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Let P\mathtt{P} be a (topological, for example) operad, and let P\vec{\mathtt{P}} be the associated bicolored operad whose algebras are triplets (A,B,f)(A,B,f) where AA and BB are P\mathtt{P}-algebras and ff is a morphism of such algebras. It has two colors aa and bb and is given by:

P(x1,,xn;a)={P(n)xi=ai,otherwise.\vec{\mathtt{P}}(x_1, \dots, x_n; a) = \begin{cases} \mathtt{P}(n) & x_i = a \, \forall i, \\ \varnothing & \text{otherwise.} \end{cases}
P(x1,,xn;b)=P(n)\vec{\mathtt{P}}(x_1, \dots, x_n; b) = \mathtt{P}(n)

Then if P\mathtt{P} is formal, so is P\vec{\mathtt{P}}. This is to be contrasted with the fact that the Swiss-Cheese operad is not formal, even though the operad of little 22-disks is. The operad governing morphisms is formal, but not the one governing actions.

Let P\mathtt{P} be a dg-operad (maybe with some finiteness assumptions), and AA be an algebra over this operad. Then the symmetric collection {An}n0\{ A^{\otimes n} \}_{n \ge 0} is a right-comodule over P\mathtt{P}^\vee. The coaction morphism

A(k+l1)iAkP(l)A^{\otimes(k+l-1)} \xrightarrow{\circ_i^*} A^{\otimes k} \otimes \mathtt{P}^\vee(l)

is dual to the map P(l)A(k+l1)Ak\mathtt{P}(l) \otimes A^{\otimes(k+l-1)} \to A^{\otimes k} given by

ρa1ak+l1a1ρ(ai,,ai+l1)ak+l1.\rho \otimes a_1 \otimes \dots \otimes a_{k+l-1} \mapsto a_1 \otimes \dots \rho(a_i, \dots, a_{i+l-1}) \dots \otimes a_{k+l-1}.

If F:C×DEF : \mathsf{C} \times \mathsf{D} \to \mathsf{E} is a bifunctor that preserves reflexive coequalizers in each variable, then it preserves reflexive coequalizers in the following sense: if M1d0,d1M0d0MM_1 \xrightarrow{d_0, d_1} M_0 \xrightarrow{d_0} M is a reflexive coequalizer in C\mathsf{C} (with reflector s0s_0), and M1d0,d1M0d0MM'_1 \xrightarrow{d_0', d_1'} M'_0 \xrightarrow{d_0} M' is a reflexive coequalizer in D\mathsf{D} (with reflector s0s'_0), then so is

F(M1,M1)F(M0,M0)F(M,M).F(M_1, M'_1) \rightarrow F(M_0, M'_0) \to F(M, M').

This is a classical result, and the proof hinges on the following trick (that I saw in Goerss and Hopkins’ André–Quillen (co)-homology for simplicial algebras over simplicial operads, but it can probably be found elsewhere): if ff equalizes F(d0,d0)F(d_0, d'_0) and F(d1,d1)F(d_1, d'_1), then it must also equalize F(1,d0)F(1, d'_0) and F(1,d1)F(1, d'_1), and it must also equalize F(d0,1)F(d_0, 1) and F(d1,1)F(d_1, 1). Indeed:

fF(d0,1)=fF(d0,d0)F(1,s0)=fF(d1,d1)F(1,s0)=fF(d1,1).f F(d_0, 1) = f F(d_0, d'_0) F(1, s'_0) = f F(d_1, d'_1) F(1, s'_0) = f F(d_1, 1).

This then implies that if P\mathtt{P} is an operad, then the free P\mathtt{P}-algebra functor preserves reflexive coequalizers. It didn’t really click in before for me that this was the reason why.

The retract argument: if f=(XiApY)f = (X \xrightarrow{i} A \xrightarrow{p} Y) is a composite and if ff has the left lifting property against pp, then ff is a retract of ii. Similarly, if it has the right lifting property against ii, then it is a retract of pp. This seems to be useful for weak factorization systems, for example if ff is an acyclic cofibration and you manage to factorize it as something followed by a fibration, then ff is a retract of that something.