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Proposition Let $F:\mathcal{C}\to\mathcal{D}$ and $G,G':\mathcal{D}\to\mathcal{C}$ be functors. If $F\dashv G$ and $F\dashv G'$ then $G\simeq G'$.

proof. Since $F\dashv G$, there exists a natural isomorphism $$\Phi_{C,D}:\mathcal{D}(FC,D)\to\mathcal{C}(C,GD)$$ Since $F\dashv G'$, there exists a natural isomorphism $$\Phi_{C,D}':\mathcal{D}(FC,D)\to\mathcal{C}(C,GD')$$ Therefore, there exists a natural isomorphism $$\Phi_{C,D}'\circ\phi_{C,D}^{-1}:\mathcal{C}(C,GD)\to\mathcal{C}(C,G'D)$$ which goes from the functor $\alpha$ to the functor $\beta$, where

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and

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Fix $D\in\mathcal{D}$. Then $\alpha$ becomes $\bar{\alpha}=\mathcal{C}(-, GD)$ and $\beta$ becomes $\bar{\beta}=\mathcal{C}(-, G'D)$. Therefore, there exists an isomorphism $\bar{\Phi}:\mathcal{C}(-, GD)\Rightarrow\mathcal{C}(-, G'D)$ in the functor category $\operatorname{Set}^{\mathcal{C}^{\operatorname{op}}}$. Yoneda then gives a natural bijection $$Y:\operatorname{Set}^{\mathcal{C}^{\operatorname{op}}}(\mathcal{C}(-,GD),\mathcal{C}(-,GD'))\to\mathcal{C}(GD, GD')$$ Therefore, we have that $Y(\bar{\Phi}):GD\to GD'$ is an isomorphism of $\mathcal{C}$, for every $D\in\mathcal{D}$.

Here comes the part I have problems with. This isomorphism is also natural. Let $h:D\to D'$ be a morphism of $\mathcal{D}$. By naturality of $\Phi_{C,D}'\circ\phi_{C,D}^{-1}$ on the second argument, we get that the following square commutes for every object $C\in\mathcal{C}$. enter image description here

Therefore, the following diagram commutes

enter image description here

I suppose I need to use that Yoneda embedding is...an embedding. Is it correct? I am confuse on what version of Yoneda embedding to use, the covariant or contravariant one? Essentially I can't prove what follows: If $D \to D'$ is a morphism, then

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

commutes iff for every $C$ the diagram

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

commutes. Can you give some help please?

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Lazily avoiding retyping, see if this helps -- the explanation of the proof of Theorem 165 (p. 275) of these notes: https://www.logicmatters.net/resources/pdfs/GentleIntro.pdf

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