Chain Complexes of Sets, Homology Made Simple

Theme: What if we can do all the kinds of things we do in homological algebra, but without using groups/other algebra?

Homological Algebra and the Set Model

Definition: Chain Complexes and Homology Groups

A Chain Complex of abelian groups is a sequence

$\begin{CD} \dots @>>> A_{i+1} @>\partial_{i+1}>> A_i @>\partial_{i}>> A_{i-1} @>>> \dots \\ \end{CD}$

s.t.

\partial_{i}\circ \partial_{i+1} = 0\,\forall i

Then, the $i$ th homology group is defined as $H_{i}A = \ker{\partial_i}/\text{im}\,\partial_{i+1}$ .

We know the homology group exists because $\partial_{i}\circ\partial_{i+1} = 0$ , so the image of $\partial_{i+1}$ is contained in the kernel of $\partial_i$ .

There's a lot of cool things we can do with chain complexes. First, we can factor each map:

$\begin{CD} \dots @>>> A_{i+1} @>epi>> \text{im}\,\partial_{i+1} @>\iota>> A_i @>epi>> \text{im}\,\partial_{i} @>\iota>> A_{i-1} @>>> \dots \\ \end{CD}$

(This is an epi-mono factorization)

Definition: Chain Complexes of Sets and Homology Sets

A Chain Complex of Sets is a sequence

$\begin{CD} X_{i+1} @<<< \overline{X}_{i+1} @>mono>> X_{i} @<<< \overline{X}_i @>mono>> X_{i-1} \end{CD}$

s.t.

X_{i}\backslash \overline{X_i} \supseteq \overline{X}_{i+1}

Then, the $i$ th homology set is given by $H_{i}X = X_i\backslash (\overline{X_i}\bigsqcup \overline{X}_{i+1})$ .

A core idea is that for each pair $X_{i+1}, X_i$ , we consider conceptually that $\overline{X}_{i+1} = X_{i+1}\bigcap X_i$ .
Thus the chain complex of sets acts as a sort of chain of venn diagrams, with each intersecting area featuring inclusion maps into the two sets.

Recall that in the case of chain complexes of groups, we can factor through the trivial group (the condition that comes from $\partial_{i}\circ\partial_{i+1} = 0$ ):

$\begin{CD} \text{im}\,\partial_{i+1} @>\iota>> A_i \\ @VepiVV @VVepiV \\ 0 @>\iota>> \text{im}\,\partial_i \\ \end{CD}$

In the case of chain complexes of sets, instead of commuting, we require that the diagram

$\begin{CD} \varnothing @>mono>> \overline{X_i} \\ @VVV @VVV \\ \overline{X_{i+1}} @>mono>> X_i \\ \end{CD}$

is a pullback, with

\varnothing

being the pullback.

^The above is an informal analogy of chain complexes on graphs, so why is it important?

With abelian groups, with inclusion we can do quotienting, and if we take surjections we can take this kerneling thing.

With sets, we are only really doing complements. Both operations are complements because we reverse the direction of the arrows.

With abelian groups, we have the following diagram:

$\begin{CD} \text{im}\,\partial_{i+1} @>\iota>> \ker{\partial_i} @>epi>> H_i \\ @| @V{\iota}VV @V{\iota}VV \\ \text{im}\,\partial_{i+1} @>\iota>> A_i @>epi>> A_i / \text{im}\,\partial_{i+1} \\ @V{epi}VV @V{epi}VV @V{epi}VV \\ 0 @>\iota>> \text{im}\,\partial_i @= \text{im}\,\partial_i \\ \end{CD}$

We can thus define the homology group in two ways:

The kernel $\ker{\partial_i}$ quotiented by the image $\text{im}\,\partial_{i+1}$
Taking the quotient of the image $A_i / \text{im}\,\partial_{i+1}$ first, then taking the kernel of the inclusion of that quotient into $\text{im}\,\partial_i$

With sets, it's much simpler:

$\begin{CD} \overline{X}_{i+1} @>mono>> X_i\backslash\overline{X}_i @<<< H_i \\ @| @V{mono}VV @V{mono}VV \\ \overline{X}_{i+1} @>mono>> X_i @<<< X_i\backslash\overline{X}_{i+1} \\ @AAA @AAA @AAA \\ \varnothing @>mono>> \overline{X}_i @= \overline{X}_i \\ \end{CD}$

where

\varnothing

and

H_i

are pullbacks.

Given a chain complex of sets $X$ and an abelian group $A$ , then we have a chain complex of groups

$\begin{CD} A^{X_{i+1}} @>epi>> A^{\overline{X}_{i+1}} @>\iota>> A^{X_i} @>epi>> A^{\overline{X}_i} @>\iota>> A^{X_{i-1}} \\ \end{CD}$

The homology group is given by

H_{i}A^{X} \cong A^{H_{i}X}

Because $\overline{X}_i \to X_{i-1}$ is a projection map, the inclusion $A^{\overline{X}_i}\hookrightarrow A^{X_{i-1}}$ will just insert 0 everywhere not in the image of $\overline{X}_i$ .

On the other hand, because $X_i \gets \overline{X}_i$ is a backwards inclusion, $A^{X_i}\twoheadrightarrow A^{\overline{X}_i}$ is the projection map where we just take the portion from $\overline{X}_i$ to $A$ .

We could think of this as picking out only the most basic fragment of chain complexes: only looking at powers of a single group + projections and inclusions between them.
But we will use this to gain an intuition of homology.

Just as we say a chain complex of groups is exact if the image of one map is exactly the kernel of the next, a chain complex of sets is exact if each homology is zero: $H_i = \varnothing \forall i$ . In this case, each circle in the "venn diagram" is covered exactly by the other two circles.

The Snake Lemma

Theorem: The Snake Lemma

Given a map of short exact sequences

$\begin{CD} 0 @>>> A @>mono>> B @>epi>> C @>>> 0 \\ @. @VfVV @VgVV @VhVV @.\\ 0 @>>> A'@>mono>> B'@>epi>> C'@>>> 0 \\ \end{CD}$

there is an exact sequence

\ker{f}\to \ker{g}\to \ker{h}\to A'/A \to B'/B \to C'/C

.
See also: Snake Lemma (Wikipedia)

Snake Lemma on Sets

For chain complexes on sets, the diagram for the snake lemma is given by

$\begin{CD} A @>epi>> B @<<< C @. A @>epi>> B @. B'' @<<< C'' \\ @AAA @AAA @AAA @AAA @AAA @VepiVV @VepiVV \\ A''@>epi>> B''@<<< C''@. A'' @>epi>> B'' @. B' @<<< C' \\ @VepiVV @VepiVV @VepiVV @. @. @. @. \\ A' @>epi>> B' @<<< C' @. @. @. @. \\ \end{CD}$

all exact, with the right two diagrams being pullbacks on

A''

and

C''

, respectively.

For the sake of convenience, we'll define the following:

$D = C\backslash (B''\backslash A'')$
$E = B''\backslash (A''\bigsqcup C'')$
$F = A'\backslash (B''\backslash C'')$

Then, we have the exact sequence $A\backslash A'' \gets A\backslash A''\rightarrowtail B\backslash B'' \gets D\rightarrowtail C\backslash C'' \gets E\rightarrowtail A'\backslash A'' \gets F\rightarrowtail B'\backslash B'' \gets C'\backslash C'' \rightarrowtail C'\backslash C''$ .

The following venn diagrams illustrate this concept:

Diagrams of B, B', and B''

Division of B into A and C, B' into A' and C', and B'' into A'' and C''

Labeling of specific subsets of A, C, A', C' and B''

Diagram of resulting simplicial complex. Red lines represent boundaries between subsequent elements in the exact sequence.

Note that the two pullbacks in the above diagram ensure that $A'' = A\bigcap B''$ and $C'' = C'\bigcap B''$ .
However, the two other squares in the diagram aren't pullbacks, which means that $A''\to B'' \gets C''$ isn't exact (i.e. $B'' \neq A''\bigsqcup C''$ ).

In the context of groups, we think of the kernel as the part we are losing/cutting out, but the venn diagram model still holds as a conceptual model for how the snake theorem works on groups.

Chain Complexes of Homologies

Theorem: Pairs of Long Exact Sequences

If we have an inclusion $A\hookrightarrow B \to B/A$ of chain complexes of groups, there exists a long exact sequence of the following form:

$H_i A \to H_i B \to H_i (A/B) \to H_{i-1}A \to H_{i-1}B \to H_{i-1}(A/B)\to\dots$

Definition: Mono Map

A mono map $X\rightarrowtail Y$ of chain complexes of sets is of the form

where any mixed square (i.e. those with

\overline{X}_i

in the top right corner) is a pullback on

\overline{X}_i

\forall i

.
If we regard each differential as a partial monomorphism, then mono maps from elements of

X

to elements of

Y

are just a commuting sequence of monos.

Importantly, the map $X\rightarrowtail Y\gets Y\backslash X$ of chain complexes of sets induces the following diagram:

$\begin{CD} X_{i+1} @<<< \overline{X}_{i+1} @>mono>> X_{i} @<<< \overline{X}_i \\ @VmonoVV @VmonoVV @VmonoVV @VmonoVV \\ Y_{i+1} @<<< \overline{Y}_{i+1} @>mono>> Y_{i} @<<< \overline{Y}_i \\ @AAA @AAA @AAA @AAA \\ Y_{i+1}\backslash X_{i+1} @<<< @>mono>> Y_i\backslash X_i @<<< \dots \\ \end{CD}$

We can imagine this effectively as a chaining-together of the venn diagrams that resulted from the snake theorem on sets.

Then, the long exact sequence of sets, resulting from our theorem above, is of the form

$H_i X \gets\rightarrowtail H_i Y\gets\rightarrowtail H_i (X\backslash Y) \gets\rightarrowtail H_{i-1}X \gets\rightarrowtail\dots$

where the bridging portions have been omitted.

Importantly, the bottom row of the commutative diagram above isn't an exact sequence.
However, the omitted elements of the bottom sequence are the elements that connect the homologies between degrees (i.e. connecting $H_{i+1}(X\backslash Y)$ to $H_i X$ ).