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).



If a category is "monadic", is it necessarily so in a unique manner?
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.).


