Let be a (topological, for example) operad, and let be the associated bicolored operad whose algebras are triplets where and are -algebras and is a morphism of such algebras. It has two colors and and is given by:
Then if is formal, so is . This is to be contrasted with the fact that the Swiss-Cheese operad is not formal, even though the operad of little -disks is. The operad governing morphisms is formal, but not the one governing actions.
Let be a dg-operad (maybe with some finiteness assumptions), and be an algebra over this operad. Then the symmetric collection is a right-comodule over . The coaction morphism
is dual to the map given by
If is a bifunctor that preserves reflexive coequalizers in each variable, then it preserves reflexive coequalizers in the following sense: if is a reflexive coequalizer in (with reflector ), and is a reflexive coequalizer in (with reflector ), then so is
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 equalizes and , then it must also equalize and , and it must also equalize and . Indeed:
This then implies that if is an operad, then the free -algebra functor preserves reflexive coequalizers. It didn’t really click in before for me that this was the reason why.
The retract argument: if is a composite and if has the left lifting property against , then is a retract of . Similarly, if it has the right lifting property against , then it is a retract of . This seems to be useful for weak factorization systems, for example if is an acyclic cofibration and you manage to factorize it as something followed by a fibration, then is a retract of that something.