symmetric monoidal (∞,1)-category of spectra
Let $T$ be a monad on a category $C$, and let $C^T$ denote the Eilenberg-Moore category of $T$. Let
be the usual underlying or forgetful functor, with left adjoint $F: C \to C^T$, unit $\eta: 1_C \to U F$, and counit $\varepsilon: F U \to 1_{C^T}$. It is well-known that $U$ reflects limits, so that if $C$ is complete, then $C^T$ is also complete and $U$ is continuous.
The situation with regard to colimits is more complicated. It is not generally true that if $C$ is cocomplete, then $C^T$ is also. In this article we collect some partial results which address these issues.
A simple but basic fact is the following. Suppose $J$ is a small category, and suppose that the monad $T$ preserves colimits over $J$, that is, suppose that for every $F: J \to C$ the canonical map
is an isomorphism.
Under these hypotheses, $U: C^T \to C$ reflects colimits over $J$.
Here are some sample applications of this proposition which arise frequently in practice. Let $J$ be the generic reflexive fork, having exactly two objects $0, 1$, generated by three non-identity arrows
and subject to the condition that the two composites from $0$ to $0$ are the identity. A colimit over $J$ is called a reflexive coequalizer. It frequently happens that a monad $T: C \to C$ preserves reflexive coequalizers; in this case, if $C$ has reflexive coequalizers, then so does $C^T$.
If $C$ is cocomplete and $C^T$ has reflexive coequalizers, then $C^T$ is cocomplete.
First observe that if $(c, \xi: T c \to c)$ is a $T$-algebra, then $\xi$ is the coequalizer of the reflexive fork
To show $C^T$ has coproducts, let $(c_i, \xi_i)$ be a collection of algebras. Then $F(\sum_i U c_i)$ is the coproduct $\sum_i F U c_i$ in $C^T$ (since $F$ preserves coproducts and $C$ has them). We have a reflexive fork
and it is not difficult to show that the coequalizer in $C^T$ of this diagram is the coproduct $\sum_i c_i$.
Finally, general coequalizers in $C^T$ are constructed from coproducts and reflexive coequalizers: given a parallel pair $f, g: c \stackrel{\to}{\to} d$ in $C^T$, the coequalizer of $f$ and $g$ is the colimit of the reflexive fork
where the first arrow is the coproduct coprojection.
If $T$ is a monad on $Set$, then $Set^T$ is cocomplete.
It is enough to show that $Set^T$ has coequalizers. Suppose given a pair of algebra maps $f, g: A \stackrel{\to}{\to} B$ whose coequalizer we wish to construct. Let $R$ be the $T$-algebra relation
and then let $E$ be the smallest $T$-congruence (equivalence relation that is a $T$-subalgebra map $E \hookrightarrow B \times B$) through which $R$ factors. (This is the intersection of all $T$-congruences through which $R$ factors, and may be calculated in $Set$, where it is reflected in $T$-$Alg$.) The coequalizer as calculated in $Set$,
is a split coequalizer, because every quotient of an equivalence relation in $Set$ is a split coequalizer. (This requires the axiom of choice. A splitting is given by any splitting $i: Q \to B$ of $p$, which picks a representative in each equivalence class, together with $\langle i p, 1 \rangle: B \to E$.) It is therefore an absolute colimit, which the monad $T$ preserves. Hence the top row in
(the first two vertical arrows being algebra structure maps) is a coequalizer in $Set^T$. The last vertical arrow making the diagram commute gives $Q$ a $T$-algebra structure, and the split coequalizer in the bottom row is reflected in $Set^T$.
If $T$ is a monad on a complete and cocomplete category $C$ that preserves reflexive coequalizers, then $C^T$ is complete and cocomplete.
The hypotheses of the preceding corollary hold when $T$ is a monad on a complete, cocomplete, cartesian closed category that is induced from a finitary algebraic theory. (The key observation being that the finitary power functors $x \mapsto x^n$ preserve reflexive coequalizers.)
Here is a more difficult result given in Toposes, Theories, and Triples (theorem 3.9, p. 267):
If $C$ has coequalizers and equalizers of arbitrary sets of parallel morphisms, and if a monad $T: C \to C$ preserves colimits of countable chains $\omega \to C$, then $C^T$ has coequalizers.
If $C$ is complete and cocomplete and $T: C \to C$ preserves filtered colimits, then $C^T$ is complete and cocomplete.
Suppose that $\theta: S \to T$ is a morphism of monads on $C$, and suppose that $C^T$ has coequalizers. Then the forgetful functor
has a left adjoint.
Since the following diagram is commutative:
(using an obvious notation), it follows immediately from a corollary to the adjoint lifting theorem that if $C^T$ has coequalizers of reflexive pairs, then $C^{\theta}$ has a left adjoint and is, in fact, monadic.
This completes the proof, but it is interesting to observe that there is a concrete description of the left adjoint in this case: The left adjoint sends an $S$-algebra $(c, \xi: S c \to c)$ to the (reflexive) coequalizer of the pair
where $\mu: T T \to T$ is the monad multiplication. (If $u: 1_C \to S$ is the unit of $S$, then $T u c: T c \to T S c$ is a common right inverse of both arrows of the pair.) This coequalizer is analogous to the construction of the left adjoint $B \otimes_A -$ to the functor $Ab^f: Ab^B \to Ab^A$ between module categories given by restricting scalar multiplication; the coequalizer will be denoted $T \circ_S c$ to underline the analogy.
To see that $T \circ_S -$ is the left adjoint, let $(d, \alpha: T d \to d)$ be a $T$-algebra. Any map $f: c \to d$ in $C$ induces a unique $T$-algebra map $\phi: T c \to d$:
and the claim is that $f: c \to d$ is an $S$-algebra map $c \to C^\theta(d)$ if and only if $\phi$ coequalizes the pair of above, i.e., if $\phi$ factors (uniquely) through a $T$-algebra map $T \circ_S c \to d$.
Indeed, the condition that $f$ is an $S$-algebra map is satisfaction of the equation
and now we have a long train of equations
which gives $\phi \circ \mu c \circ T \theta c = \phi \circ T \xi$. This completes one direction. In the other direction, suppose $\phi \circ \mu c \circ T \theta c = \phi \circ T \xi$. Then
as desired. This completes the proof.