representation, ∞-representation?
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 $T$, i.e., the category of $T$-algebras. 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. (See Adámek & Koubek, III.10, for an example when $C =$ Pos.) In this article we collect various results that guarantee existence of colimits of algebras.
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 $C^T$.
The following very useful observation was first made by Linton.
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 $f, g: c \stackrel{\longrightarrow}{\longrightarrow} d$ 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 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 $C$ is a complete, cocomplete, cartesian closed category and $T$ is the monad corresponding to a finitary algebraic theory. (The key observation being that the finitary power functors $x \mapsto x^n$ preserve reflexive coequalizers if $C$ is cartesian closed.)
If $T$ is a monad on Set, then $Set^T$ is cocomplete (under the axiom of choice). Similarly upon replacing $Set$ by a slice $Set/X \simeq Set^X$, or by Vect.
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 $T$-$Alg$ since $U: Set^T \to Set$ reflects arbitrary intersections.) 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 $i: Q \to U B$ of $p$, which picks a representative in each equivalence class, together with $\langle i p, 1 \rangle: U B \to U E$.) The proof is completed by the following lemma.
In a category $\mathbf{C}$, given a pair $\pi_1, \pi_2: E \rightrightarrows B$ in $\mathbf{C}^T$ such that $U\pi_1, U\pi_2: U E \rightrightarrows U B$ has a split coequalizer $U B \to Q$ in $\mathbf{C}$, the pair $\pi_1, \pi_2$ has a coequalizer in $\mathbf{C}^T$ (reflected by the split coequalizer).
(Cf. monadicity theorem.) A split coequalizer is an absolute colimit, which the functor $T$ preserves. Hence the top row in
(the first two vertical arrows being algebra structure maps) is a coequalizer in $\mathbf{C}^T$. The last vertical arrow making the diagram commute gives $Q$ a $T$-algebra structure, and the split coequalizer in the bottom row is thereby reflected in $\mathbf{C}^T$ (i.e., lifts to a coequalizer in $\mathbf{C}^T$, albeit not necessarily to one that is itself split).
If $\mathbf{C}$ is a regular category or exact category in which regular epimorphisms split, and $T$ is any monad on $\mathbf{C}$, then $\mathbf{C}^T$ is a regular category (or exact category, respectively).
For regularity, we first construct coequalizers of kernel pairs in $\mathbf{C}^T$. So suppose $\pi_1, \pi_2: E \rightrightarrows B$ is the kernel pair of some $f: B \to C$ in $\mathbf{C}^T$. The kernel pair $U\pi_1, U\pi_2: U E \to U B$ of $U f$ in $\mathbf{C}$ has a coequalizer $q: U B \to Q$ in $\mathbf{C}$, and of course $U\pi_1, U\pi_2$ is the kernel pair of $q$ as well. It follows from the splitting hypothesis that the fork
splits in $\mathbf{C}$, hence by Lemma this diagram lifts to a coequalizer for $\pi_1, \pi_2: E \rightrightarrows B$ in $\mathbf{C}^T$. Thus kernel pairs in $\mathbf{C}^T$ have coequalizers.
That regular epis in $\mathbf{C}^T$ are stable under pullback follows a similar line of reasoning: let $p: B \to P$ be a regular epi in $\mathbf{C}^T$. It is the coequalizer of its kernel pair $\pi_1, \pi_2: E \rightrightarrows B$. We just calculated that the coequalizer $q: U B \to Q$ of $U\pi_1, U\pi_2$ in $\mathbf{C}$ lifts to $\mathbf{C}^T$, so that $Q$ is identified with $U P$ and $q$ with $U p$. Thus $U p$ is a regular epi in $\mathbf{C}$. Now if $f: A \to P$ is a map in $\mathbf{C}^T$, and $b = f^\ast p$ is the pullback of $p$ along $f$ (with kernel pair $\ker(b)$), then $U b$ is the pullback of $U p$ along $U f$ since $U$ preserves pullbacks, and so $U b$ is a regular epi since $\mathbf{C}$ is regular. This $U b$ is the coequalizer of its kernel pair, and splits, so by Lemma , its lift $b$ is the coequalizer of $\ker(b)$. Thus regular epis in $\mathbf{C}^T$ are stable under pullback.
For Barr-exactness, suppose $\pi_1, \pi_2: E \rightrightarrows B$ is an equivalence relation (or congruence) in $\mathbf{C}^T$. Then $U\pi_1, U\pi_2: U E \rightrightarrows U B$ is an equivalence relation in $\mathbf{C}$, and hence a kernel pair since $\mathbf{C}$ is exact. It is the kernel pair of its coequalizer $q$ in $\mathbf{C}$. By Lemma , the split coequalizer
lifts to a coequalizer diagram in $\mathbf{C}^T$, and since kernel pairs are preserved and reflected by $U: \mathbf{C}^T \to \mathbf{C}$, we conclude that $\pi_1, \pi_2$ is the kernel pair of the lifted regular epi over $q$.
If $T$ is a monad on a slice category $Set/X$, then the category of $T$-algebras is (Barr-)exact. If $T$ is a monad on $Vect$, then $Vect^T$ is exact.
Returning now to existence of general coequalizers, here is a more difficult and arcane 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, or even just colimits of $\omega$-chains, then $C^T$ is complete and cocomplete.
If $C$ is a locally presentable category and $T$ is an accessible monad (aka a bounded monad, aka a monad with rank) on $C$, then $C^T$ is also locally presentable and in particular cocomplete. Details may be found in Locally presentable and accessible categories.
Suppose that $\theta: S \to T$ is a morphism of monads on $C$, and suppose that $C^T$ has coequalizers. Then the relative “forgetful” functor
(pulling back a $T$-algebra $(c, \xi: T c \to c)$ to the $S$-algebra $(c, S c \stackrel{\theta c}{\longrightarrow} T c \stackrel{\xi}{\longrightarrow} c)$, thus remembering only underlying $S$-algebra structure) 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 actually completes the proof, but here is a concrete description of the left adjoint to $C^\theta$: it 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 “restriction” functor $Ab^f: Ab^B \to Ab^A$ between module categories (restricting scalar multiplication on a $B$-module along a ring map $f: A \to B$); given an $A$-module $(M, \alpha_M: A \otimes M \to M)$, the $B$-module $B \otimes_A M$ is the coequalizer in $Ab^B$ of
and so the coequalizer of (1) 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 in (1), i.e., if $\phi$ factors (uniquely) through a $T$-algebra map $T \circ_S c \to d$.
Indeed, assume $f$ is an $S$-algebra map, so we have a commutative diagram
That $\phi = \alpha \circ T f$ coequalizes the pair of (1) follows by expanding the diagram
to
where using (2), the path along the top may be replaced by $T f \circ T \xi: T S c \to T d$, reducing the desired coequalizing of (3) to the tautology $\alpha \circ T f \circ T \xi = \alpha \circ T f \circ T \xi$.
Conversely, assuming the coequalizing of (3), the perimeter of (4) commutes, and on top of that we stack naturality diagrams for the monad unit $\eta$ of $T$:
The vertical composite on the right is $1_d$ by a unit equation for a $T$-algebra, and thus we may simplify the perimeter. Retaining the (simplified) perimeter of (5), and inserting some naturality squares and a unit diagram inside, we arrive at the commutative diagram
where the commutativity of the unlabeled polygonal region is just the commutativity of (2). This completes the proof of the claim.
Last revised on January 31, 2018 at 20:01:31. See the history of this page for a list of all contributions to it.