Monthly Archives: October, 2011

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 ith 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