If $R$ doesn’t preserve limits, then I don’t see how to show that $(a\downarrow R) \to C$ reflects them.

]]>Ok yes, that’s excellent. Thank you. So $a\downarrow R$ is definitely not monadic. But even in this case, the functor $a\downarrow R\to C$ reflects limits (vacuously).

]]>To answer the title question: no, the comma category need not be monadic. One can easily contrive an example where $C$ is non-empty but $(a \downarrow R)$ is empty. (For instance, take $C$ to be the category of non-trivial rings, $D$ the category of all rings, $R$ the inclusion, and $a$ a trivial ring.)

]]>In overcategory, it is shown that the forgetful functor $a\downarrow C\to C$ reflects limits, and it is mentioned that this is a consequence of the undercategory being the category of algebras for the monad $b\mapsto a\coprod b.$ What about the comma category $a\downarrow R$, where $R\colon C\to D$ and $a\in D$? On the one hand, it seems like the diagrammatic proof still goes through and the forgetful functor $a\downarrow R\to C$ which takes $(a\to Rx) \mapsto x$ reflects limits. On the other hand, I cannot see a monad (or even just an endofunctor) for which $a\downarrow R$ comprises the algebras. Seems like it wants to be the functor $x\mapsto a\coprod Rx$, but this is not an endofunctor. Is there a better way to understand how $a\downarrow R\to C$ behaves with respect to limits?

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