I predict that someone such as Steve Lack or Mike Shulman will tell you about the existence of (co)limits in **Mon**, and they'll do it better than I would, so instead I'll address a question in the last paragraph: do $M(0)$, $M(1)$ and $M(0 \to 1)$ tell you much about the rest of $M$? The answer is basically no.

To see this -- and to understand monads -- it's helpful to observe that if $M$ is regarded as an algebraic theory then $M(n)$ is the set of words in $n$ letters, or equivalently $n$-ary operations in the theory. For example, if $M$ is the monad for groups then $M(n)$ is the set of words-in-the-group-theory-sense in $n$ letters, which are the same as the $n$-ary operations in the theory of groups. For example, $x^3 y^2 x^{-1}$ is a typical word in two letters, and $(x, y) \mapsto x^3 y^2 x^{-1}$ is a typical binary operation (way of turning a pair of elements of a group into a single element). Similarly, if $M$ is the monad for rings then $M(n)$ is the set of polynomials over $\mathbb{Z}$ in $n$ variables, which are the $n$-ary operations in the theory of rings.

(Personally I'm happy to think of *any* monad on **Set** as an algebraic theory, although others prefer a more restrictive definition of theory. Anyway, this point of view doesn't matter in what follows.)

In particular, $M(0)$ is the set of nullary operations, or constants, in the theory, and $M(1)$ is the set of unary operations. Now let's consider on the one hand the identity monad $I$, and on the other the monad $M$ corresponding to the theory generated by a single binary operation $\cdot$ subject to the equation $x\cdot x = x$. (Or if you want a more familiar but more complicated example, take $M$ to be the monad for [not necessarily bounded] lattices or semilattices; the point is that $x \vee x = x$.) We then have $I(0) = 0 = M(0)$ and $I(1) = 1 = M(1)$, hence also
$$
I(0 \to 1) = (0 \to 1) = M(0 \to 1).
$$
But $I$ and $M$ are very different monads.

On the positive side, I think you might be interested in the following result; it seems in the spirit of your question. Suppose there exists an $M$-algebra with more than one element. Then:

- $M$ reflects isomorphisms
- $M$ is faithful
- the unit $\eta$ of $M$ is monic (i.e. the free $M$-algebra on a set of generators contains the generators as a subset -- it doesn't identify any of them).

What's more, all but two monads on **Set** have this property that there exists an algebra with more than one element. One of the exceptions is the monad $M$ with $M(A) = 1$ for all sets $A$; it's the theory generated by a single constant $e$ and the equation $x = e$. The other is the monad $M$ with $M(A) = 1$ for all nonempty sets $A$ and $M(0) = 0$; that's the theory generated by no operations and the equation $x = y$.

**Edit** David asked where a proof of this "positive" result could be found. I learned it from a Cambridge Part III problem sheet by Peter Johnstone, and it's probably out there somewhere in the literature (e.g. maybe in Johnstone's *Sketches of an Elephant*). But since consulting the literature means getting up from the couch, I'll type out a proof instead.

So, let $M = (M, \eta, \mu)$ be a monad on **Set** such that there exists at least one $M$-algebra with more than one element. Write $F: \mathrm{Set} \to \mathrm{Set}^M$ for the free algebra functor, and $U: \mathrm{Set}^M \to \mathrm{Set}$ for the underlying set functor; thus, $M = UF$.

First we show that $\eta: id \to M$ is monic. This means for each set $A$ and each $a, b \in A$ with $a \neq b$, we have $\eta_A(a) \neq \eta_A(b)$. By the universal property of $\eta_A$, this is equivalent to the assertion that for some $M$-algebra $X$ and map $f: A \to U(X)$ of sets, we have $f(a) \neq f(b)$. (I'd like to draw a commutative triangle to illustrate this.) Choose any $M$-algebra $X$ with more than one element: then such an $f$ can indeed be constructed.

Next we show that $F$ is faithful. It will follow that $M = UF$ is faithful, since $U$ (being monadic) is also faithful. Faithfulness of $F$ is in fact *equivalent* to each $\eta_A$ being monic, by general properties of adjunctions; I think that's in *Categories for the Working Mathematician*. Anyway, the proof is that if $f, g: A \to B$ are maps of sets with $Ff = Fg$ then $\overline{Ff} = \overline{Fg}$ (where the bar indicates transpose); but $\overline{Ff} = \eta_B \circ f: A \to UF(B)$ and similarly for $g$, and $\eta_B$ is monic, so $f = g$.

Finally we show that $F$ reflects isos. It will follows that $M = UF$ reflects isos, since $U$ (being monadic) also reflects isos. A faithful functor always reflects both epis and monos, and any map in **Set** that's both epi and mono is an iso. Hence any faithful functor out of **Set** reflects isos, and the result follows from the previous part.

nothave that property: it preserves colimits on one side, but not the other. $\endgroup$ – Mike Shulman Feb 14 '11 at 5:10