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    • CommentRowNumber1.
    • CommentAuthorjoe.hannon
    • CommentTimeSep 26th 2014

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

    • CommentRowNumber2.
    • CommentAuthorZhen Lin
    • CommentTimeSep 26th 2014
    • (edited Sep 27th 2014)

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

    • CommentRowNumber3.
    • CommentAuthorjoe.hannon
    • CommentTimeSep 27th 2014

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

    • CommentRowNumber4.
    • CommentAuthorMike Shulman
    • CommentTimeSep 29th 2014

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

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