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

and

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 $$C\in\mathcal{C}$$.

Therefore, the following diagram commutes

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?