TITLE: Simple explanation of a monad
QUESTION [48 upvotes]: I have been learning some functional programming recently and I so I have come across monads. I understand what they are in programming terms, but I would like to understand what they are mathematically. Can anyone explain what a monad is using as little category theory as possible?

REPLY [15 votes]: Monads in Haskell and monads in category theory are very much the same: A monad consists of a functor $T: C \to C$ and two natural transformations $\eta_X : X \to T(X)$ (return in Haskell) and $\mu_X : T(T(X)) \to T(X)$ (join in Haskell) subject to the following laws
$\mu_X \circ T(\eta_X) = \mu_X \circ \eta_{T(X)} = 1_{T(X)}$ (left and right unit laws)
$\mu_X \circ \mu_{T(X)} = \mu_X \circ T(\mu_X)$ (associativity)
So, compared to Haskell, the monad is defined in terms of return, join and fmap instead of return and (>>=). For more details on this, see also the Haskell wikibook.
Two examples may illuminate this definition.
The powerset functor

$\mathcal{P} = X \mapsto \mathcal{P}(X)$ maps a set to the set of its subsets.
Functions $f:X \to Y$ are extended point-wise to $\mathcal{P}(f):\mathcal{P}(X) \to \mathcal{P}(Y)$
$\eta_X : X \to \mathcal{P}(X)$ is the function $x \mapsto \left\{x\right\}$
$\mu_X : \mathcal{P}(\mathcal{P}(X)) \to \mathcal{P}(X)$ flattens the inner layer of subsets: $\mu_X(A) = \left\{ b | a \in A, b \in a \right\}$.
This is similar to the list monad in Haskell.

The closure operation on the subsets of a topological space $S$ is a monad, too.

The objects of the category $C$ are the subsets of a given topological space $S$.
There is an unique arrow $X \to Y$ between to objects $X$ and $Y$ exactly when $X \subseteq Y$.
The monad is given by the functor that maps each object $X$ to its topological closure $\bar X$ and the arrow $X \subseteq Y$ to the arrow $\bar{X}\subseteq \bar{Y}$.
Clearly, we have $X \subseteq \bar X$; this is $\eta_X$.
Also, we know that $\bar{\bar X} = \bar X$, in particular $\bar{\bar X} \subseteq \bar X$; this is $\mu_X$.