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


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




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


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\}


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


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


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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