# 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

$H(X)=[x_0]-\left(d(X)-\{x_0\}\right).$

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 $i$th homology of the complex as

$H_i(X)=[x_i]-(d_{i+1}(X_{i+1})-\{x_i\})$

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

$H(X,A)=[x_0]\cap(X-(A-\{x_0\}))-d(X-A).$

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

$H_1=H(A)-(i(H(A))-\{x_0\})=H(A)-(H(A)-\{x_0\})=\{x_0\}.$

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

$j(H(X))=[x_0]\cap(X-(A-\{x_0\}))-(d(X)-\{x_0\}).$

So

$\begin{array}{lcl}H_0&=&H(X,A)-\left([x_0]\cap(X-(A-\{x_0\}))-(d(X)-\{x_0\})\right)\\&=&H(X,A)-\left([x_0]\cap(X-(A-\{x_0\}))-(d(X-A)-\{x_0\})\right)\\&=&H(X,A)-(H(X,A)-\{x_0\})\\&=&\{x_0\}.\end{array}$

$\square$