Let be a monoid. Then is a pointed set with point Let be a map satisfying

and

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

since

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 ().