Let be a monoid. Then is a pointed set with point Let be a map satisfying
Such a map is called a monoidal action, and we correspondingly call an -pointed set. Note that since it is a pointed morphism. We will also write for short.
Suppose acts on and that is a differential pointed set (so ). Now define the pointed set
with point defined by for all Then this is a differential pointed set with differential defined by
for all so We then define the cohomology of with coefficients in as
where of course is the equivalence class of maps under the relation iff
Let be an -pointed set. Consider the two sets
where iff or for some nonzero (Note the “or” gives us symmetry). belongs to both of these (as a class of elements in the latter case such that (since is never true for nonzero and )). The point (or trivial class) in is called the torsion subpointed set of with respect to
Proposition 1. Let and the pointed natural numbers be -pointed monoids. Then
Proof. Let and and define be defined by (since ). The map is injective since if then for all but this just implies As is a monoid with one generator, any morphism is determined by the image of But such a map induces a preimage under And we have
So is a morphism.
We don’t however have that without some way of moving natural numbers between components of (i.e. a tensor product). We can define a tensor product of two -pointed sets and which are monoids, denoted as with the usual bilinear relations. We will denote simple elements by It’s easy to verify that this is a monoid. We then have the result:
Proposition 2. Let the hypotheses of Proposition 1 hold, then
Proof. Define by Suppose Then
Also if then since action for all we have a preimage And lastly so it is a morphism.
If the functors and are left (right) exact functors in the category of pointed sets and has an injective (projective) resolution, then we can define the -pointed cohomology (homology) via the right (left) derived functors ().