most recent 30 from http://mathoverflow.net 2013-05-25T14:20:17Z http://mathoverflow.net/feeds/question/15435 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/15435/if-a-category-is-monadic-is-it-necessarily-so-in-a-unique-manner Sridhar Ramesh 2010-02-16T07:32:21Z 2010-02-16T15:35:17Z

<p>Just a minor curiosity that's flitted across my mind, but that's (part of) what this site's for, right?:</p> <p>Is it possible for Hom(a, -) and Hom(b, -) to both be monadic functors from C to Set, for non-isomorphic objects a and b in C? Ideally, the answer would come with either a nice example or an outline of a nice proof of impossibility (i.e., proof that all monadic representable functors on a category are isomorphic).</p>

http://mathoverflow.net/questions/15435/if-a-category-is-monadic-is-it-necessarily-so-in-a-unique-manner/15454#15454 Charles Rezk 2010-02-16T14:26:17Z 2010-02-16T15:35:17Z

<p>You can certainly have non-equivalent monadic functors. Here's one example: Let $\mathcal{V}_k$ be the category of $k$-vector spaces. For a vector space $V$, let $H_V: \mathcal{V}_k\to \mathcal{V}_k$ be the functor $$ H_V(W) = hom_k(V,W). $$ Such a functor is always monadic, as long as $V$ is non-zero and finite dimensional. The associated monad is $$ T_V(W) = hom_k(V, V\otimes_k W) = End_k(V)\otimes_k W, $$ so this is presenting a Morita equivalence: $k$-vector spaces are equivalent to modules over the matrix ring $End_k(V)$.</p> <p>You wanted functors to set; let $U_V:\mathcal{V}_k\to Set$ be given by the same formula as $H_V$. Then again, this will be monadic, as long as $V$ is non-zero and finite dimensional (and I'm not sure you even really need the finite dimensionality condition for either of these examples; <strong>added:</strong> you certainly don't in the first example, since $H_V$ is an exact functor, so the hypotheses of the Barr-Beck theorem certainly hold, though $T_V$ is not tensoring with an endomorphism ring if $V$ is infinite.).</p>