List is known to be a monad. It takes a set and maps it to lists of elements of that set. The natural transformations are, singleton and flatten, whereby we map a set to a set of singleton lists each with one element, and we take lists of lists and flatten them into lists (somewhat like concatenating). Suppose we define two more natural transformations, copy and delete. Copy takes a list and maps it to a list of two lists, they being copies of the original list. Delete takes a list and maps it to the underlying set of the List functor. The Delete natural transformation seems a bit dubious. Does this form a monad that is also a comonad? [Here we have a paper](http://homepages.inf.ed.ac.uk/s1225336/papers/fossacs12.pdf) that seems to suggest that non-empty list is a comonad. Does this apply to the example I am talking about?



List is a monad, but is it a comonad with these natural transformations?
Todd's comment provides an important limitation on what you can do here, but here's what I think is the most interesting way to answer your question. Define an endofunctor $L^+$ on the category of sets by
$$
L^+(X) = \sum\_{n \geq 1} X^n
$$
for sets $X$, where $\sum$ means coproduct (disjoint union). Thus, an element of $L^+(X)$ is a *nonempty* finite list of elements of $X$.


Then $L^+$ can be given the structure of both a monad and a comonad. 


The monad structure is the same as the one you described for possibly-empty lists: the unit sends $x \in X$ to the single-element list $x$, and the multiplication is concatenation ("flattening"). Its algebras are semigroups.


The comonad structure is less well-known (at least, to mathematicians; it seems better known in computer science). The counit map $L^+(X) \to X$ sends a list $(x\_1, \ldots, x\_n)$ to $x\_1$. The comultiplication $L^+(X) \to L^+(L^+(X))$ forms the "tails" of a list; for instance,
$$
(x\_1, x\_2, x\_3, x\_4)
\mapsto
((x\_1, x\_2, x\_3, x\_4), (x\_2, x\_3, x\_4), (x\_3, x\_4), (x\_4))
$$
and the general definition is what you'd guess from this.


(In fact, there's a relationship between the monad structure and the comonad structure on $L^+$, or more exactly a mixed distributive law between them, with $L^+$ itself as its underlying functor. In the terminology that some people use, this makes $L^+$ into a "bimonad".)


More generally, it sounds like you're looking for examples of monads that are also comonads. There are many of these: e.g. for any monoid $M$, the endofunctor $M \times -$ on the category of sets is both a monad and a comonad in a natural way.


