# Cohomology of Pointed Sets

Let $M$ be a monoid.  Then $M$ is a pointed set with point $1.$  Let $\varphi:(M,0)\times(X,x_0)\to (X,x_0)$ be a map satisfying

$\varphi(1,x)=x,$ and

$\varphi(mn,x)=\varphi(m,\varphi(n,x)).$

Such a map is called a monoidal action, and we correspondingly call $X$ an $M$-pointed set.  Note that since $\varphi(1,x_0)=x_0,$ it is a pointed morphism.  We will also write $mx:=\varphi(m,x)$ for short.

Suppose $M$ acts on $X$ and that $X$ is a differential pointed set (so $d^2(x)=x_0$).  Now define the pointed set

$\displaystyle Hom_M(X,M)=\{f:X\to M:f(x_0)=1\}$

with point $1:X\to M$ defined by $1(x)=1$ for all $x\in X.$  Then this is a differential pointed set with differential defined by

$(df)(x)=f(dx)$

since

$(d^2f)(x)=d\left(f(dx)\right)=(df)(dx)=f(d^2x)=f(x_0)=1$

for all $x\in X,$ so $d^2f=1.$  We then define the cohomology of $X$ with coefficients in $M$ as

$\dot{H}(M,X)=[1]-(dHom(X,M)-\{1\})$

where of course $[1]$ is the equivalence class of maps under the relation $f\sim g$ iff $df=dg=1.$

Let $X$ be an $M$-pointed set.  Consider the two sets

$X^M=\{x\in X:mx=x\}$

$X_M=X/\sim$

where $x\sim y$ iff $mx=y$ or $x=my$ for some nonzero $m\in M.$  (Note the “or” gives us symmetry).  $x_0$ belongs to both of these (as a class of elements in the latter case such that $mx=x_0$ (since $mx_0=x$ is never true for nonzero $m$ and $x\neq x_0$)).  The point (or trivial class) in $X_M$ is called the torsion subpointed set of $X$ with respect to $M.$

Proposition 1.  Let $X$ and the pointed natural numbers $(\mathbb{N},0)$ be $M$-pointed monoids.  Then

$X^M=Hom_M(\mathbb{N},X).$

Proof.  Let $m\in M$ and $n\in\mathbb{N}$ and define $\phi:X^M\to Hom_M(\mathbb{N},X)$ be defined by $m\phi(x)(n)=\sum_{i=1}^nx:=nx$ (since $mx=x$).  The map is injective since if $\phi(x)=\phi(y),$ then $nx=ny$ for all $n,$ but this just implies $x=y.$  As $\mathbb{N}$ is a monoid with one generator, any morphism $\varphi:\mathbb{N}\to X$ is determined by the image of $1.$  But such a map induces a preimage under $\phi:$ $\phi^{-1}(\varphi)=\varphi(1)\in X_M.$  And we have

$\phi(0_X)(n)=n0_X=0_X.$

So $\phi$ is a morphism.

We don’t however have that $X_M=\mathbb{N}\times X$ without some way of moving natural numbers between components of $\mathbb{N}\times X$ (i.e. a tensor product).  We can define a tensor product of two $M$-pointed sets $X$ and $Y$ which are monoids, denoted $X\odot_M Y,$ as $X\times Y/\sim$ with the usual bilinear relations.  We will denote simple elements by $x\odot y.$  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

$X_M=\mathbb{N}\odot_{\mathbb{N}[M]} X.$

Proof.     Define $\phi:\mathbb{N}\odot_M X\to X_M$ by $\phi(n\odot x)=nx.$  Suppose $nx=n'y.$  Then

$\begin{array}{lcl}1\odot nx=1\odot n'y&\Rightarrow&1\odot n(1_Mx)=1\odot n'(1_My)\\&\Rightarrow&1\odot (n1_M)x=1\odot (n'1_M)y\\&\Rightarrow&n1_M\odot x=n'1_M\odot y\\&\Rightarrow&n\odot x=n'\odot y.\end{array}$

Also if $x\in X_M,$ then since action $mx=x$ for all $m\in M,$ we have a preimage $\phi^{-1}(x)=1\odot x.$  And lastly $\phi(0,0_X)=00_X=0_X,$ so it is a morphism.

If the functors $\cdot^M$ and $\cdot_M$ are left (right) exact functors in the category of pointed sets and $X$ has an injective (projective) resolution, then we can define the $M$-pointed cohomology (homology) via the right (left) derived functors $D^n(F,M)=H^n(F(I)^M)$ ($D_n(F,M)=H_n(F(P)_M)$).

# 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 $H(x_0)=x_0$ where $x_0$ denotes the trivial pointed set.

If $(X,x_0)$ and $(Y,y_0)$ are pointed sets, then we define the intuitive product pointed set as the pointed set $(X\times Y,(x_0,y_0)).$  If $X$ and $Y$ are also differential sets with differentials $d_1$ and $d_2,$ then we define the product differential component-wise: $d_1\times d_2 (x,y)=(d_1(x),d_2(y)).$  This is clearly a differential on $X\times Y,$ which we in turn call the product differential set.

If we then define the relation $\sim$ on $X\times Y$ where $(x,y)\sim (x',y')$ iff $(d_1(x),d_2(y))=(d_1(x'),d_2(y')),$ then we have $[(x,y)]_{d_1\times d_2}=[x]_{d_1}\times [y]_{d_2}.$  Hence we have

$\begin{array}{lcl}H(X\times Y)&=&[(x,y)]-(d(X\times Y)-(x_0,y_0))\\&=&[x]\times[y]-\left((d_1(X),d_2(Y))-(x_0,y_0)\right)\\&=&\left([x]-(d_1(X)-\{x_0\}),[y]-(d_2(Y)-\{y_0\})\right)\\&=&H(X)\times H(Y).\end{array}$

We can similarly define a formal sum of pointed spaces $\sqcup_i(X_i,x_i)$ whose corresponding homology is just defined as

$H\left(\bigsqcup_i(X_i,x_i)\right)=\bigsqcup_iH(X_i).$

So we have two forms of additivity.  We also have excision, for if $(X,A)$ is a pair of sets with $A$ a sub-differential set of $X$ and $U\subseteq A,$ then

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

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 $A$).  Let $f:(X,A)\to (Y,B)$ be a map between paired sets.  We say this is a paired set morphism if $f(A)\subseteq B.$  If we furthermore have that these are paired pointed sets $(X,A,x_0)$ and $(Y,B,y_0)$ with $A$ and $B$ sub-pointed sets, a paired set morphism $f$ between them is a paired pointed morphism if we also have that $f(x_0)=y_0.$  If both are differential sets as well, we can define a paired differential morphism if we further have $f\circ d_1=d_2\circ f.$  Let $f$ be a paired differential morphism between such sets.  Then we would like for its restriction $f:H(X,A)\to H(Y,B)$ 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 $f,g:(X,A)\to (Y,B)$ to be homotopic mod $A$ provided they agreed on $H(X,A).$

# 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

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

$\square$