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