Homology on Pointed Sets

Let (X,x_0) be a pointed set (i.e. there is a nullary operation x_0:\varnothing\to X where x_0(\,)=x_0).

Definition 1.  A pointed set is a differential set if there is an endomorphism d:X\to X (i.e. d(x_0)=x_0) such that d^2(x)=x_0 for all x\in X.  We say two elements x,y\in X are homologous, denoted x\sim y, if d(x)=d(y).

Proposition 2.  The homologous relation is an equivalence relation.

Definition 3.  Elements of the equivalence class [x_0] are called cycles.  Elements of the set d(X) are called boundaries.  The homology of X is defined to be the set


Note that all boundaries are cycles, and x_0\in H(X), so H(X) can have a pointed set structure with point x_0.  Now let the quotient set X/\sim be pointed as (X/\sim,[x_0]), and note that (d(X),x_0) can be pointed since d(x_0)=x_0.

Proposition 4.  (X/\sim,[x_0])\approx (d(X),x_0) as pointed sets.

Proof.     Since the equivalence class [x]=\{y:d(x)=d(y)\}, it follows that the map [x]\mapsto d(x) is injective.  This map is also clearly surjective since d is defined on X, which is partitioned via \sim.  And of course we also have [x_0]\mapsto x_0.

 Now suppose we have a sequence \{X_i,x_i\} of pointed sets and a sequence \{d_{i+1}:X_{i+1}\to X_i\} of pointed set homomorphisms such that d_i\circ d_{i+1}(x)=x_{i-1} for all x\in X_{i+1}.  We will denote such a collection by \{(X_i,x_i),d_i\} and call it a pointed set complex.  We can also define the ith homology of the complex as


where we have the intuitive equivalence relation on each set X_i with x\sim y\Leftrightarrow d_i(x)=d_i(y).  We will also call a pointed set complex exact if H_i(X)=(\{x_i\},x_i)=:0_i for all i.

Let (A,x_0) be a sub-differential set of a differential set (X,x_0).  That is, d(A)\subseteq A.  We will call A a normal sub-differential set if for x\in X we have d(x)\in A\Rightarrow x\in A.

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


Theorem 6.  If A is a normal sub-differential set of X, then there are homomorphisms i,j such that

x_0\longrightarrow H(A)\stackrel{i}{\longrightarrow}H(X)\stackrel{j}{\longrightarrow}H(X,A)\to x_0

is an exact complex, where x_0 denotes the trivial pointed set (\{x_0\},x_0).

Proof.     Let x\in H(A), then x\in A, d(x)=x_0, and x is not the boundary of any element in A (excluding x_0).  Since x\in A, let us just define i(x)=x.  This map is fine provided x is not the boundary of any element in X-A.  But this is not possible since A is normal, so the map is well-defined.

Now let x\in H(X).  Then x\in X, d(x)=x_0, and x is not the boundary of another element in X.  We want to send it to a cycle in X-(A-\{x_0\}) which is not the boundary of an element in X-A.  We already have that x can’t be the boundary of something in X-A since it’s not the boundary of anything in X.  So we will define j by

j(x)=\left\{\begin{array}{lcl}x&\mbox{if}&x\in X-A\\x_0&\mbox{if}&x\in A\end{array}\right.

The image points are of course boundaries since the final map sends everything to x_0.  It follows that j\circ i(x)=x_0 for x\in H(A), 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 H_2,H_1 and H_0.  Since the homology pointed sets will all have x_0 as their point, we will simply find the underlying sets.  H_2=x_0 since i is just the identity map, that is, [x_0]_i=\{x_0\}.  For H_1 note that [x_0]_j=H(A), so we have that


Lastly we have [x_0]_j=H(X,A) and






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