# 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 $D^n(F,M)=H^n(F(I)^M)$ ($D_n(F,M)=H_n(F(P)_M)$).

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