# Cohomology of Pointed Sets

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 ().

# Homology on Pointed Sets 2

We wish to address four of the Eilenberg-Steenrod axioms for this category. First note that for *dimension* we have where denotes the trivial pointed set.

If and are pointed sets, then we define the intuitive **product pointed set** as the pointed set If and are also differential sets with differentials and then we define the **product differential** component-wise: This is clearly a differential on which we in turn call the **product differential set**.

If we then define the relation on where iff then we have Hence we have

We can similarly define a formal sum of pointed spaces whose corresponding homology is just defined as

So we have two forms of *additivity*. We also have *excision*, for if is a pair of sets with a sub-differential set of and then

We lastly address *homotopy* and omit the *long exact sequence* as we have not developed a dimensional concept (although recall we constructed a short exact sequence under an assumption on ). Let be a map between paired sets. We say this is a **paired set morphism** if If we furthermore have that these are **paired pointed sets** and with and sub-pointed sets, a paired set morphism between them is a **paired pointed morphism** if we also have that If both are differential sets as well, we can define a **paired differential morphism** if we further have Let be a paired differential morphism between such sets. Then we would like for its restriction to be well-defined. Hopefully some combination of requirements like normality and being a paired differential morphism will work, but I have not yet convinced myself. If it worked, we would in turn just define two maps to be **homotopic mod** provided they agreed on

# Homology on Pointed Sets

Let be a pointed set (i.e. there is a nullary operation where ).

**Definition 1.** A pointed set is a **differential set** if there is an endomorphism (i.e. ) such that for all We say two elements are **homologous**, denoted if

**Proposition 2.** The homologous relation is an equivalence relation.

**Definition 3.** Elements of the equivalence class are called **cycles**. Elements of the set are called **boundaries**. The **homology** of is defined to be the set

Note that all boundaries are cycles, and so can have a pointed set structure with point Now let the quotient set be pointed as and note that can be pointed since

**Proposition 4.** as pointed sets.

*Proof.* Since the equivalence class it follows that the map is injective. This map is also clearly surjective since is defined on which is partitioned via And of course we also have

Now suppose we have a sequence of pointed sets and a sequence of pointed set homomorphisms such that for all We will denote such a collection by and call it a **pointed set complex**. We can also define the **th homology** of the complex as

where we have the intuitive equivalence relation on each set with We will also call a pointed set complex **exact** if for all

Let be a sub-differential set of a differential set That is, We will call a **normal** sub-differential set if for we have

**Definition 5.** We define the **relative homology** of with respect to as the set

**Theorem 6.** If is a normal sub-differential set of then there are homomorphisms such that

is an exact complex, where denotes the trivial pointed set .

*Proof.* Let then and is not the boundary of any element in (excluding ). Since let us just define This map is fine provided is not the boundary of any element in But this is not possible since is normal, so the map is well-defined.

Now let Then and is not the boundary of another element in We want to send it to a cycle in which is not the boundary of an element in We already have that can’t be the boundary of something in since it’s not the boundary of anything in So we will define by

The image points are of course boundaries since the final map sends everything to It follows that for so we have a differential complex of homology pointed sets. To prevent confusion, we will denote the homology sets of this complex of homology sets by and Since the homology pointed sets will all have as their point, we will simply find the underlying sets. since is just the identity map, that is, For note that so we have that

Lastly we have and

So