Let $\text{Ob}:\textbf{Cat}\rightarrow\textbf{Set}$ be the forgetful functor mapping a small category to its set of objects. Consider the functor $R:\mathbf{Set}\rightarrow\textbf{Cat}$ mapping a set $X$ to the category having $X$ as its set of objects and a single morphism between each pair of objects. I am trying to show that $R$ is right adjoint to $\text{Ob}$.

For $x,y\in X$, should the single morphism between $x$ and $y$ in $R(X)$ be unique? Or, can I use the same morphism for each pair of objects in $R(X)$?–Does it matter?


Giving unique arrows would complicate the matter because I would have to specify the composition for each pair; whereas, in giving a constant morphism $*$, I can merely specify $*\circ*:=*$. Right?


1 Answer 1


All that matters is that the hom-set $R(X)(x, y)$ is a singleton for each $x, y \in X$. Identities and composition are therefore trivial, because there's only a single choice. It doesn't matter exactly what the singleton sets are (e.g. whether they are equal or not). You should try to avoid thinking of sets up to equality, and instead consider them only up to isomorphism.

  • $\begingroup$ You're correct. I thought that it would matter in the proof of $\theta_{X,\mathcal{C}}:X^{\text{Ob}(\mathcal{C})}\cong[\mathcal{C},R(X)]$, when trying to show that $\theta_{X,\mathcal{C}}(f)$ is a functor for $f:\text{Ob}(\mathcal{C})\rightarrow X$...but it doesn't since the singleton property implies $\theta_{X,\mathcal{C}}(f)(c'\circ c)=\theta_{X,\mathcal{C}}(f)(c')\circ\theta_{X,\mathcal{C}}(f)(c)$ without relying on the actual values of $\theta_{X,\mathcal{C}}(f)$ on these morphisms. Thanks! $\endgroup$
    – alf262
    Jul 22, 2020 at 18:31

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