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Every pair $F \dashv G$ of adjoint functors $F: \mathcal C \to \mathcal D$, $G: \mathcal D \to \mathcal C$ induces a monad $\mathbb T = (T,\eta,\mu)$ on $\mathcal C$. Given a monad $\mathbb T = (T,\eta,\mu)$ on $\mathcal C$, we define $\operatorname{Adj}(\mathbb T)$ to be the category of adjunctions inducing $\mathbb T$. Its objects are adjoint pairs of functors $F \dashv G$ between $\mathcal C$ and some category $\mathcal D$ such that the monad induced by the adjunction is $\mathbb T$, i.e. $GF = T$, the unit is $\eta$, and $G\varepsilon F = \mu$ where $\varepsilon$ is the counit. The morphisms between two such adjunctions $F \dashv G$, $F' \dashv G'$ are functors $H: \mathcal D \to \mathcal D'$ such that $HF=F'$ and $G' H = G$.

I am now worried whether this is really a category. Consider the example where $\mathcal C$ is a one-object category with only the identity morphism and $\mathbb T = (T,\eta,\mu)$ is the trivial monad. For every category $\mathcal D$ which has an initial object $I$, we get an adjunction $F \dashv G$, where $F: \mathcal C \to \mathcal D$ sends the one object of $\mathcal C$ to $I$, and $G: \mathcal D \to \mathcal C$ sends everything to the one object (it's easy to check that this is really an adjunction). Its induced monad on $\mathcal C$ is necessarily $\mathbb T$. So $\operatorname{Adj}(\mathbb T)$ contains at least one object for each category having an initial object. However, $\operatorname{Adj}(\mathbb T)$ itself has an initial object, namely the Kleisli adjunction, which in this case is just $1_{\mathcal C} \dashv 1_{\mathcal C}$. If $\operatorname{Adj}(\mathbb T)$ is a category, it seems that in a way it contains itself, in the form of the adjunction $\mathcal C \rightleftarrows \operatorname{Adj}(\mathbb T)$ defined by the Kleisli adjunction. (How) Is that possible?

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Well, either one works in a framework where self-containing categories are permitted (but I do not know of any), or else one works more carefully. So I would say something like, fix a universe $\mathbf{U}$ such that $\mathcal{C}$ is a $\mathbf{U}$-small category, and let $\mathbf{Adj}(\mathcal{C})$ be the category of $\mathbf{U}$-small categories equipped with an adjunction etc.; then $\mathbf{Adj}(\mathcal{C})$ will not be $\mathbf{U}$-small, and so will not member of itself.

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