[proofplan] We prove the left-adjoint statement by using the universal property encoded by the unit of an adjunction: for each object $c\in\mathcal C$, the pair $(Fc,\eta_c)$ is initial among arrows from $c$ to $G$. Since $(Fc,\eta_c)$ and $(F'c,\eta'_c)$ solve the same universal problem, they are uniquely isomorphic, and these objectwise isomorphisms define the components of $\alpha$. Naturality and compatibility with the counits follow by applying the same uniqueness property to carefully chosen morphisms. The right-adjoint statement is obtained by the dual argument, replacing initial universal arrows by terminal universal arrows. [/proofplan] [step:Construct the component $\alpha_c$ from the universal property of $F\dashv G$] Fix an object $c\in\mathcal C$. Since $F\dashv G$ with unit $\eta:\operatorname{id}_{\mathcal C}\Rightarrow GF$, the [adjunction bijection](/page/Adjunction) sends a morphism $\bar f:Fc\to d$ in $\mathcal D$ to $G(\bar f)\circ \eta_c:c\to Gd$. Hence the following universal property holds: for every object $d\in\mathcal D$ and every morphism $f:c\to Gd$ in $\mathcal C$, there exists a unique morphism $\bar f:Fc\to d$ in $\mathcal D$ such that \begin{align*} G(\bar f)\circ \eta_c=f. \end{align*} Apply this universal property with $d:=F'c$ and $f:=\eta'_c:c\to GF'c$. There exists a unique morphism \begin{align*} \alpha_c:Fc\to F'c \end{align*} such that \begin{align*} G(\alpha_c)\circ \eta_c=\eta'_c. \end{align*} [guided] Fix an object $c\in\mathcal C$. The [adjunction](/page/Adjunction) $F\dashv G$ gives a natural bijection between morphisms $Fc\to d$ in $\mathcal D$ and morphisms $c\to Gd$ in $\mathcal C$. In unit form, this bijection sends $\bar f:Fc\to d$ to $G(\bar f)\circ \eta_c$. Therefore, whenever $d\in\mathcal D$ and $f:c\to Gd$ is a morphism in $\mathcal C$, there is a unique morphism $\bar f:Fc\to d$ in $\mathcal D$ with \begin{align*} G(\bar f)\circ \eta_c=f. \end{align*} We want a morphism from $Fc$ to $F'c$. The second adjunction supplies exactly the arrow from $c$ into the image under $G$ of $F'c$, namely \begin{align*} \eta'_c:c\to GF'c. \end{align*} Therefore we apply the universal property of $(Fc,\eta_c)$ with $d:=F'c$ and $f:=\eta'_c$. This gives a unique morphism \begin{align*} \alpha_c:Fc\to F'c \end{align*} such that \begin{align*} G(\alpha_c)\circ \eta_c=\eta'_c. \end{align*} This equation is the promised compatibility of $\alpha_c$ with the two units. [/guided] [/step] [step:Construct the inverse component and prove $\alpha_c$ is an isomorphism] Using the universal property of the adjunction $F'\dashv G$, applied with $d:=Fc$ and $f:=\eta_c:c\to GFc$, there exists a unique morphism \begin{align*} \gamma_c:F'c\to Fc \end{align*} such that \begin{align*} G(\gamma_c)\circ \eta'_c=\eta_c. \end{align*} We prove that $\gamma_c$ and $\alpha_c$ are inverse morphisms. First, \begin{align*} G(\gamma_c\circ \alpha_c)\circ \eta_c &=G(\gamma_c)\circ G(\alpha_c)\circ \eta_c \\ &=G(\gamma_c)\circ \eta'_c \\ &=\eta_c. \end{align*} The identity morphism $\operatorname{id}_{Fc}:Fc\to Fc$ also satisfies \begin{align*} G(\operatorname{id}_{Fc})\circ \eta_c=\eta_c. \end{align*} By the uniqueness part of the universal property of $(Fc,\eta_c)$ applied to the arrow $\eta_c:c\to GFc$, we get \begin{align*} \gamma_c\circ \alpha_c=\operatorname{id}_{Fc}. \end{align*} Similarly, \begin{align*} G(\alpha_c\circ \gamma_c)\circ \eta'_c &=G(\alpha_c)\circ G(\gamma_c)\circ \eta'_c \\ &=G(\alpha_c)\circ \eta_c \\ &=\eta'_c. \end{align*} Since $\operatorname{id}_{F'c}:F'c\to F'c$ also satisfies \begin{align*} G(\operatorname{id}_{F'c})\circ \eta'_c=\eta'_c, \end{align*} the uniqueness part of the universal property of $(F'c,\eta'_c)$ gives \begin{align*} \alpha_c\circ \gamma_c=\operatorname{id}_{F'c}. \end{align*} Thus $\alpha_c$ is an isomorphism with inverse $\gamma_c$. [guided] The construction of $\alpha_c$ used the universal property of $F\dashv G$. To prove that $\alpha_c$ is an isomorphism, we repeat the same construction in the opposite direction using the adjunction $F'\dashv G$. The unit $\eta_c:c\to GFc$ is an arrow from $c$ into the $G$-image of the object $Fc$. Applying the universal property of $(F'c,\eta'_c)$ to this arrow gives a unique morphism \begin{align*} \gamma_c:F'c\to Fc \end{align*} such that \begin{align*} G(\gamma_c)\circ \eta'_c=\eta_c. \end{align*} We now show that $\gamma_c$ is the inverse of $\alpha_c$. The composite $\gamma_c\circ \alpha_c:Fc\to Fc$ satisfies \begin{align*} G(\gamma_c\circ \alpha_c)\circ \eta_c &=G(\gamma_c)\circ G(\alpha_c)\circ \eta_c \\ &=G(\gamma_c)\circ \eta'_c \\ &=\eta_c. \end{align*} Here the first equality is functoriality of $G$, the second equality is the defining property of $\alpha_c$, and the third equality is the defining property of $\gamma_c$. The identity morphism $\operatorname{id}_{Fc}:Fc\to Fc$ has the same property: \begin{align*} G(\operatorname{id}_{Fc})\circ \eta_c=\operatorname{id}_{GFc}\circ \eta_c=\eta_c. \end{align*} The universal property of $(Fc,\eta_c)$ says that there is only one morphism $Fc\to Fc$ inducing the arrow $\eta_c:c\to GFc$. Therefore \begin{align*} \gamma_c\circ \alpha_c=\operatorname{id}_{Fc}. \end{align*} The other composite is handled by the same universal-property argument, now for $(F'c,\eta'_c)$: \begin{align*} G(\alpha_c\circ \gamma_c)\circ \eta'_c &=G(\alpha_c)\circ G(\gamma_c)\circ \eta'_c \\ &=G(\alpha_c)\circ \eta_c \\ &=\eta'_c. \end{align*} The identity morphism $\operatorname{id}_{F'c}:F'c\to F'c$ also induces $\eta'_c$, since \begin{align*} G(\operatorname{id}_{F'c})\circ \eta'_c=\eta'_c. \end{align*} Uniqueness for $(F'c,\eta'_c)$ gives \begin{align*} \alpha_c\circ \gamma_c=\operatorname{id}_{F'c}. \end{align*} Thus $\alpha_c$ is an isomorphism. [/guided] [/step] [step:Prove the components assemble into a natural transformation] Let $u:c\to c'$ be a morphism in $\mathcal C$. We must prove \begin{align*} F'(u)\circ \alpha_c=\alpha_{c'}\circ F(u). \end{align*} Both sides are morphisms $Fc\to F'c'$ in $\mathcal D$. It is enough to show that they induce the same morphism $c\to GF'c'$ under the universal property of $(Fc,\eta_c)$. For the left-hand side, using functoriality of $G$, the defining property of $\alpha_c$, and naturality of $\eta'$, we obtain \begin{align*} G(F'(u)\circ \alpha_c)\circ \eta_c &=G(F'(u))\circ G(\alpha_c)\circ \eta_c \\ &=G(F'(u))\circ \eta'_c \\ &=\eta'_{c'}\circ u. \end{align*} For the right-hand side, using functoriality of $G$, the defining property of $\alpha_{c'}$, and naturality of $\eta$, we obtain \begin{align*} G(\alpha_{c'}\circ F(u))\circ \eta_c &=G(\alpha_{c'})\circ G(F(u))\circ \eta_c \\ &=G(\alpha_{c'})\circ \eta_{c'}\circ u \\ &=\eta'_{c'}\circ u. \end{align*} The two morphisms $Fc\to F'c'$ therefore induce the same arrow $c\to GF'c'$. By uniqueness in the universal property of $(Fc,\eta_c)$, \begin{align*} F'(u)\circ \alpha_c=\alpha_{c'}\circ F(u). \end{align*} Thus $\alpha:F\Rightarrow F'$ is a natural isomorphism. [guided] To prove naturality, fix a morphism $u:c\to c'$ in $\mathcal C$. Naturality of $\alpha$ means that the square with vertices $Fc,Fc',F'c,F'c'$ commutes; explicitly, we must prove \begin{align*} F'(u)\circ \alpha_c=\alpha_{c'}\circ F(u). \end{align*} Both sides are morphisms from $Fc$ to $F'c'$. The universal property of $(Fc,\eta_c)$ gives a way to prove equality of such morphisms: two morphisms $h_1,h_2:Fc\to F'c'$ are equal if \begin{align*} G(h_1)\circ \eta_c=G(h_2)\circ \eta_c. \end{align*} So we compute the two induced arrows from $c$ to $GF'c'$. For the left-hand side, \begin{align*} G(F'(u)\circ \alpha_c)\circ \eta_c &=G(F'(u))\circ G(\alpha_c)\circ \eta_c \\ &=G(F'(u))\circ \eta'_c \\ &=\eta'_{c'}\circ u. \end{align*} The first equality uses functoriality of $G$, the second is the defining equation for $\alpha_c$, and the third is naturality of the unit $\eta'$. For the right-hand side, \begin{align*} G(\alpha_{c'}\circ F(u))\circ \eta_c &=G(\alpha_{c'})\circ G(F(u))\circ \eta_c \\ &=G(\alpha_{c'})\circ \eta_{c'}\circ u \\ &=\eta'_{c'}\circ u. \end{align*} Here the first equality is functoriality of $G$, the second is naturality of $\eta$, and the third is the defining equation for $\alpha_{c'}$. Both morphisms $Fc\to F'c'$ induce the same morphism $\eta'_{c'}\circ u:c\to GF'c'$. The uniqueness part of the universal property of $(Fc,\eta_c)$ therefore forces \begin{align*} F'(u)\circ \alpha_c=\alpha_{c'}\circ F(u). \end{align*} Hence the family $(\alpha_c)_{c\in\mathcal C}$ is a natural transformation, and since each component is an isomorphism, it is a natural isomorphism. [/guided] [/step] [step:Verify the counit compatibility of $\alpha$] Let $d\in\mathcal D$. The morphism $\varepsilon'_d\circ \alpha_{Gd}:FGd\to d$ satisfies \begin{align*} G(\varepsilon'_d\circ \alpha_{Gd})\circ \eta_{Gd} &=G(\varepsilon'_d)\circ G(\alpha_{Gd})\circ \eta_{Gd} \\ &=G(\varepsilon'_d)\circ \eta'_{Gd} \\ &=\operatorname{id}_{Gd}, \end{align*} where the last equality is one of the [triangle identities](/page/Adjunction) for the adjunction $F'\dashv G$. The counit $\varepsilon_d:FGd\to d$ satisfies \begin{align*} G(\varepsilon_d)\circ \eta_{Gd}=\operatorname{id}_{Gd} \end{align*} by the [triangle identity](/page/Adjunction) for $F\dashv G$. By uniqueness in the universal property of $(FGd,\eta_{Gd})$ applied to $\operatorname{id}_{Gd}:Gd\to Gd$, we obtain \begin{align*} \varepsilon'_d\circ \alpha_{Gd}=\varepsilon_d. \end{align*} [/step] [step:Prove uniqueness from unit compatibility] Now let $\tilde\alpha:F\Rightarrow F'$ be any natural transformation satisfying the unit compatibility \begin{align*} G(\tilde\alpha_c)\circ \eta_c=\eta'_c \end{align*} for every $c\in\mathcal C$. For each $c$, both $\tilde\alpha_c$ and $\alpha_c$ are morphisms $Fc\to F'c$ inducing $\eta'_c:c\to GF'c$. The uniqueness part of the universal property of $(Fc,\eta_c)$ gives \begin{align*} \tilde\alpha_c=\alpha_c. \end{align*} Thus $\tilde\alpha=\alpha$. [/step] [step:Prove uniqueness from counit compatibility] It remains to prove that the counit compatibility alone also characterizes $\alpha$. Let $\tilde\alpha:F\Rightarrow F'$ be any natural transformation satisfying \begin{align*} \varepsilon'_d\circ \tilde\alpha_{Gd}=\varepsilon_d \end{align*} for every object $d\in\mathcal D$. Fix $c\in\mathcal C$. By naturality of $\tilde\alpha$ applied to the morphism $\eta'_c:c\to GF'c$, we have \begin{align*} F'(\eta'_c)\circ \tilde\alpha_c=\tilde\alpha_{GF'c}\circ F(\eta'_c). \end{align*} Composing on the left with $\varepsilon'_{F'c}$ and using the counit compatibility at $d:=F'c$, we obtain \begin{align*} \tilde\alpha_c &=\varepsilon'_{F'c}\circ F'(\eta'_c)\circ \tilde\alpha_c \\ &=\varepsilon'_{F'c}\circ \tilde\alpha_{GF'c}\circ F(\eta'_c) \\ &=\varepsilon_{F'c}\circ F(\eta'_c). \end{align*} The first equality is the triangle identity for $F'\dashv G$. The morphism $\varepsilon_{F'c}\circ F(\eta'_c):Fc\to F'c$ induces $\eta'_c$ under the universal property of $(Fc,\eta_c)$, since \begin{align*} G(\varepsilon_{F'c}\circ F(\eta'_c))\circ \eta_c &=G(\varepsilon_{F'c})\circ GF(\eta'_c)\circ \eta_c \\ &=G(\varepsilon_{F'c})\circ \eta_{GF'c}\circ \eta'_c \\ &=\eta'_c, \end{align*} where the second equality is naturality of $\eta$ and the third equality is the triangle identity for $F\dashv G$. Hence the uniqueness part of the universal property of $(Fc,\eta_c)$ gives \begin{align*} \tilde\alpha_c=\alpha_c. \end{align*} Since this holds for every $c\in\mathcal C$, the counit compatibility also uniquely characterizes $\alpha$. This proves uniqueness in both equivalent senses. [guided] We next check that the same natural isomorphism is compatible with the counits. Fix an object $d\in\mathcal D$. We compare two morphisms from $FGd$ to $d$: \begin{align*} \varepsilon'_d\circ \alpha_{Gd}:FGd\to d \end{align*} and \begin{align*} \varepsilon_d:FGd\to d. \end{align*} The universal property of $(FGd,\eta_{Gd})$ says that these morphisms are equal if they induce the same morphism $Gd\to Gd$ after applying $G$ and precomposing with $\eta_{Gd}$. For the first morphism, \begin{align*} G(\varepsilon'_d\circ \alpha_{Gd})\circ \eta_{Gd} &=G(\varepsilon'_d)\circ G(\alpha_{Gd})\circ \eta_{Gd} \\ &=G(\varepsilon'_d)\circ \eta'_{Gd} \\ &=\operatorname{id}_{Gd}. \end{align*} The first equality is functoriality of $G$, the second is the defining property of $\alpha_{Gd}$, and the third is the triangle identity for the adjunction $F'\dashv G$. For the original counit of $F\dashv G$, the triangle identity gives \begin{align*} G(\varepsilon_d)\circ \eta_{Gd}=\operatorname{id}_{Gd}. \end{align*} Thus both morphisms $FGd\to d$ induce the same arrow $\operatorname{id}_{Gd}:Gd\to Gd$. By uniqueness in the universal property of $(FGd,\eta_{Gd})$, \begin{align*} \varepsilon'_d\circ \alpha_{Gd}=\varepsilon_d. \end{align*} Finally, suppose $\tilde\alpha:F\Rightarrow F'$ is another natural transformation satisfying the same unit compatibility: \begin{align*} G(\tilde\alpha_c)\circ \eta_c=\eta'_c \end{align*} for every $c\in\mathcal C$. For each fixed $c$, the morphisms $\tilde\alpha_c:Fc\to F'c$ and $\alpha_c:Fc\to F'c$ both induce the same arrow $\eta'_c:c\to GF'c$. The uniqueness part of the universal property of $(Fc,\eta_c)$ forces \begin{align*} \tilde\alpha_c=\alpha_c. \end{align*} Since this holds for every object $c\in\mathcal C$, we have $\tilde\alpha=\alpha$. Now suppose instead that $\tilde\alpha:F\Rightarrow F'$ satisfies only the counit compatibility \begin{align*} \varepsilon'_d\circ \tilde\alpha_{Gd}=\varepsilon_d \end{align*} for every object $d\in\mathcal D$. We prove that this already forces $\tilde\alpha=\alpha$. Fix $c\in\mathcal C$. Naturality of $\tilde\alpha$ for the morphism $\eta'_c:c\to GF'c$ gives \begin{align*} F'(\eta'_c)\circ \tilde\alpha_c=\tilde\alpha_{GF'c}\circ F(\eta'_c). \end{align*} Composing with $\varepsilon'_{F'c}:F'GF'c\to F'c$ and using the triangle identity $\varepsilon'_{F'c}\circ F'(\eta'_c)=\operatorname{id}_{F'c}$, we get \begin{align*} \tilde\alpha_c &=\varepsilon'_{F'c}\circ F'(\eta'_c)\circ \tilde\alpha_c \\ &=\varepsilon'_{F'c}\circ \tilde\alpha_{GF'c}\circ F(\eta'_c) \\ &=\varepsilon_{F'c}\circ F(\eta'_c), \end{align*} where the last equality is the assumed counit compatibility at $d:=F'c$. It remains to identify this morphism with $\alpha_c$. The morphism $\varepsilon_{F'c}\circ F(\eta'_c):Fc\to F'c$ induces $\eta'_c$ under the universal property of $(Fc,\eta_c)$ because \begin{align*} G(\varepsilon_{F'c}\circ F(\eta'_c))\circ \eta_c &=G(\varepsilon_{F'c})\circ GF(\eta'_c)\circ \eta_c \\ &=G(\varepsilon_{F'c})\circ \eta_{GF'c}\circ \eta'_c \\ &=\eta'_c. \end{align*} The second equality is naturality of the unit $\eta$ applied to $\eta'_c:c\to GF'c$, and the third equality is the triangle identity for $F\dashv G$. Since $\alpha_c$ is the unique morphism $Fc\to F'c$ inducing $\eta'_c$, we obtain \begin{align*} \tilde\alpha_c=\alpha_c. \end{align*} Thus the counit compatibility alone also uniquely characterizes $\alpha$. This proves uniqueness in both equivalent senses. [/guided] [/step] [step:Obtain the uniqueness of right adjoints by applying the left-adjoint result in opposite categories] For the dual statement, suppose $F:\mathcal C\to\mathcal D$ has right adjoints $G,G':\mathcal D\to\mathcal C$. Thus $F\dashv G$ and $F\dashv G'$. Passing to opposite categories reverses each [adjunction](/page/Adjunction): the functor \begin{align*} F^{\mathrm{op}}:\mathcal C^{\mathrm{op}}\to\mathcal D^{\mathrm{op}} \end{align*} is the common right adjoint of the two functors \begin{align*} G^{\mathrm{op}},(G')^{\mathrm{op}}:\mathcal D^{\mathrm{op}}\to\mathcal C^{\mathrm{op}}. \end{align*} Equivalently, \begin{align*} G^{\mathrm{op}}\dashv F^{\mathrm{op}} \quad\text{and}\quad (G')^{\mathrm{op}}\dashv F^{\mathrm{op}}. \end{align*} Applying the left-adjoint uniqueness result already proved, with $F^{\mathrm{op}}$ as the common right adjoint, gives a unique natural isomorphism between $G^{\mathrm{op}}$ and $(G')^{\mathrm{op}}$ compatible with the opposite adjunctions. Taking opposites, and reversing the resulting isomorphism if necessary, gives a unique natural isomorphism \begin{align*} \beta:G\Rightarrow G' \end{align*} such that \begin{align*} \beta_{Fc}\circ \eta_c=\eta'_c \end{align*} for every object $c\in\mathcal C$, equivalently \begin{align*} \varepsilon'_d\circ F(\beta_d)=\varepsilon_d \end{align*} for every object $d\in\mathcal D$. This proves the right-adjoint uniqueness statement and completes the proof. [/step]