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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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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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  • $\begingroup$ Thanks a lot for the clear-cut answer! $\endgroup$ – Sakif Khan May 13 '13 at 1:23
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This seems to be one of those cases where the proof from scratch is the easy, and you don't need to know (explictly) anything about Yoneda.

1). You know there is a natural isomorphism given by $\varphi_{A,D} :$hom$(A,GD)\rightarrow $hom$(A,G'D)$

2). Set $A=GD$, $\varphi _{GD,D}(1 _{GD})=\tau _{D}$. Then, by what follows, the $\tau _{D}$ are the components of the required natural isomorphism $G\overset{\cdot }{\rightarrow}G'$

3) Write

\begin{array}{ccc} \hom(GD,GD) & \rightarrow & \hom(GD,G'D) \\ \downarrow & & \downarrow \\ \hom(GD,GD_{1}) & \rightarrow & \hom(GD,G'D_{1} ) \end{array} Then if $f:D\rightarrow D_{1}$, follow $1_{GD}$ around the square. This gives $G'f\circ \tau _{D}=\varphi _{GD,D_{1}}(Gf)$.

4). Write \begin{array}{ccc} \hom(GD_{1},GD_{1}) & \rightarrow & \hom(GD_{1},G'D_{1}) \\ \downarrow & & \downarrow \\ \hom(GD,GD_{1}) & \rightarrow & \hom(GD,G'D_{1} ) \end{array}

and follow $1_{GD_{1}}$ around the square. This gives $\varphi _{GD,D_{1}}(Gf)= \tau _{D_{1}}\circ Gf$

5). Combining the previous two items, you get $G'f\circ \tau _{D}=\tau _{D_{1}}\circ Gf$, so $\tau $ is a natural transformation. Now, since $\varphi$ is an isomorphism, natural in $A$ and $D$, for each $D$, $\varphi _{GD,D}(1 _{GD})=\tau_{D}$ is an isomorphism, which implies $\tau $ itself is.

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    $\begingroup$ I'm confused by the last sentence. I don't see how $\varphi$ being an isomorphism implies that $\tau_D$ is. $\endgroup$ – prt13463 May 9 '18 at 17:10
  • $\begingroup$ Can we still use the same $f:D\rightarrow D_1$ in step (4)? $\endgroup$ – uk62116 May 18 '18 at 14:37
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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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  • $\begingroup$ Yes, but I'm not quite sure how that implies that the isomorphism $GD\cong G'D$ is natural. $\endgroup$ – Sakif Khan May 12 '13 at 23:51
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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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    $\begingroup$ I think this is a nice answer. $\endgroup$ – Sakif Khan May 13 '13 at 0:59
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A very compact proof:

We have $\text{Hom}(\cdot,G\cdot)\cong\text{Hom}(\cdot,G'\cdot)$ in $[{\cal C}^\text{op}\times{\cal D},\text{Set}]$ and want to show $G\cong G'$.

The Yoneda embedding $y:{\cal C}\rightarrow[{\cal C}^\text{op},\text{Set}]$ is full and faithfull, this property transmits to its pushforward $y_*:[{\cal D},{\cal C}]\rightarrow[{\cal D},[{\cal C}^\text{op},\text{Set}]]\cong[{\cal C}^\text{op}\times{\cal D},\text{Set}]$, which means $G\cong G'$ iff $y_*G\cong y_*G'$. But the last equation is exactly our premise when written out.

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