symmetric monoidal (∞,1)-category of spectra
An algebra over an endofunctor is like an algebra over a monad, but without a notion of associativity (which would not make sense).
For a category $C$ and endofunctor $F$, an algebra (or module) of $F$ is an object $X$ in $C$ and a morphism $\alpha\colon F(X) \to X$. ($X$ is called the carrier of the algebra)
A homomorphism between two algebras $(X, \alpha)$ and $(Y, \beta)$ of $F$ is a morphism $m\colon X \to Y$ in $C$ such that the following square commutes:
Composition of such morphisms of algebras is given by composition of the underlying morphisms in $C$. This yields the category of $F$-algebras, which comes with a forgetful functor to $C$.
The dual concept is a coalgebra for an endofunctor. Both algebras and coalgebras for endofunctors on $C$ are special cases of algebras for bimodules.
To a category theorist, algebras over a monad may be more familiar than algebras over just an endofunctor. In fact, when $C$ and $F$ are well-behaved, then algebras over an endofunctor $F$ are equivalent to algebras over a certain monad, the algebraically-free monad generated by $F$ (Maciej, Gambino-Hyland 04, section 6).
This is analogous to the relationship between an action $M \times B \to B$ of a monoid $M$ and a binary function $A \times B \to B$ (an action of a set): such a function is the same thing as an action of the free monoid $A^*$ on $B$.
Returning to the endofunctor case, the general statement is:
The category of algebras of the endofunctor $F\colon \mathcal{C} \to \mathcal{C}$ is equivalent to the category of algebras of the algebraically-free monad on $F$, should such exist.
Actually, this proposition is merely a definition of the term “algebraically-free monad”. If $F$ has an algebraically-free monad, denoted say $F^*$, then in particular the forgetful functor $F Alg \to C$ has a left adjoint, and $F^*$ is the monad on $C$ generated by this adjunction. Conversely, if such a left adjoint exists, then the monad it generates is algebracially-free on $F$; for the straightforward proof, see for instance (Maciej).
Algebraically-free monads exist in particular when $C$ is a locally presentable category and $F$ is an accessible functor; see transfinite construction of free algebras.
It turns out that an algebraically-free monad on $F$ is also free in the sense that it receives a universal arrow from $F$ relative to the forgetful functor from monads to endofunctors. The converse, however, is not necessarily true: a free monad in this sense need not be algebraically-free. It is true when $C$ is complete, however.
A textbook account of the basic theory is in chapter 10 of
The relation to free monads is discussed in
Last revised on January 8, 2014 at 17:11:55. See the history of this page for a list of all contributions to it.