MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Just a minor curiosity that's flitted across my mind, but that's (part of) what this site's for, right?:

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).

share|cite|improve this question
up vote 6 down vote accepted

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 $End_k(V)$.

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; added: 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.).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.