}\rightarrow\mathcal{D}$ and $G,G':\mathcal{D}\rightarrow\mathcal{C}$ such that $G$ and $G'$ are both right adjoint to $F$. To show that $G$ and $G'$ are isomorphic, we wish to come up with a natural isomorphism between the two functors. This is how Awodey does it:

For any $D\in\mathcal{D}$ and any $C\in\mathcal{C}$, we have that $Hom(C,GD)\cong Hom(FC,D)\cong Hom(C,G'D)$ simply using the hom-set definition of adjunctions. By the Yoneda principle, this means that $GD\cong G'D$ for all $D\in\mathcal{D}$.

Awodey then states that

But this isomorphism is natural in $D$, again by adjointness.

I dont' quite understand how the naturality follows from adjointness. Any help is appreciated.

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3 Answers

up vote 5 down vote accepted

The bijection $\hom(C,GD) \cong \hom(C,G'D)$ is natural in $C$, thus it is induced by an isomorphism $GD \cong G'D$ (Yoneda Lemma). But it is also natural in $D$, and since the Yoneda embedding is faithful, this means that $GD \cong G'D$ is natural in $D$. More details:

If $D \to E$ is a morphism, then

$\begin{array}{ccc} GD & \rightarrow & G'D \\ \downarrow & & \downarrow \\ GE & \rightarrow & G'E \end{array}$

commutes iff for every $C$ the diagram

$\begin{array}{ccc} \hom(C,GD) & \rightarrow & \hom(C,G'D) \\ \downarrow & & \downarrow \\ \hom(C,GE) & \rightarrow & \hom(C,G'E) \end{array}$

commutes.

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Thanks a lot for the clear-cut answer! –  Sakif Khan May 13 '13 at 1:23
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We actually have a composite of natural isomorphisms $\tau:\hom(\bullet,G\bullet)\to \hom(F\bullet,\bullet)\to \hom(\bullet,G'\bullet)$.

In more details, $\tau$ has components $\tau_{C,D}:\hom(C,GD)\to\hom(C,G'D)$ satisfying the naturality condition: for arrows $\gamma:C_1\to C$, $\ \delta:D\to D_1\,$ and $\, f:C\to GD$ we have $$\tau(G\delta\circ f\circ\gamma)=G'\delta\circ\tau(f)\circ\gamma\,. $$ Now applying the Yoneda lemma basically means to consider the case when $C=GD$ and apply $\tau$ to $1_{GD}$. It leads to a $GD\to G'D$ arrow, verify (using the above equation) that it's natural in $D$, and find its inverse (e.g. by applying the same to $C=G'D$ and $\tau^{-1}$.

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Yes, but I'm not quite sure how that implies that the isomorphism $GD\cong G'D$ is natural. –  Sakif Khan May 12 '13 at 23:51
    
Thanks for the reference. –  Sakif Khan May 13 '13 at 0:58
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I don't sure that I answer precisely on your question, but I want to share my idea.

Let's view the category $C$ as a full subcategory of its presheaves(image of the Yoneda embedding $C\to\mathbf{Set}^{C^{op}}$).

Adjunction gives you a natural isomorphism $hom_C(C,G(D))\to hom_C(C,G'(D))$, which is actually a natural isomorphism between the following two functors: $$ X=(hom_C\circ(I_{C^{op}}\times G))\colon C^{op}\times D\to\mathbf{Set}; $$ $$ Y=(hom_C\circ(I_{C^{op}}\times G'))\colon C^{op}\times D\to\mathbf{Set}, $$ which are objects in the category $\mathbf{Set}^{C^{op}\times D}$. Consider an exponential functor: $$ E\colon\mathbf{Set}^{C^{op}\times D}\to(\mathbf{Set}^{C^{op}})^D $$ So we have two isomorphic functors $E(X),E(Y)\colon D\to\mathbf{Set}^{C^{op}}$. It's easy to check that their images are in $C$, and constriction on $C$ yields $E(X)|_C=G, E(Y)|_{C}=G'$. Thus, the natural isomorphism between $E(X)$ and $E(Y)$ induce a natural isomorphism between $G$ and $G'$.

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I think this is a nice answer. –  Sakif Khan May 13 '13 at 0:59
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