[proofplan] We use the universal property twice: first with target $(V,\eta_V)$ to obtain a compatible morphism $\alpha: U \to V$, and then with target $(U,\eta_U)$ to obtain a compatible morphism $\beta: V \to U$. The assumed functoriality of compatibility makes the composites $\beta \circ \alpha$ and $\alpha \circ \beta$ compatible endomorphisms of the corresponding universal objects. The assumed identity-compatibility makes the identity morphisms compatible as well, so uniqueness in the universal property forces the composites to be identities. Thus $\alpha$ and $\beta$ are inverse isomorphisms, and the same uniqueness clause gives uniqueness of $\alpha$. [/proofplan] [step:Use the universal property to construct the two compatible morphisms] Apply the universal property of $(U,\eta_U)$ to the admissible pair $(V,\eta_V)$. This gives a unique morphism \begin{align*} \alpha: U \to V \end{align*} compatible with $\eta_U$ and $\eta_V$. Apply the universal property of $(V,\eta_V)$ to the admissible pair $(U,\eta_U)$. This gives a unique morphism \begin{align*} \beta: V \to U \end{align*} compatible with $\eta_V$ and $\eta_U$. [/step] [step:Show that the composite $\beta \circ \alpha$ is the identity on $U$] The morphism \begin{align*} \beta \circ \alpha: U \to U \end{align*} is compatible with $\eta_U$ and $\eta_U$ by the assumed closure of compatible morphisms under composition, applied to the compatible morphisms $\alpha: U \to V$ and $\beta: V \to U$. The identity morphism \begin{align*} \operatorname{id}_U: U \to U \end{align*} is compatible with $\eta_U$ and $\eta_U$ by the assumed compatibility of identity morphisms. By the uniqueness clause in the universal property of $(U,\eta_U)$, there is only one morphism $U \to U$ compatible with $\eta_U$ and $\eta_U$. Therefore \begin{align*} \beta \circ \alpha = \operatorname{id}_U. \end{align*} [/step] [step:Show that the composite $\alpha \circ \beta$ is the identity on $V$] The morphism \begin{align*} \alpha \circ \beta: V \to V \end{align*} is compatible with $\eta_V$ and $\eta_V$ by the assumed closure of compatible morphisms under composition, applied to the compatible morphisms $\beta: V \to U$ and $\alpha: U \to V$. The identity morphism \begin{align*} \operatorname{id}_V: V \to V \end{align*} is compatible with $\eta_V$ and $\eta_V$ by the assumed compatibility of identity morphisms. By the uniqueness clause in the universal property of $(V,\eta_V)$, there is only one morphism $V \to V$ compatible with $\eta_V$ and $\eta_V$. Therefore \begin{align*} \alpha \circ \beta = \operatorname{id}_V. \end{align*} [/step] [step:Conclude that the compatible morphism is a unique isomorphism] The equalities \begin{align*} \beta \circ \alpha = \operatorname{id}_U, \qquad \alpha \circ \beta = \operatorname{id}_V \end{align*} show that $\alpha$ is an isomorphism with inverse $\beta$. Finally, if $\alpha': U \to V$ is any morphism compatible with $\eta_U$ and $\eta_V$, then $\alpha'$ and $\alpha$ are both compatible morphisms from $(U,\eta_U)$ to $(V,\eta_V)$. By the uniqueness clause in the universal property of $(U,\eta_U)$, we have \begin{align*} \alpha' = \alpha. \end{align*} Hence the compatible isomorphism $U \to V$ is unique. [/step]